Python String Exercise with Solutions

Updated on: March 13, 2025 | 234 Comments

As you know, the stings are widely used to hold textual data. To perform any programming tasks in Python, a good understanding of string manipulation is necessary.

These string exercises will help Python developers to learn and practice string operations, manipulations, slicing, and string functions.

Also Read:

This String Exercise includes the following: –

It contains 18 Python string programs, questions, problems, and challenges to practice.
The solution is provided for all questions.
All string programs are tested on Python 3

Use Online Code Editor to solve exercise questions. Let us know if you have any alternative solutions in the comment section below.

Exercise 1A: Create a string made of the first, middle and last character
Exercise 1B: Create a string made of the middle three characters
Exercise 2: Append new string in the middle of a given string
Exercise 3: Create a new string made of the first, middle, and last characters of each input string
Exercise 4: Arrange string characters such that lowercase letters should come first
Exercise 5: Count all letters, digits, and special symbols from a given string
Exercise 6: Create a mixed String using the following rules
Exercise 7: String characters balance Test
Exercise 8: Find all occurrences of a substring in a given string by ignoring the case
Exercise 9: Calculate the sum and average of the digits present in a string
Exercise 10: Write a program to count occurrences of all characters within a string
Exercise 11: Reverse a given string
Exercise 12: Find the last position of a given substring
Exercise 13: Split a string on hyphens
Exercise 14: Remove empty strings from a list of strings
Exercise 15: Remove special symbols / punctuation from a string
Exercise 16: Removal all characters from a string except integers
Exercise 17: Find words with both alphabets and numbers
Exercise 18: Replace each special symbol with # in the following string

Exercise 1A: Create a string made of the first, middle and last character

Write a program to create a new string made of an input string’s first, middle, and last character.

Given:

str1 = "James"Code language: Python (python)

Expected Output:

Jms

Show Hint

String index always starts with 0
Use string indexing to get the character present at the given index
Get the index of the middle character by dividing string length by 2

Show Solution

Use string indexing to get the character present at the given index.
Use str1[0] to get the first character of a string and add it to the result variable
Next, get the middle character’s index by dividing string length by 2. x = len(str1) /2. Use str1[x] to get the middle character and add it to the result variable
Use str1[len(str1)-1] to get the last character of a string and add it to the result variable
print result variable to display new string

str1 = 'James'
print("Original String is", str1)

# Get first character
res = str1[0]

# Get string size
l = len(str1)
# Get middle index number
mi = int(l / 2)
# Get middle character and add it to result
res = res + str1[mi]

# Get last character and add it to result
res = res + str1[l - 1]

print("New String:", res)Code language: Python (python)

Exercise 1B: Create a string made of the middle three characters

Write a program to create a new string made of the middle three characters of an input string.

Given:

Case 1

str1 = "JhonDipPeta"Code language: Python (python)

Output

Dip

Case 2

str2 = "JaSonAy"Code language: Python (python)

Output

Son

Show Hint

First, get the middle index number by dividing string length by 2.
Use string slicing to get the middle three characters starting from the middle index to the next two character.

Show Solution

Get the middle character’s index using x = len(str1) /2.
Use string slicing to get the middle three characters starting from the middle index to the next two character str1[middle_index-1:middle_index+2]

def get_middle_three_chars(str1):
    print("Original String is", str1)

    # first get middle index number
    mi = int(len(str1) / 2)

    # use string slicing to get result characters
    res = str1[mi - 1:mi + 2]
    print("Middle three chars are:", res)

get_middle_three_chars("JhonDipPeta")
get_middle_three_chars("JaSonAy")Code language: Python (python)

Exercise 2: Append new string in the middle of a given string

Given two strings, s1 and s2. Write a program to create a new string s3 by appending s2 in the middle of s1.

Given:

s1 = "Ault"
s2 = "Kelly"Code language: Python (python)

Expected Output:

AuKellylt

Show Hint

Use built-in function len(s1) to get the string length.
Next, get the middle index number by dividing string length by 2.

Show Solution

First, get the middle index number of s1 by dividing s1’s length by 2
Use string slicing to get the character from s1 starting from 0 to the middle index number and store it in x
concatenate x and s2. x = x + s2
concatenate x and remeaning character from s1
print x

def append_middle(s1, s2):
    print("Original Strings are", s1, s2)

    # middle index number of s1
    mi = int(len(s1) / 2)

    # get character from 0 to the middle index number from s1
    x = s1[:mi:]
    # concatenate s2 to it
    x = x + s2
    # append remaining character from s1
    x = x + s1[mi:]
    print("After appending new string in middle:", x)

append_middle("Ault", "Kelly")
Code language: Python (python)

Exercise 3: Create a new string made of the first, middle, and last characters of each input string

Given two strings, s1 and s2, write a program to return a new string made of s1 and s2’s first, middle, and last characters.

Given:

s1 = "America"
s2 = "Japan"Code language: Python (python)

Expected Output:

AJrpan

Show Hint

String index starts with index 0. The first character is present at index 0, and the last character is at the index string’s length -1
Use built-in function len(s1) to get the string length.
Next, get the middle index number by dividing string length by 2.

Show Solution

Get the first character from both strings, concatenate them, and store them in variable x
Get the middle character from both strings, concatenate them, and store them in variable y
Get the last character from both strings, concatenate them, and store them in variable x
In the end, join x, y, and z and save it in the result variable
print the result

def mix_string(s1, s2):
    # get first character from both string
    first_char = s1[0] + s2[0]

    # get middle character from both string
    middle_char = s1[int(len(s1) / 2):int(len(s1) / 2) + 1] + s2[int(len(s2) / 2):int(len(s2) / 2) + 1]

    # get last character from both string
    last_char = s1[len(s1) - 1] + s2[len(s2) - 1]

    # add all
    res = first_char + middle_char + last_char
    print("Mix String is ", res)

s1 = "America"
s2 = "Japan"
mix_string(s1, s2)
Code language: Python (python)

Exercise 4: Arrange string characters such that lowercase letters should come first

Given string contains a combination of the lower and upper case letters. Write a program to arrange the characters of a string so that all lowercase letters should come first.

Given:

str1 = PyNaTive

Expected Output:

yaivePNT

Show Hint

Iterate each character from a string and check if the current character is the lower or upper case using islower() string function

Show Solution

Create two lists lower and upper
Iterate a string using a for loop
In each loop iteration, check if the current character is the lower or upper case using the islower() string function.
If a character is the lower case, add it to the lower list, else add it to the upper list
to join the lower and upper list using a join() function.
convert list to string
print the final string

str1 = "PYnAtivE"
print('Original String:', str1)
lower = []
upper = []
for char in str1:
    if char.islower():
        # add lowercase characters to lower list
        lower.append(char)
    else:
        # add uppercase characters to lower list
        upper.append(char)

# Join both list
sorted_str = ''.join(lower + upper)
print('Result:', sorted_str)Code language: Python (python)

Exercise 5: Count all letters, digits, and special symbols from a given string

Given:

str1 = "P@#yn26at^&i5ve"Code language: Python (python)

Expected Outcome:

Total counts of chars, digits, and symbols 

Chars = 8 
Digits = 3 
Symbol = 4

Show hint

Use the following string functions

isalpha(): To check if a string/character is an alphabet
isdigit(): To check if a string/character is a digit.

Show Solution

Iterate each character from a string using a for loop
In each loop iteration, check if the current character is the alphabet using an isalpha() function. If yes, increase the character counter. Check if it is a digit using the isdigit() function and increase the digit counter; otherwise, increase the symbol counter.
Print the value of each counter

def find_digits_chars_symbols(sample_str):
    char_count = 0
    digit_count = 0
    symbol_count = 0
    for char in sample_str:
        if char.isalpha():
            char_count += 1
        elif char.isdigit():
            digit_count += 1
        # if it is not letter or digit then it is special symbol
        else:
            symbol_count += 1

    print("Chars =", char_count, "Digits =", digit_count, "Symbol =", symbol_count)

sample_str = "P@yn2at&#i5ve"
print("total counts of chars, Digits, and symbols \n")
find_digits_chars_symbols(sample_str)
Code language: Python (python)

Exercise 6: Create a mixed String using the following rules

Given two strings, s1 and s2. Write a program to create a new string s3 made of the first char of s1, then the last char of s2, Next, the second char of s1 and second last char of s2, and so on. Any leftover chars go at the end of the result.

Given:

s1 = "Abc"
s2 = "Xyz"Code language: Python (python)

Expected Output:

AzbycX

Show Solution

s1 = "Abc"
s2 = "Xyz"

# get string length
s1_length = len(s1)
s2_length = len(s2)

# get length of a bigger string
length = s1_length if s1_length > s2_length else s2_length
result = ""

# reverse s2
s2 = s2[::-1]

# iterate string 
# s1 ascending and s2 descending
for i in range(length):
    if i < s1_length:
        result = result + s1[i]
    if i < s2_length:
        result = result + s2[i]

print(result)
Code language: Python (python)

Exercise 7: String characters balance Test

Write a program to check if two strings are balanced. For example, strings s1 and s2 are balanced if all the characters in the s1 are present in s2. The character’s position doesn’t matter.

Given:

Case 1:

s1 = "Yn"
s2 = "PYnative"Code language: Python (python)

Expected Output:

True

Case 2:

s1 = "Ynf"
s2 = "PYnative"Code language: Python (python)

Expected Output:

False

Show Hint

Iterate each character from a string s1 and check if the current character is present in the string s2.

Show Solution

def string_balance_test(s1, s2):
    flag = True
    for char in s1:
        if char in s2:
            continue
        else:
            flag = False
    return flag


s1 = "Yn"
s2 = "PYnative"
flag = string_balance_test(s1, s2)
print("s1 and s2 are balanced:", flag)

s1 = "Ynf"
s2 = "PYnative"
flag = string_balance_test(s1, s2)
print("s1 and s2 are balanced:", flag)
Code language: Python (python)

Exercise 8: Find all occurrences of a substring in a given string by ignoring the case

Write a program to find all occurrences of “USA” in a given string ignoring the case.

Given:

str1 = "Welcome to USA. usa awesome, isn't it?"Code language: Python (python)

Expected Outcome:

The USA count is: 2

Show Hint

Use the string function count()

Show Solution

str1 = "Welcome to USA. usa awesome, isn't it?"
sub_string = "USA"

# convert string to lowercase
temp_str = str1.lower()

# use count function
count = temp_str.count(sub_string.lower())
print("The USA count is:", count)
Code language: Python (python)

Exercise 9: Calculate the sum and average of the digits present in a string

Given a string s1, write a program to return the sum and average of the digits that appear in the string, ignoring all other characters.

Given:

str1 = "PYnative29@#8496"Code language: Python (python)

Expected Outcome:

Sum is: 38 Average is  6.333333333333333

Show Hint

Iterate each character from a string s1 and check if the current character is digit using the isdigit() function

Show Solution

Solution 1: Use string functions

Iterate each character from a string s1 using a loop
In the body of a loop, check if the current character is digit using the isdigit() function
If it is a digit, then add it to the sum variable
In the end, calculate the average by dividing the total by the count of digits

input_str = "PYnative29@#8496"
total = 0
cnt = 0
for char in input_str:
    if char.isdigit():
        total += int(char)
        cnt += 1

# average = sum / count of digits
avg = total / cnt
print("Sum is:", total, "Average is ", avg)
Code language: Python (python)

Solution 2: Use regular expression

import re

input_str = "PYnative29@#8496"
digit_list = [int(num) for num in re.findall(r'\d', input_str)]
print('Digits:', digit_list)

# use the built-in function sum
total = sum(digit_list)

# average = sum / count of digits
avg = total / len(digit_list)
print("Sum is:", total, "Average is ", avg)Code language: Python (python)

Exercise 10: Write a program to count occurrences of all characters within a string

Given:

str1 = "Apple"Code language: Python (python)

Expected Outcome:

{'A': 1, 'p': 2, 'l': 1, 'e': 1}

Show Hint

Use the string function count()

Show Solution

create an empty dictionary to store the result. character is the key, and the count is the value
Iterate each character from a string s1 using a loop
In the body of a loop, use the count() function to find how many times a current character appeared in a string
Add key-value pair in a dictionary

str1 = "Apple"

# create a result dictionary
char_dict = dict()

for char in str1:
    count = str1.count(char)
    # add / update the count of a character
    char_dict[char] = count
print('Result:', char_dict)
Code language: Python (python)

Exercise 11: Reverse a given string

Given:

str1 = "PYnative"Code language: Python (python)

Expected Output:

evitanYP

Show Hint

Use negative slicing
Or use the built-in function reversed().

Show Solution

Solution 1: Negative String slicing

str1 = "PYnative"
print("Original String is:", str1)

str1 = str1[::-1]
print("Reversed String is:", str1)Code language: Python (python)

Solution 2: Using the reversed() function

str1 = "PYnative"
print("Original String is:", str1)

str1 = ''.join(reversed(str1))
print("Reversed String is:", str1)Code language: Python (python)

Exercise 12: Find the last position of a given substring

Write a program to find the last position of a substring “Emma” in a given string.

Given:

str1 = "Emma is a data scientist who knows Python. Emma works at google."Code language: Python (python)

Expected Output:

Last occurrence of Emma starts at index 43

Show Hint

Use the string function rfind()

Show Solution

str1 = "Emma is a data scientist who knows Python. Emma works at google."
print("Original String is:", str1)

index = str1.rfind("Emma")
print("Last occurrence of Emma starts at index:", index)Code language: Python (python)

Exercise 13: Split a string on hyphens

Write a program to split a given string on hyphens and display each substring.

Given:

str1 = Emma-is-a-data-scientistCode language: Python (python)

Expected Output:

Displaying each substring

Emma
is
a
data
scientist

Show Hint

Use the string function split()

Show Solution

str1 = "Emma-is-a-data-scientist"
print("Original String is:", str1)

# split string
sub_strings = str1.split("-")

print("Displaying each substring")
for sub in sub_strings:
    print(sub)
Code language: Python (python)

Exercise 14: Remove empty strings from a list of strings

Given:

str_list = ["Emma", "Jon", "", "Kelly", None, "Eric", ""]Code language: Python (python)

Expected Output:

Original list of sting
['Emma', 'Jon', '', 'Kelly', None, 'Eric', '']

After removing empty strings
['Emma', 'Jon', 'Kelly', 'Eric']

Show Hint

Use the built-in function filter() to remove empty strings from a list
Or use the for loop and if condition to remove the empty strings from a list

Show Solution

Solution 1: Using the loop and if condition

str_list = ["Emma", "Jon", "", "Kelly", None, "Eric", ""]
res_list = []
for s in str_list:
    # check for non empty string
    if s:
        res_list.append(s)
print(res_list)Code language: Python (python)

Solution 2: Using the built-in function filter()

str_list = ["Emma", "Jon", "", "Kelly", None, "Eric", ""]

# use built-in function filter to filter empty value
new_str_list = list(filter(None, str_list))

print("After removing empty strings")
print(new_str_list)Code language: Python (python)

Exercise 15: Remove special symbols / punctuation from a string

Given:

str1 = "/*Jon is @developer & musician"Code language: Python (python)

Expected Output:

"Jon is developer musician"

Show Hint

Use string functions translate() and maketrans()

Use the translate() function to get a new string where specified characters are replaced with the character described in a dictionary or a mapping table.
Use the maketrans() function to create a mapping table.

Or Use the regex in Python. See Python regex replace.

Show Solution

Solution 1: Use string functions translate() and maketrans().

The string.punctuation constant contain all special symbols.

import string

str1 = "/*Jon is @developer & musician"
print("Original string is ", str1)

new_str = str1.translate(str.maketrans('', '', string.punctuation))

print("New string is ", new_str)Code language: Python (python)

Solution 2: Using regex replace pattern in a string

import re

str1 = "/*Jon is @developer & musician"
print("Original string is ", str1)

# replace special symbols with ''
res = re.sub(r'[^\w\s]', '', str1)
print("New string is ", res)
Code language: Python (python)

Exercise 16: Removal all characters from a string except integers

Given:

str1 = 'I am 25 years and 10 months old'Code language: Python (python)

Expected Output:

Show Hint

Use the string function isdigit()

Show Solution

str1 = 'I am 25 years and 10 months old'
print("Original string is", str1)

# Retain Numbers in String
# Using list comprehension + join() + isdigit()
res = "".join([item for item in str1 if item.isdigit()])

print(res)Code language: Python (python)

Exercise 17: Find words with both alphabets and numbers

Write a program to find words with both alphabets and numbers from an input string.

Given:

str1 = "Emma25 is Data scientist50 and AI Expert"Code language: Python (python)

Expected Output:

Emma25
scientist50

Show Hint

Use the built-in function any() with the combination of string functions isalpha() and isdigit()

Show Solution

str1 = "Emma25 is Data scientist50 and AI Expert"
print("The original string is : " + str1)

res = []
# split string on whitespace
temp = str1.split()

# Words with both alphabets and numbers
# isdigit() for numbers + isalpha() for alphabets
# use any() to check each character

for item in temp:
    if any(char.isalpha() for char in item) and any(char.isdigit() for char in item):
        res.append(item)

print("Displaying words with alphabets and numbers")
for i in res:
    print(i)
Code language: Python (python)

Exercise 18: Replace each special symbol with # in the following string

Given:

str1 = '/*Jon is @developer & musician!!'Code language: Python (python)

Expected Output:

##Jon is #developer # musician##

Show Hint

Use string function replace()

Show Solution

Use the string.punctuation constant to get the list of all punctuations
Iterate each symbol from a punctuations
Use string function replace() to replace the current special symbol in a string with #

import string

str1 = '/*Jon is @developer & musician!!'
print("The original string is : ", str1)

# Replace punctuations with #
replace_char = '#'

# string.punctuation to get the list of all special symbols
for char in string.punctuation:
    str1 = str1.replace(char, replace_char)

print("The strings after replacement : ", str1)
Code language: Python (python)

Comments

Vaidik says

September 7, 2025 at 12:24 pm

# Exercise 16: Removal all characters from a string except integers
import string
str1 = ‘I am 25 years and 10 months old’
word=””.join([x for x in str1 if x.isdigit()])
print(word)

Reply
Vaidik says

September 7, 2025 at 12:15 pm

Exercise 17: Find words with both alphabets and numbers

def alpdit(text):
word=[x for x in text.split() if not x.isalpha()]
return word
print(alpdit(“Emma25 is Data scientist50 and AI Expert”))

Reply
harry says

July 11, 2025 at 7:11 am

# Exercise 10: Write a program to count occurrences of all characters within a string
str1=”abcdabcdabcd”
print(dict((char, str1.count(char)) for char in str1))

Reply
harry says

July 11, 2025 at 7:07 am

Exercise 9: Calculate the sum and average of the digits present in a string
str1=”123daf45sdfdsf”
nums=[int(num) for num in str1 if num.isdigit()]
print(sum(nums), sum(num)/len(nums))

Reply
harry says

July 11, 2025 at 7:03 am

Exercise 8: Find all occurrences of a substring in a given string by ignoring the case
import re
str1=”Welcome to Python To Python too”
str2=”to”
pattern = re.compile(‘\\b’+str2+’\\b’, re.IGNORECASE)
print(len(pattern.findall(str1)))

Reply
harry says

July 11, 2025 at 6:38 am

Exercise 7: String characters balance Test

str1 = “Yellow”
str2 = “Yell”
balanced = False

if all(i in str1 for i in str2) or all(i in str2 for i in str1):
balanced = True

print(f”{str1} and {str2} are balanced: {balanced}”)

Reply
- Rieri says
  
  July 26, 2025 at 9:58 pm
  
  I think u are right
  
  Reply
Stella says

May 27, 2025 at 6:46 am

#task 7
a=input(“enter a word: “)
b=input(“enter b word: “)
rule=””
if(len(a)==len(b)):
for i in range(len(a)):
rule+=a[i]
rule+=b[len(b)-1-i]
print(rule)
else:
print(“MUst be the same length !”)

Reply
Stella says

May 27, 2025 at 6:45 am

# task 19
txt=input(“enter a text: “)
newTxt=””
for l in txt:
if l==” “:
newTxt+=” ”
elif not l.isalnum():
l=”#”
newTxt+=l
print(“result:”,newTxt )

Reply
hamta says

April 20, 2025 at 8:49 pm

s1 = “sdf”
s2 = “gth”
# AzbycX
first=s1[0]+s2[-1]
middle=s1[int(len(s1)/2)]+s2[int(len(s2)/2)]
last=s1[-1]+s2[0]
print(first+middle+last)

Reply
Okbazghi says

January 28, 2025 at 5:43 pm

Exercise 1.A
str1=”James”
print(f'{str1[0]}{str1[len(str1)//2]}{str1[-1]}’)

Reply
karthik says

July 3, 2024 at 6:59 pm

Last exercise
str1 = “/*Jon is @developer & musician!!”
str=” ”

for i in range(0,len(str1)):
if str1[i].isalnum() or str1[i].isspace():
str+=str1[i]
else:
str+=”#”

print(str)

Reply
fasulu says

January 22, 2024 at 11:37 pm

Exercise 1A:

This will give “J M S” for 5 character long string and “N A” for 7 character long string

# str1 = “James”
str1 = “Banana”

length_ = round(len(str1)) // 2

print(str1[0]) # first

if len(str1) % 2:
print((str1[length_])) # middle
else:
print((str1[length_ – 1]), (str1[length_])) # middle

print(str1[len(str1) – 1]) # last

Reply
- Lokesh varma somula says
  
  June 9, 2024 at 6:08 am
  
  Here is the best answer
  1A. Write a program to create a new string made of an input string’s first, middle, and last character.
  
  str1 = “James”
  str1[0::2]
  
  Reply
  - Lokesh varma somula says
    
    June 11, 2024 at 3:27 am
    
    3. Given two strings, s1 and s2, write a program to return a new string made of s1 and s2’s first, middle, and last characters
    
    s1 = “America”
    s2 = “Japan”
    s1[0]+s2[0]+s1[len(s1)//2]+s2[len(s2)//2]+s1[-1]+s2[-1]
    
    Reply
  - Bogdan says
    
    October 4, 2024 at 11:47 am
    
    This one works only for a string with 5 characters like James, this answer is universal: str1 = “abcdefghj”
    print(str1[0::int(len(str1)/2)]) , if you assigned to str1 a string with an even number of characters it will output the first middle and last character of string.
    
    Reply
bigapple says

January 11, 2024 at 1:54 pm

# 17
def stay_alnum(string):
for item in string.split():
if not(item.isalpha()) and not(item.isdigit()):
print(item)

Reply
- dlinhares says
  
  March 28, 2024 at 8:34 pm
  
  #17
  i think i had a similar solution:
  
  str1 = “Emma25 is Data scientist50 and AI Expert”
  
  for i in str1.split():
  if i.isalpha() or i.isdigit():
  continue
  else:
  print(i)
  
  Reply
Walker says

December 4, 2023 at 10:56 am

Exercise 7:

def balance(s1, s2):
ct = s2.count(s1)
if ct > 0:
return True
else:
return False

x = “Ynf”
y = “PYnative”
z = balance(x, y)
print(z)

Reply
panos says

November 12, 2023 at 2:27 am

exercise 10

str1 = “Apple”

dict_str1 = {}

for i in str1:
dict_str1[i] = 0

for k in dict_str1:
count = str1.count(k)
dict_str1[k] += count

print(dict_str1)

Reply
panos says

November 12, 2023 at 1:31 am

exercise 8

str1 = “Welcome to USA. usa awesome, isn’t it?”

spl = str1.split()
for pos in range(len(spl)):
pos1 = spl[pos]
if spl[pos] == “USA.”:
print(f”USA pos is: {pos}”)

Reply
panos says

November 12, 2023 at 1:02 am

exercise 6

s1 = “Abc”
s2 = “Xyz”

for i in range(len(s2)):
rev = s2[-1-i]
s1.split()
j = “”.join((s1[i] + rev))
print(j, end=”)

Reply
panos says

November 7, 2023 at 1:11 am

Exercise 1

str1 = “James”

for i in range(len(str1)):
if i % 2 == 0:
print(str1[i], end=””)

Reply
- fasulu says
  
  January 21, 2024 at 6:04 pm
  
  This will produce answer for the word “Python” as “Pto, and “Strings” as “Srns”
  
  Reply
Irina says

October 26, 2023 at 12:27 am

Q15
def rem_sym(str1):
for i in str1:
for j in i:
if j in “!@#$%^&*?/”:
str1=str1.replace(i,””)
print(str1)

str1 = “/*Jon is @developer & musician”
rem_sym(str1)

Reply
m.Usman Momin says

October 24, 2023 at 10:10 pm

my solution to question 6:
def mixing(s1, s2):
s3 = ”
l1 = len(s1)
l2 = len(s2)
l = min(l1, l2)
print(l)
for i in range(l):
s3 = s3 + s1[i] + s2[l1-1-i]
print(s3)
if l1 > l2:
s3 += s1[l:]
elif l2 > l1:
s3 += s2[l:]
return s3

s1 = “Abc”
s2 = “Xyz”
s3 = mixing(s1, s2)
print(s3)

Reply
Millie says

October 24, 2023 at 6:25 pm

the 12th one can also be done by using
n=str1.index(“Emma”)
print(str1.index(“Emma”,n+1,len(str1)))

Reply
ऋषित रघुवंशी says

October 10, 2023 at 1:22 am

question2
str1 = “auly”
str2 = “kelly”
middle = int(len(str1)/2)
x=str1[:middle]+str2+str1[middle:]
print(x)

Reply
ऋषित रघुवंशी says

October 10, 2023 at 1:17 am

Question2
str1= “Auly”
middle = int(len(str1)/2)
str2= “kelly”
print(str1[:middle],str2,str1[middle:])

Reply
- ऋषित रघुवंशी says
  
  October 10, 2023 at 1:23 am
  
  question4
  str1 = “auly”
  str2 = “kelly”
  middle = int(len(str1)/2)
  x=str1[:middle]+str2+str1[middle:]
  print(x)
  
  Reply
Viswanath says

September 25, 2023 at 11:26 am

Ans for #17

str1 = “Emma25 is Data scientist50 and AI Expert”

for item in str1.split():
if not item.isdigit or not item.isalpha():
print(item,end=’\n’)

Reply
hosein says

August 23, 2023 at 3:30 am

A1راه حل بسیار کوتاه و عالی برای سوال

str1 = “Jemes”

print(str1)

res = str1[:: 2]

print(“new string:”, res)

Reply
- Mohammad says
  
  December 20, 2023 at 7:47 am
  
  str=’james’
  print(“Original string is: “,str)
  res=str[0]+str[2]+str[-1]
  print(“New string is: “,res)
  
  Reply
Ritabrata says

June 14, 2023 at 8:32 pm

Q.16 Using Regex.

import re
str1 = 'I am 25 years and 10 months old'
res=re.findall('\d+',str1) str_val="".join(res) print(str_val)

OUTPUT:
2510

Reply
Ritabrata says

June 14, 2023 at 8:27 pm

Q.17 Using Regex.

import re
str1 = "Emma25 is Data scientist50 and AI Expert" res=re.findall('\w+\d+',str1) for val in res: print(val)

OUTPUT:
Emma25
scientist50

Reply
Habibur Rahman says

May 29, 2023 at 12:57 am

my solution of q15
str1 = "/*Jon is @developer & musician" str = '' for i in str1: if i.isalnum() or i.isspace(): str += i
print(str)

Reply
Leigh says

April 21, 2023 at 8:28 pm

Firstly… this site is incredible and thank you so much for the time and effort you’ve put into this so we can all enjoy .^_^.

Exercise 4:
def lower_order(a:str)->str: new = "" for letter in a: if letter.islower(): new += letter for letter in a: if letter.isupper(): new += letter return(new)
s1 = "PyNaTive" print(lower_order(s1))

Reply
Anurag says

March 23, 2023 at 2:12 am

Question 2

str1 = input("Enter 1st word: ") str2 = input("Enter 2nd word: ") len1 = (len(str1))//2 print(str1[0:len1]+str2+str1[len1:])

Reply
Anurag says

March 22, 2023 at 7:10 pm

str1 = input("Enter 1st word: ") len1 = (len(str1))//2 print(str1[len1-1:len2+2])

Reply
Mahesh Reddy says

March 6, 2023 at 4:13 pm

exercise 18:

str1 = '/*Jon is @developer & musician!!'
old="/ * @ & ! !" new="# # # # # #"
print(str1.translate(str.maketrans(old,new)))

Reply
Mahesh Reddy says

March 6, 2023 at 4:07 pm

Exercise 18:

simple answer compared to exiting answer..

I hope this answer is helpful..

str1 = '/*Jon is @developer & musician!!' old="/ * @ & ! !" new="# # # # # #" change=str.maketrans(old,new) print(str1.translate(change))

Reply
- mohammad basheer says
  
  August 18, 2023 at 8:41 pm
  
  if you have hundreds of punctuations it is hard to replace like this
  
  Reply
0xAH says

March 2, 2023 at 7:30 pm

# 1B
def exe(x): mid = int(len(x)/2) print(x[mid-1:mid+2])

Reply
0xAH says

March 2, 2023 at 7:27 pm

# 1A

print(x[::int(len(x)/2)])

Reply
kalai says

February 8, 2023 at 12:35 pm

question 1:
a="james" a1=a[0] a2=a[2] a3=a[-1] b=a1+a2+a3 print(b)

Reply
Nermen says

October 17, 2022 at 5:00 pm

Question 7 :

s1 = "Yn" s2 = "PYnative"
print(all(item in s2 for item in s1))

Reply
- Abhishek says
  
  February 28, 2023 at 1:33 pm
  
  str1 = "PyNaTive" lowercase = "" uppercase = "" for char in str1: if char.islower(): lowercase = lowercase + char else: uppercase = uppercase + char
  result = lowercase + uppercase print(result)
  
  Reply

vini1955 says

October 10, 2022 at 2:10 am

Exercise 4:

def lower_first(str1):
    for c in str1:
        if c == c.upper():
            str1 = str1[1:] + str1[0]
    print(str1)

Shubham Yadav says

October 8, 2022 at 9:27 pm
Solution for 1A.
```
str1='james'
print(str[0:2])
```
Reply
- Onirom says
  
  January 28, 2023 at 5:13 pm
  
  str='james' print(str[0:6:2])
  
  Reply

lily fullery says

October 1, 2022 at 3:37 pm

question 17

str1 = "emma25 is Data scientist and AI Expert30"
m = str1.split()
for i in m :
    for j in i:
        if(j.isalpha()):
            continue
        elif(j.isdigit()):
            print(i)
            break

VijayKumar says

October 7, 2022 at 11:12 am

m = []
str1 = "Emma25 is Data scientist50 and AI Expert"
s = str1.split()
for i in s:
    if i.isalnum():
        m.append(i)
print(m[0:4:3])

DJ says

October 20, 2022 at 3:50 pm

I think you can omit the first condition just like this:

str1 = "Emma25 is Data scientist50 and AI Expert"
str1_list = str1.split()
for i in str1_list: for j in i: if j.isdigit(): print(i) break

Reply

Ali says

September 27, 2022 at 2:43 am
(Pretty) Short Solution for Exercise 6:
```
s1 = "Abc"
s2 = "Xyz"
for index in range(0,4):
    s3 = s1[index] + s2[-index-1]
    print(s3,end='')
```
Reply
- malleswari says
  
  December 7, 2022 at 1:03 pm
  
  AzbycX Traceback (most recent call last):
  s3 = s1[index] + s2[-index-1] IndexError: string index out of range
  
  here I got the index error can you please check the code once
  
  Reply

Amir says

August 29, 2022 at 1:03 pm

Short Solution of Exercise 14

str_list = ["Emma", "Jon", "", "Kelly", None, "Eric", ""]
for i in str_list:
    if i=="":
        continue
    else:
        print(i)

chethan says

October 4, 2022 at 4:33 pm

None will come

Reply

Chinuogu David Enweani says

August 21, 2022 at 9:01 pm

This is an alternate solution for Exercise 5

 str1 = "P@#yn26at^&i5ve"
print('Total counts of chars, digits and symbols')
print()
count = 0
count2 = 0
count3 = 0
for i in str1:
	if i.isalpha() == True:
		count += 1
	elif i.isdigit() == True:
		count2 += 1
	else:
		count3 += 1
		
print('Chars = ', count)
print('Digits = ', count2)
print('Digits = ', count3)

Chinuogu David Enweani says

August 21, 2022 at 9:10 pm

This is another more concise alt solution to exercise 5

 str1 = "P@#yn26at^&i5ve"
print('Total counts of chars, digits and symbols')
print()
count, count2, count3 = 0, 0, 0
for i in str1:
	if i.isalpha():
		count += 1
	elif i.isdigit():
		count2 += 1
	else:
		count3 += 1
		
print('Chars = ', count, 'Digits = ', count2, 'Symbols = ', count3)

sruthy kannath says

August 6, 2022 at 3:36 am

s1 = "Abc"
s2 = "Xyz"

s3=''
s2_rev = s2[::-1]
for i in range(len(s1)):
    s3 += s1[i]
    s3 += s2_rev[i]
print(s3)

luke says

August 2, 2022 at 10:23 pm

Solution for exercise number 1, take into a count in case the length of the string when divided by 2 does not =0

string = input("type string")
str_len = len(string)
sl = 0
if str_len%2!=0:
    sl = str_len-1
    print("First letter is: ", string[0], " middle letter is: ", string[int(sl/2)], " last letter is: ", string[-1])
elif str_len%2==0:
    sl = int(str_len)
    print("First letter is: ", string[0], " middle letter is: ", string[int(sl/2)], " last letter is: ", string[-1])

Prakash says

June 30, 2022 at 12:14 pm

Exercise 3:

def mix_string(s1, s2):
    print('Original String: ', s1, s2)
    first_s1 = s1[0]
    last_s1 = s1[-1]
    l1 = int(len(s1)/2)
    len1 = len(s1)
    if len1 % 2 == 0:
        mid_s1 = s1[l1 - 1]
    else:
        mid_s1 = s1[l1 - 0]
    first_s2 = s2[0]
    last_s2 = s2[-1]
    l2 = int(len(s2)/2)
    len2 = len(s2)
    if len2 % 2 == 0:
        mid_s2 = s2[l2 - 1]
    else:
        mid_s2 = s2[l2 - 0]
    final_string = first_s1 + first_s2 + mid_s1 + mid_s2 + last_s1 + last_s2
    return final_string

print('Mix String: ', mix_string("America", "Japan"))

Output:

Original String:  America Japan
Mix String:  AJrpan

Prakash says

June 30, 2022 at 10:36 am

def append_mid(s1 , s2):

    mid = int(len(s1)/2)
    x1 = s1[:mid]
    x2 = s1[mid:]
    x = x1 + s2 + x2
    return x

print( append_mid("Ault" , "Kelly"))

Output:

AuKellylt

Swati Sharma says

June 21, 2022 at 5:03 am

for exercise 17:

str1 ="Emma25 is Data scientist50 and AI Expert"
my_list=[]
temp=str1.split()
for i in temp:
    flagd=False
    flaga=False
    for char in i:
        if char.isalpha():
            flaga=True
        if char.isdigit():
            flagd=True
    if flaga==True and flagd==True:
        my_list.append(i)
print(my_list)

Sasidhar Puthineedi says

August 24, 2022 at 11:33 am

#Optimize using break statement in inner for loop:

str1 ="Emma25 is Data scientist50 and AI Expert"
my_list=[]
temp=str1.split()
for i in temp:
    flagd=False
    flaga=False
    for char in i:
        if char.isalpha():
            flaga=True
        if char.isdigit():
            flagd=True
        if flaga and flagd:
            my_list.append(i)
            break
print(my_list)

lily fullery says

October 1, 2022 at 3:35 pm

hey, thanks your code helped me solve this question.
please have a look at how I modified ur code, and it optimized

str1 = "emma25 is Data scientist and AI Expert30"
m = str1.split()
for i in m :
    for j in i:
        if(j.isalpha()):
            continue
        elif(j.isdigit()):
            print(i)
            break

lily fullery says

October 1, 2022 at 3:36 pm

str1 = "emma25 is Data scientist and AI Expert30"
m = str1.split()
for i in m :
    for j in i:
        if(j.isalpha()):
            continue
        elif(j.isdigit()):
            print(i)
            break

Mahendra says

May 28, 2025 at 6:39 pm

Very good champ

Reply

kalyan says

June 14, 2022 at 3:18 pm

s1 = "Ault"
s2 = "Kelly"
print(s1[0]+s2[0]+s1[2]+s2[2]+s1[-1]+s2[-1])

ex 3 answer

ram mote says

June 20, 2022 at 12:55 pm

Q 01:

str2 = "JaSonAy"
for i in range(len(str2)//2):
    y=str2[i:i+3]
print(y)

kalyan says

June 14, 2022 at 3:14 pm
```
s1 = "Ault"
s2 = "Kelly"
print(s1[:2]+s2[::]+s1[2:])
```
add string s2 in middle of s1 string
Reply

SinaJ says

May 27, 2022 at 7:23 pm

for exercise 7:

s1 = "Yn"
s2 = "PYnative"
print(bool(s2.count(s1)))
s1 = "Ynf"
s2 = "PYnative"
print(bool(s2.count(s1)))

lily fullery says

September 30, 2022 at 12:07 am

flag = True
s1 = "Ynf"
s2 = "PYnative"
for i in s1:
    if(i in s2):
        continue
    else:
        flag = False
        print(flag)
print(flag)

lily fullery says

September 30, 2022 at 12:13 am

please delete my previous reply

s1 = "Ynf"
s2 = "PYnative"
a = 0 
for i in s1:
    if(i not in s2):
        print(False)
        break
    elif(i in s2):
        a = a + 1
if(len(s1)==a):
    print(True)

Alvin says

May 5, 2022 at 1:58 pm
for number 14
```
s1 = ["Emma", "Jon", "", "Kelly", None, "Eric", ""]
while "" in s1:
    s1.remove("")
print(s1)
```
Reply
- Alvin says
  
  May 5, 2022 at 2:00 pm
  
  i forgot to add s1.remove(None) though
  
  Reply

Arnav saini says

April 14, 2022 at 9:56 pm

Alternate solution of Exercise 15: Remove special symbols/punctuation from a string

s = "/*Jon is @developer & musician"
 
 for item in s:
     if item.isalnum()!=True and ord(item)!=32:
         continue
     else:
         print(item,end="")

Arnav saini says

April 14, 2022 at 11:56 am

s = "PYnative29@#8496"
l=[]
for item in s:
    if item.isnumeric()==True:
        l.append(item)
print(l)
l = list(map(int,l))
print(sum(l))

bro says

April 27, 2022 at 3:56 am

Can you pleas explain to me this code

Reply

VijayKumar says

September 28, 2022 at 3:19 pm

str1 = "PYnative29@#8496"
m = []
sum = 0
for i in str1:
    if i.isdigit():
        m.append(i)
        sum = sum+int(i)
print(sum/6)

Rohit Chandoriya says

March 21, 2022 at 2:26 pm

For Problem 16


num1=input("Enter the starting number: ")
#num2=input("Enter the starting number: ")
#num1=[1,2,300,'',4,5,67,8,81,89,9,100,'',]


#num2=[]
def flo(a):
    num2=[]
    for word in a:
        if word.isdigit():
            num2.append(word)
    x=''.join(num2)
    print(x)

print(flo(num1))

Rishikesh Shah says

March 21, 2022 at 12:23 pm

Q 15

 import re
str1 = "/*John is @developer & musician"
new_str = ''
regex = re.compile('[@_!#$%^&*()?/\|}{~:]')
for i in str1:
    if regex.search(i) == None:
        new_str = new_str + i
print(new_str)

Vyom says

March 1, 2022 at 2:05 am

Example-18

str1 = '/*Jon is @developer & musician!!'
str2=list(split(str1))
lt=['/','*','@','&','!']
for i in range (len(lt)):
    str1=str1.replace(lt[i],'#')
print(str1)

Rishikesh Shah says

March 22, 2022 at 12:07 pm

Ex- 18

 import re
str1 = "/*John is @developer & musician"
new_str = ''
regex = re.compile('[@_!#$%^&*()?/\|}{~:]')
for i in str1:
    if regex.search(i):
        new_str = new_str + '#'
    else:
        new_str = new_str + i    
print(new_str)

Ulvy says

October 31, 2023 at 2:09 pm

str1 = “/*Jon is @developer!@#,><"
for i in str:
if i not in specialSymbols:
newStrList += i
return newStrList
print(removeSpecialSymbol(str1))

Reply

Olasunkanmi says

January 15, 2022 at 10:26 am

Question 6 alternative solution:

 def mixed_srting(x, y):
                  new_y = y[::-1]
                  for char in range(len(y)):
                         u = x[char] + new_y[char]
                         print(u, end="')
 
s1 = "Abc"
s2 = "Xyz"
mixed_string(s1, s2)

Olaniyi Peter Olasunkanmi says

January 14, 2022 at 10:56 am

Exercise 6

S1 = "Abc"
S2 = "Xyz"
b = 0
while b < len(s1):
        u = S2[::-1]
        x = S1[b] + S2[b]
        b += 1
        print(x, end="")

Haarini R says

December 30, 2021 at 3:54 pm

Exercise 10:

str1 = "Apple"
def my_new_func(each_char):
     return each_char,str1.count(each_char)

final_dict = dict(map(my_new_func,str1))
print(final_dict)

Giorgos says

December 22, 2021 at 12:53 pm
So, here’s the thing that I simply can’t understand. I tried the “if not i.isalpha” in order to filter the letters and also “if not i.isdigit” so it can be considered a complete filter for showing the symbols of a string. I printed i and I saw that it works. But then, the replacement worked only for the “!” and not the rest. So, I made a third-string outside of the for loop and I replaced “@” with “#” and it worked. So, why doesn’t the replace work completely inside the for loop, but when I use it outside of it with symbols that I had taken, it does?
```
str1 = '/*Jon is @developer & musician!!'
rep = "#"
for i in str1:
    if not i.isalpha() and not i.isdigit():
        print(i)
        str2 = str1.replace(i, "#")

str3 = str1.replace("@", "#")
print(str2)
print(str3)
```
Reply

Prakash A says

December 8, 2021 at 6:15 pm

QN 17

str1 = "emma25 is Data scientist50 and AI Expert"
m=(str1.split())
q=[]
for i in m:
    for j in i:
        if j.isdigit():
            for x in i:
                if x.isalpha():
                  q.append(i)
w=(set(q))
m=list(w)
for i in range(len(w)):
 print(m[i])

prakash says

December 8, 2021 at 6:17 pm

  
str1 = "emma25 is Data scientist50 and AI Expert"
m=(str1.split())
q=[]
for i in m:
    for j in i:
        if j.isdigit():
            for x in i:
                if x.isalpha():
                  q.append(i)
w=(set(q))
m=list(w)
for i in range(len(w)):
 print(m[i])

Preetam says

March 7, 2022 at 3:29 pm

str1 = "Emma25 is Data scientist50 and AI Expert"
import re
data = re.findall(pattern=r"(\w+\d)", string=str1)
print(data)

Gokul says

January 10, 2023 at 9:15 pm

str1 = "Emma25 is Data scientist50 and AI Expert"
str2=str1.split() l='' for i in str2: if i.isalpha(): None else: l=l+'\n'+i
print(l)

#Output
##Emma25
##scientist50

Reply

Inbar says

December 4, 2021 at 4:57 pm

Q13:
using list comprehension:

str_list = ["Emma", "Jon", "", "Kelly", None, "Eric", ""]
new_list =[item for item in str_list if item]
print(new_list)

Danilo says

November 18, 2021 at 3:43 pm

Exercise 3

s1 = "America"
s2 = "Japan"

# get middle index
x, y = int(len(s1)), int(len(s2))

# concatenate
mixed_string = s1[0] + s2[0] + s1[x] + s2[y] + s1[-1] + s2[-1]

print(mixed_string)

Danilo says

November 18, 2021 at 1:41 pm

Exercice 1A

str1 = "James"
print("String is:", str1)
print("New string is: ", str[0::2])

Prem Varma says

November 11, 2021 at 4:11 pm

Exercise 3

s1 = "America"
s2 = "Japan"
a = len(s1)//2
b = len(s2)//2
print(s1[0],s2[0],s1[a],s2[b],s1[-1],s2[-1])

sama71 says

October 25, 2021 at 7:26 pm

Q15

str1 = "/*Jon is @developer & musician"
for char in str1:
    if not char.isalpha():
        str1 = str1.replace(char, ' ')
print(str1)

output:
Jon is developer musician

Baraa says

September 23, 2021 at 11:38 pm
Q9
```
str1 = "English = 78 Science = 83 Math = 68 History = 65"

str1 = str1.split()
total =0
avg = 0
for num in str1:
  if num.isnumeric():
    num = int(num)
    total += num
    avg += 1
print(total)
print(total/avg)
```
Reply
- Olasunkanmi says
  
  September 28, 2021 at 11:13 am
  
  You’re an amazing thinker bro! I was very close to doing it on my own… But this piece has really broadened my horizon! God bless you
  
  Reply

Mazhar Khilji says

September 1, 2021 at 4:48 pm

Exercise 16:

str1 = 'I am 25 years and 10 months old'

for char in str1:
    if char.isdigit():
        print(char,end="")

amani says

August 9, 2021 at 12:44 am

str_list = ["Emma", "Jon", "", "Kelly", None, "Eric", ""]
def verif(char):
      if char is not None :
            return char
res = list(filter(verif, str_list))
print(res)

P. V. CHOWDARY says

August 3, 2021 at 10:12 pm

 def counter(x):
    output={}
    for i in x:
        output[i]=x.count(i)
    return output
counter(str1)

P. V. CHOWDARY says

August 3, 2021 at 9:47 pm

Q.9:

 def sum_average(x):
    sum=0
    count=0
    x=x.split()
    for i in range(0,len(x)):
        if x[i].isnumeric():
            count+=1
            sum=sum+int(x[i])
    avg=sum/count
    print(f'sum is {sum}\naverage is {avg}')

sum_average(str1)

P. V. CHOWDARY says

August 3, 2021 at 9:23 pm
Q.8:
```
 def usa_count(x):
    return x.lower().count('usa')
usa_count(str1)
```
Reply
- ashokkumar says
  
  September 6, 2023 at 9:55 pm
  
  can you give the code without using count function or any other inbuilt pythons libraries
  
  Reply

P. V. CHOWDARY says

August 3, 2021 at 8:28 pm

 def isbalance(x,y):
    output=True
    for i in x:
        if i in y:
            continue
        else:
            output=False
    return output

Chowdary P V says

August 1, 2021 at 9:40 pm

 def string_generator(x,y):
    a=[i for i in x]
    b=[j for j in y]
    b.reverse()
    c=[]
    for k in range(0,len(a)):
        for l in range(0,len(b)):
            if k==l:
                c.append(a[k])
                c.append(b[l])
    print(''.join(c))
string_generator('Abc','Xyz')

P. V. CHOWDARY says

August 1, 2021 at 7:44 pm

Question-5:

 def counter(x):
    digits=0
    chars=0
    symbols=0
    spaces=0
    for i in x:
        if i.isalpha():
            chars+=1
        elif i.isnumeric():
            digits+=1
        elif i.isspace():
            spaces+=1
        else:
            symbols+=1
    print(f'Chars={chars}\nDigits={digits}\nSymbols={symbols}\nSpaces={spaces}')
counter(s)#function calling

Output:

Chars=8
Digits=3
Symbols=4
Spaces=0

P.Raja Chowdary Polepalli says

July 21, 2021 at 1:57 pm

Exercise:4

 def string_manipulator(x):
       list1=[i for i in x if i==i.lower()]
       list2=[j for j in x if j==j.upper()]
       list3=list1 + list2
       print("".join(list3))

Raja Chowdary Polepalli says

July 21, 2021 at 1:24 pm

Exercise:3

def new_string(x,y):
    if len(x)%2 != 0 and len(y)%2 !=0:
        list1=[i for i in x]
        mid_val1 = list1.pop(len(list1) // 2)  #first ,we have to pop the middle value before popping 1st and last values
        first_val1=list1.pop(0)
        last_val1=list1.pop()
        list2 = [i for i in y]
        mid_val2 = list2.pop(len(list2) // 2)
        first_val2 = list2.pop(0)
        last_val2 = list2.pop()
        list3=[first_val1,first_val2,mid_val1,mid_val2,last_val1,last_val2]
    print("".join(list3))
new_string(s1, s2)

Raja Chowdary Polepalli says

July 21, 2021 at 12:38 pm

Exercise 2:

s1 = "Ault"
s2 = "Kelly"

def new_string(x,y):
    list=[i for i in x]
    list.insert(len(x)//2,y)
    print(''.join(list))
new_string(s1,s2)

Output: AuKellylt

Rishikesh Shah says

March 16, 2022 at 11:04 am

Exercise 2:


s1 = "Ault"
s2 = "Kelly"
# expected output AuKellylt
s3 = s1[0:len(s1)//2] + s2 + s1[-len(s1)//2:]
print(s3)

Raja Chowdary Polepalli says

July 21, 2021 at 12:14 pm

Question 1:

def middle_three(x):
    n=len(x)//2-1
    return x[n:n+3]
middle_three("JhonDipPeta") # 'Dip'
middle_three("JaSonAy") # 'Son'

Subin jerin says

July 20, 2021 at 11:03 pm

Anyone explain Q2 , I can’t understand that …? Pls help me

Reply
- Raja Chowdary Polepalli says
  
  July 21, 2021 at 12:39 pm
  Exercise 2:
```
s1 = “Ault”
s2 = “Kelly”

def new_string(x,y):
list=[i for i in x]
list.insert(len(x)//2,y)
print(”.join(list))
new_string(s1,s2)
```
  Output: AuKellylt
  Reply

Meghna says

July 9, 2021 at 4:28 pm

Q6 Alternate Solution

def mixString(s1,s2):
  lengthS1 = len(s1)
  lengthS2 = len(s2)
  length  = lengthS1 if lengthS1 < lengthS2 else lengthS2 
  resultString=""
  for i in range(length):
    resultString=resultString+s1[i]+s2[-(i+1)]
  if i < lengthS1-1 :
    resultString=resultString+s1[i+1:]
  elif i < lengthS2-1:
    resultString=resultString+s2[i-1:-(lengthS2)-1 :-1]
  print(resultString)
  
s1 = "Abcjjjj"
s2 = "Xyzss"
mixString(s1, s2)

Karyse says

July 8, 2021 at 9:13 am

Wow, where have you been all my life? Thank you ???? You have made the world a better place.

Narendra Yadav says

July 14, 2021 at 12:23 pm

17.

str1 = "Emma25 is Data scientist50 and AI Expert"
isAlnum = []
temp = str1.split()
# print(temp)

for data in temp:
	if not data.isalpha():
		isAlnum.append(data)
output = ' '.join(isAlnum)
print(output)

Danilo says

December 9, 2021 at 2:01 pm

I like the way you put it.
I would suggest the code ends like this to match the expected output.


str1 = "Emma25 is Data scientist50 and AI Expert"
isAlnum = []
temp = str1.split()
# print(temp)

for data in temp:
	if not data.isalpha():
		isAlnum.append(data)
# output = ' '.join(isAlnum)

for item in isAlnum:
    print(item)

Meghna says

July 6, 2021 at 9:54 pm

Question 1
with string as user input and length and odd length checking

def getMiddleThreeChars(sampleStr):
  middleIndex = int(len(sampleStr) /2)
  print("Original String is", sampleStr)
  middleThree = sampleStr[middleIndex-1:middleIndex+2]
  print("Middle three chars are", middleThree)
str1=input("enter a string of odd length greater than 7  ")
str1=str1.strip()
length=len(str1)
while ((length < 7) or ((length%2) == 0)):
  str1=input("enter a string of odd length greater than 7")
  str1=str1.strip()
  length=len(str1)
getMiddleThreeChars(str1)

Soe Wai Yan Aung says

April 1, 2021 at 10:08 pm

For Q 17


str1 = "Emma25 is Data scientist50 and AI Expert"
for i in str1.split(' '):
    if not i.isalpha() and not i.isnumeric():
        print(i)

Tushar says

April 24, 2022 at 5:09 pm

Thanks awesome way brilliant thinking!!

Reply

Renegade says

March 27, 2021 at 6:30 am

Alternative for Q. 13

def display(given):

  for i in given:
    replace = given.replace("-", "\n")
  print(replace)

str = "Emma-is-a-data-scientist"
display(str)

Attiya says

March 16, 2021 at 12:11 am

Another solution for question 4:

str1 = 'PyNaTive'
x = len(str1)

for y in range(x):
    if str1[y].islower() == True:
        print(str1[y],end='')

for i in range(x):
    if str1[i].isupper() == True:
        print(str1[i],end='')

Sat says

February 25, 2021 at 3:35 pm

For, Question 6,

s1 = "ABC"
s2 = "XYZ"
list1 = list(s1)
list2 = list(s2[::-1])
result = "".join(["".join(list(i)) for i in zip(list1, list2)])
print(result)

satheesh says

February 17, 2021 at 7:59 pm

For Question 15: (could be a simple solution)

str1 = "/*Jon is @developer$ &musician"
result = []
for i in str1:
    if i.islower() or i.isupper() or i == " ":
        result.append(i)
print("".join(result))

HT says

January 29, 2021 at 3:05 am
another solution to question 18:
```
import re
str2 = re.sub(r"[^\w\s]", "#", str1)
print(str2)
```
Reply
HT says

January 29, 2021 at 2:36 am
another solution to question 16
```
import re
str2 = re.sub(r"[^\d]", "", str1)
print(str2)
```
Reply
Farish says

January 28, 2021 at 9:05 am
I Just Wanna Know, Is this method is Correct or Not

# Question 1
```
str1 = "Jhon#Dip#Peta"
x=str1.split('#')
print(x)
print(x[1])
```
Reply

HT says

January 28, 2021 at 12:51 am

Another solution to question 7:


def string(s1,s2):
    set1 = set(s1)
    set2 = set(s2)
    result = set1.issubset(set2)
    return result

Rishikesh Shah says

March 17, 2022 at 11:51 am

Exercise 7


def fun_balanced_str(s1, s2):
    length_s1 = len(s1)
    length_s2 = len(s2)
    counter = 0    

    if length_s1 > length_s2:
        length_big = length_s1
        length_small =length_s2
    else:
        length_big = length_s2
        length_small= length_s1

    for i in range(length_small):
        for j in range(length_big):
            if s1[i] == s2[j]:                
                counter += 1          
        
    if counter >= length_small:
        print(True)
    else:
        print(False)

fun_balanced_str("Ynf", "PYnative")

Khanyi says

January 15, 2021 at 1:06 am

Another solution to question 1

str1 = "JaSonAy"
str2 = "JhonDipPeta"

hello = str1[2:5]
bye = str2[4:7]
print(hello)
print(bye)

Rajkumar says

January 11, 2021 at 1:33 pm

A solution to question 14

str_list = ["Emma", "Jon", "", "Kelly", None, "Eric", ""]

while '' in str_list:
    str_list.remove('')
print(str_list)

Rishikesh Shah says

March 21, 2022 at 11:56 am


str_list = ["Emma", "jon", "", "Kelly", None, "Eric", ""]
print("Original list of string", str_list)
new_list = []

for i in str_list:
    if '' != i:    
        new_list.append(i)       

print("After removing empty strings", new_list)

Rajkumar says

January 7, 2021 at 10:59 am

A solution to Question 5

str1 = "P@#yn26at^&i5ve"
lower = []
upper = []
digits = []
symb = []

for letters in str1:
    if letters.islower():
        lower.append(letters)
    elif letters.isupper():
        upper.append(letters)
    elif letters.isdigit():
        digits.append(letters)
    else:
        symb.append(letters)
result = (''.join(lower+upper+digits+symb))
print('Aligned String: ',result)
print('No. of lower case :', len(lower))
print('No. of upper case :', len(upper))
print('No. of digits     :', len(digits))
print('No. of Spl symbols:', len(symb))

Adham Jubran says

January 6, 2021 at 9:14 pm

The solution to Question 10

str1 = "Apple"

adict = {char : str1.count(char) for char in str1}

print(adict)

Adham Jubran says

January 6, 2021 at 8:56 pm

The solution to Question 9

str1 = "English = 78 Science = 83 Math = 68 History = 65"

str1 = str1.split()
str1 = str1[2: :3]
nSum = 0
for i in str1:
    nSum+= int(i)
print(nSum)

Adham Jubran says

January 6, 2021 at 8:38 pm

The solution to Question 7

s1 = "Yn"
s2 = "PYnative"
s3 = "Ynf"

print(set(s1).issubset(set(s2)))
print(set(s3).issubset(set(s2)))

Adham Jubran says

January 6, 2021 at 8:29 pm

The solution to Question 6:

s1 = "Abc"
s2 = "Xyz"

aset = set(s1 + s2)
while aset:
    print(aset.pop(), end= '')

sarika says

January 4, 2021 at 12:44 pm

A solution to question 3

def add(s1,s2):
    a=int(len(s1)/2)
    b=int(len(s2)/2)
    c=s1[0]+s2[0]+s1[a:a+1]+s2[b:b+1]+s1[-1]+s2[-1]
    print(c)
add("America","Japan")

Amrita Mehta says

December 30, 2020 at 10:18 am

Question 3


def name(s1,s2):
    new_name = s1[0]+s2[0]+s1[int(len(s1)/2)]+s2[int(len(s2)/2)]+s1[-1]+s2[-1]
    return new_name
name("America","Japan")

AndyLeeParker says

June 26, 2021 at 7:56 am

Beautiful!

Reply

Rajkumar says

December 24, 2020 at 5:36 pm

Question 3

def string(str1):
    
    for char1 in str1:
        if char1.islower():
            print(char1, end='')
    for char2 in str1:
        if char2.isupper():
            print(char2, end='')

str1 = "PyNaTive"
string(str1)

Rajkumar says

December 24, 2020 at 4:17 pm

Question 2

s1 = "Ault"
s2 = "Kelly"

s3 = s1[:2]+s2[:]+s1[2:]
print(s3)

Darrell Wright says

December 16, 2020 at 9:40 pm

Hi, total newbie here. with Exercise Question 6: Given two strings, s1 and s2, create a mixed String, there is reference made to the length of s1 and s2 in the function. Why is the length of the string relevant to the loop and in turn mixing of the strings? Thanks in advance.

Reply

ShwethaNagaraj says

December 16, 2020 at 4:16 pm

The answer of question no 12


str1 = "Emma is a data scientist who knows Python. Emma woks at google."
print("Last occurrence of Emma starts at index ",str1.rindex("Emma"))

Nik says

December 10, 2020 at 8:08 am

This is really helpful. Thanks

from string import punctuation

str1 = '/*Jon is @developer & musician!!'
print("The original string is: ", str1)

# Replace punctuations with #
replace_char = '#'

# Using string.punctuation to get the list of all punctuations
# use string function replace() to replace each punctuation with #

for char in punctuation:
    str1 = str1.replace(char, replace_char)

print("The strings after replacement : ", str1)

hemlata mahajan says

November 3, 2020 at 6:17 pm
q17:
```
import re
str1 = "Emma25 is Data scientist50 and AI Expert"
str2=re.findall('\w+\d+',str1)
p="\n".join(str2)
print(p)
```
Reply
- neha says
  
  November 26, 2020 at 9:34 am
  
  where is ques
  
  Reply
- Danilo says
  
  October 20, 2023 at 10:12 am
  
  Alternative:
  
  str1 = “Emma25 is Data scientist50 and AI Expert”
  for i in str1.split():
  if not i.isalpha():
  print(i)
  
  Reply

Aishwary saxena says

October 31, 2020 at 4:09 pm

str1 = "Emma is a data scientist who knows Python. Emma works at google."
count = 0
for key,value in enumerate(str1):
    if value == 'E' and key > 0:
        print('Emma at index index:',(key))

Aishwary saxena says

October 31, 2020 at 3:27 pm

ques 9 better methods to use regex than the listed above

str1 = "English = 78 Science = 83 Math = 68 History = 65"
emails = re.findall('[0-9]+', str1)
sumd = 0
for email in emails:
    sumd += int(email)
print(f"the sum is {sumd}")

percentage = sumd / len(emails)
print('average is:',percentage)

Aishwary saxena says

October 30, 2020 at 2:15 pm

MY WAY OF SOLVING QUESTION NUMBER 1

 str3 = input("enter the string:")
print(str3)
print("Write index.......")
startin = int(input("ENTER START INDEX:"))
endin = int(input("ENTER ENDING INDEX:"))
res = str3[startin:endin]
print(res)

Abdu says

October 27, 2020 at 10:22 pm

question 5

chars = 0
digits = 0
symbol = 0

str1 = "P@#yn26at^&i5ve"

for x in str1:
    if x.isalpha():
        chars+=1
    elif x.isdigit():
        digits+=1
    else:
        symbol+=1


print("chars: ",chars)
print("digits: ",digits)
print("symbol: ",symbol)

ur_bantyness says

October 4, 2020 at 11:35 pm

def middle(str_1):
    mid= None

    for i in range(0,len(str_1)-1):   #iterating from start index(0) to last index)
        mid = (0 + (len(str_1)-1) ) // 2   #finding the mid of the String

    return mid


str_1 = "JaSonAy"
op = middle(str_1)

print("the mid char are:{}{}{}".format(str_1[op-1],str_1[op],str_1[op+1])) # placing them using .format()

NerdAlert says

October 4, 2020 at 9:03 am

Q6: The solution didn’t account for the remaining letters to be added at the end.
My solution works for strings of any length, and also adds remaining letters at the end.

word1 = "abcdefih"
word2 = "xyzg"
str1 = word1[0]+word2[-1]
l1 = word1[1:]  # bcdefih
l2 = word2[0:-1]  # xyz
str2 = ""
str3 = ""
index = 0
n = 1
while True:
    if index < len(l1) and index < len(l2):
        str2 += l1[index]+l2[index-n]  #
        index += 1  # 2
        n += 2
    else:
        False
        break

final = str1+str2
res = list(word1+word2)
for item in res:
    if item not in list(final):
        str3 += item
final += str3

print(final)

Anita says

October 22, 2020 at 1:49 am

 
def mixString(s1, s2):
    newstr = ""
    length1 = int(len(s1))
    length2 = int(len(s2))
    length_min = min(length1, length2)
    length_max = max(length1, length2)
    for i in range(length_min):
        if i < length1:
            newstr = newstr + s1[i]
        if i < length2:
            newstr = newstr + s2[-i-1]
            
    for i in range(length_min, length_max):
        if i < length1:
            newstr = newstr + s1[i]
        if i < length2:
            newstr = newstr + s2[i-length_min]
    print(newstr)
    
mixString("xyzabce", "ABC")

DMS KARUNARATNE says

September 7, 2020 at 12:26 pm

Exercise Question 18

import re
str1 = "/*Jon is @developer & musician"
str2 = re.sub(r'[^\w\s]', '#', str1)
print( str2)

DMS KARUNARATNE says

September 4, 2020 at 10:25 am

Exercise Question 9:

str1 = "English = 78 Science = 83 Math = 68 History = 65"
strlist=[]
digilist=[]
total=0
digicount=0
digiav=0
strlist=str1.split()
for item in strlist:
    if item.isnumeric():
        digilist.append(item)
for i in digilist:
    total+=int(i)
    digicount+=1
    digiav=float(total/digicount)
print(total)
print(digiav)

apurv says

September 7, 2020 at 7:39 pm

thanks a lot

Reply

DMS KARUNARATNE says

September 3, 2020 at 4:03 pm

Exercise Question 7: String characters balance Test

s1 = "Yn"
s2 = "PYnative"
counter=len(s1)
cctr=0
for chr in s1:
    if chr in s2:
       cctr+=1
if counter==cctr:
    print("True")
else:
    print("False")

DMS KARUNARATNE says

September 3, 2020 at 1:24 pm

exercise-4

str1 = "PyNaTive"

for item in str1:
    if (item.islower()):
        print(item,end="")
for item in str1:
    if (item.isupper()):
        print(item,end="")

Akshata Gaekwad says

August 30, 2020 at 5:58 pm


def getMiddleThreeChars(sampleStr):
  middleIndex = int(len(sampleStr) /2)
  print("Original String is", sampleStr)
  start = middleIndex-1
  end = -start
  middleThree = sampleStr[start:end]

  print("Middle three chars are", middleThree)
  
getMiddleThreeChars("JhonDipPeta")
getMiddleThreeChars("Jasonay")

Galbatrollix says

August 13, 2020 at 10:55 pm

Q12: Alternative solution with for loop and count() function.


def where_last_Emma(string):
    for number in range(len(string)+1,0,-1):
        if string[number:].count("Emma") != 0:
            return number


print(where_last_Emma("Emma is a data scientist who knows Python. Emma works at google."))

Galbatrollix says

August 13, 2020 at 10:51 pm

Q12: Alternative solution


def where_last_Emma(string):
    for number in range(len(string)+1,0,-1):
        if string[number:].count("Emma")!=0:
            return number

print(where_last_Emma("Emma is a data scientist who knows Python. Emma works at google."))

Bhavya says

August 13, 2020 at 5:19 pm

incoming question 7.

def balance_test(str1, str2):
    if str1 in str2:
        print("str1 and str2 are balances", True)
    else:
        print("str1 and str2 are not balanced", False)


balance_test("ynz", "Pynative")

By the way, bro this site is THE BEST for beginners like me.

bhavya says

August 13, 2020 at 9:53 pm

for question number 9

def sum_and_average(english, science, math, history):
    sum = english + science + math + history
    average = (english + science + math + history) / 2
    print(f"The sum is",( sum ), "and average is",( average ), "of all the marks")


sum_and_average(78, 83, 68, 65)

can anyone please point out anything that I may have done wrong.

HOKAISEIG says

August 11, 2020 at 2:45 pm

QUESTIONS 4 and 5
why use split()?
i found no purpose.
is it to remove space ?

Reply
- Vishal says
  
  August 11, 2020 at 3:33 pm
  
  As of now, I have removed as there is a single word in the string. If you have a string with multiple words we need to use split().
  
  Reply

Galbatrollix says

August 11, 2020 at 12:21 am

Q6: I made an alternative solution that doesn’t require reversing one of the strings, and uses a while loop instead of for loop.


def mixstring2(s1,s2):
    len1=len(s1)
    len2=len(s2)
    finalstring=""
    number=0
    while True:
        if number<len1:
            finalstring+= s1[number]
        if number=len2 and number>=len1:
            break
    return finalstring
print(mixstring2("ABC", "XYZ"))

Galbatrollix says

August 11, 2020 at 12:23 am

So my solution was pasted incorrectly… Here is the correct version:


def mixstring2(s1,s2):
    len1=len(s1)
    len2=len(s2)
    finalstring=""
    number=0
    while True:
        if number<len1:
            finalstring+= s1[number]
        if number=len2 and number>=len1:
            break
    return finalstring
print(mixstring2("ABC", "XYZ"))

Galbatrollix says

August 11, 2020 at 12:26 am
For some reason I can’t post it correctly…
this part is supposed to be between first and second if statement:
```
        if number<len2:
            finalstring+= s2[-number-1]
        number+=1
```
Reply

Dalvir says

July 29, 2020 at 6:12 pm
Hello

Solution for Q. 9
```
str1 = "Welcome to USA. usa awesome, isn't it?"
string = str1.lower().count("usa")
print("The USA count is: ", string)
```
Reply
- Dalvir says
  
  July 29, 2020 at 6:15 pm
  
  Above solution is for Q8
  
  Reply
Mark says

July 26, 2020 at 10:48 pm

For question 7, as of 07-26-2020, the expected output should be “True” for Case 1. This is according to your code and description. s1 does exist within s2 for Case 1. This threw me when I was trying to solve the exercise, as I thought maybe you meant that s1 and s2 must be identical except for ordering. As you can imagine, the code for that would be different.

Reply
- Vishal says
  
  July 28, 2020 at 11:49 am
  
  You are right, Mark. I have updated the expected output
  
  Reply

shiva says

July 15, 2020 at 8:47 pm

Exercise Question 9

word = "English = 78 Science = 83 Math = 68 History = 65"

word_list = word.split()

total_sum = 0
subject_count = 0
for marks in word_list:
    if marks.isdigit():
        total_sum+=int(marks)
        subject_count+=1
print('sum is:\t', total_sum)
print('Avg is:\t', (total_sum/subject_count))

Mike Mace says

July 9, 2020 at 12:54 pm

None of the answers to Q 4, including the solution given, produce the required answer
i.e. aeiNPTvy

Reply
- Vishal says
  
  July 10, 2020 at 11:27 am
  
  Yes, it was mistakenly written I have removed the condition
  
  Reply

Aswathi Nair says

July 6, 2020 at 8:54 pm

Answer for first could be;


str1 = input("Enter the string: ") 
n = len(str1)//2
print(str1[n-1:n+2])

Angshumaan Basumatary says

June 25, 2020 at 10:11 am

Q: 7 Alternative solution hope it helps to understand better.

 def stringBalanceCheck(a, b):
    if a in b:
        return True
    else:
        return False


s1 = "yn"
s2 = "Pynative"
flag = stringBalanceCheck(s1, s2)
print("s1 and s2 are balanced", flag)

s1 = "ynf"
s2 = "Pynative"
flag = stringBalanceCheck(s1, s2)
print("s1 and s2 are balanced", flag)

SH Sourov says

June 19, 2020 at 2:26 pm


def twostr(s1,s2):
    midle = int(len(s1)/2)
    print(s1[0:midle-1]+" " + s2 + " " +s1[midle-1:len(s1)])
twostr("Chrisdem","IamNewString")

JeromeLille says

June 18, 2020 at 7:28 am

Hi,

My solution for Q10:


def count(str1):
    setstr1 = set(str1)
    dic = {}
    for letter in setstr1:
        dic[letter] = str1.count(letter)
    return dic
dic = count("pynativepynvepynative")   
print(dic)

Danilo says

November 19, 2021 at 12:43 pm

Question 5


str1 = "P@#yn26at^&i5ve"
char_list = []
digit_list = []
symbol_list = []
for i in str1:
    if i.isalpha():
        char_list.append(i)
        chars = len(char_list)
    elif i.isdigit():
        digit_list.append(i)
        digits = len(digit_list)
    else:
        symbol_list.append(i)
        symbols = len(symbol_list)
print("Total counts of chars, digits and symbols", "\n")
print("Chars =", chars)
print("Digits =", digits)
print(Symbols =", symbols)

JeromeLille says

June 18, 2020 at 7:25 am

Hi,

My solution for Q9:


def sumAndAverage(str1):
    sumNum = sum([int(s) for s in str1.split() if s.isnumeric()])
    countNum = len([int(s) for s in str1.split() if s.isnumeric()])
    print("Total Marks is:", sumNum, "Average is ", sumNum / countNum)

sumAndAverage("English = 78 Science = 83 Math = 68 History = 65")

JeromeLille says

June 18, 2020 at 7:20 am

Hi,

My solution for Q5:


def findDigitsCharsSymbols(inputString):
    charCount = len([i for i in inputString if i.isalpha()])
    digitCount = len([i for i in inputString if i.isdigit()])
    symbolCount = len(inputString) - charCount - digitCount
    print("Chars = ", charCount, "Digits = ", digitCount, "Symbol = ", symbolCount)
      
inputString = "P@#yn26at^&i5ve"
print("total counts of chars, digits,and symbols")

JeromeLille says

June 18, 2020 at 7:10 am

Hi,
My solution for Q4:


inputStr = "PyNaTive"

inputStr2 = [i for i in inputStr if i.islower()] + [i for i in inputStr if i.isupper()]

print("\n arranging characters giving precedence to lowercase letters:")
print(''.join(inputStr2))

swagat kamthe says

May 30, 2020 at 12:38 pm

The answer of Q:2 is wrong.
The correct answer is:

def appendMiddle(s1, s2):
  middleIndex = int(len(s1) /2)
  print("Original Strings are", s1, s2)
  middleThree = s1[:middleIndex:]+ s2 +s1[middleIndex:]
  print("After appending new string in middle", middleThree)
  
appendMiddle("Chrisdem", "IamNewString")

Faizan Ahmed says

May 27, 2020 at 9:36 am

Really Helpful thanks a lot

Reply

SHINU says

May 14, 2020 at 2:52 pm

Q8 character count

str1 = "Welcome to USA. usa awesome, isn't it?"
count_char = str1.lower().count("usa")
print(count_char)

output : 2

SHINU says

May 14, 2020 at 2:50 pm

str1 = "Welcome to USA. usa awesome, isn't it?"
count_char = str1.lower().count("usa")
print(count_char)

Nitish Tomar says

May 4, 2020 at 11:13 am

you should have used __contains__() function also

Reply

Nikhil K says

April 28, 2020 at 9:43 pm

Q9:

sumAndAverage = ("English = 78 Science = 83 Math = 68 History = 65")
sum = 0
count = 0
for i in sumAndAverage.split():
    if i.isnumeric():
        sum += int(i)
        count +=1
print("Total Marks are: ", sum)
print("Percentage is: ", sum/count)

Nikhil K says

April 28, 2020 at 8:42 pm

Q8: simplest way.

s1 = "Welcome to USA. usa awesome, isn't it?"
s2 = "USA"
print(s1.lower().count(s2.lower()))

Nikhil K says

April 28, 2020 at 8:35 pm

Q7: The simplest solution.

flag = True
if set(input("Enter the string: ")).issubset(input("Enter the string: ")):
    print("s1 and s2 are balanced: ", flag)
else:
    flag = False
    print("s1 and s2 are balanced ", flag)

Nikhil K says

April 28, 2020 at 8:16 pm

Q6 : without function.

s1 = "Pynative"
s2 = "Website"
s2 = s2[::-1]
ls1 = len(s1)
ls2 = len(s2)
len = ls1 if ls1 > ls2 else ls2
res = ""
for i in range(len):
    if i < ls1:
        res = res + s1[i]
    if i < ls2:
        res = res + s2[i]
print(res)

Nikhil K says

April 28, 2020 at 7:55 pm

Q5:

s1 = "P@#yn26at^&i5ve"
low = 0
upp = 0
num = 0
spl = 0
for i in s1:
    if i.islower():
       low += 1
    elif i.isupper():
        upp += 1
    elif i.isnumeric():
        num += 1
    else:
        spl += 1
print("Lower Chars = ",low,"\nUpper Chars = ",upp,"\nNumbers = ",num,"\nSpecial chars = ",spl)

Nikhil K says

April 28, 2020 at 7:26 pm

Q4:

s1 = 'PyNaTive'
low = []
upp = []
for i in s1:
    if i.islower():
       low.append(i)
    else:
        upp.append(i)
s2 = ''.join(low + upp)
print("arranging characters giving precedence to lowercase letters: \n" + s2)

Nikhil K says

April 28, 2020 at 5:15 pm
Question 3:
```
s1 = "America"
s2 = "Japan"
print(s1[0]+s2[0]+s1[int(len(s1)/2)]+s2[int(len(s2)/2)]+s1[-1]+s2[-1])
```
Reply
- Alec Nigh says
  
  June 8, 2020 at 12:45 am
  
  Hi Nikhil, thanks for posting these solutions. I have a question regarding getting the middle character for both strings – do you know why you would need the string variable (s1, s2) in front of the equation to obtain the middle variable? It seems redundant to me as you already have the variable in the equation. Thanks!
  
  Reply
  - Durga Sravanthi says
    
    October 27, 2020 at 12:32 pm
    middle_char = s1[(len(s1) / 2)] + s2[len(s2) /2]
    Reply

Nikhil K says

April 28, 2020 at 3:44 pm

Q2: The simplest solution.

s1 = "Chrisdem"
s2 = "IamNewString"
s3 = int(len(s1)/2)
print(s1[:s3-1]+s2+s1[s3-1:])

Nikhil K says

April 28, 2020 at 3:27 pm
Q1:
```
s1 = 'JhonDipPeta'
s2 = 'Jasonay'
mid1 = int(len(s1)/2)
print(s1[mid1-1:mid1+2])
mid2 = int(len(s2)/2)
print(s2[mid2-1:mid2+2])
```
Reply
- Yogi says
  
  April 25, 2022 at 12:11 pm
  
  i like this answer as a beginner i am getting this quick and easy.
  
  Reply
Prakash P Sinha says

April 14, 2020 at 10:11 pm

Dear Vishal, Its a time to appreciate you for your entire efforts to publish the site. This is really very helpful to understand from very beginning. I am based at Bangalore India and can help you out in case of any further business thoughts like Data Science implementations or more.

Of course yes, I have not seen such types of stuffs and the exercise with the detailed solutions. !!! Highly Appreciated .. 🙂

Reply
- Vishal says
  
  April 15, 2020 at 12:19 am
  
  Hey, Prakash, Thank you very much, It means a lot to me. Thanks for taking the time to let me know. When I start covering Data Science, I will let you know.
  
  Reply
AbduLLAH Zulfiqar says

April 7, 2020 at 1:00 pm

you didn’t explain arguments in Q9 for the func of re.findall. I get that but I suggest to explain them all for the new visitors to understand easily.

Reply
- Vishal says
  
  April 8, 2020 at 2:55 pm
  
  Thank you, Abdullah, for letting us know. Yes, we are planning to add a description to each question. You will see the changes in the coming days
  
  Reply
  - Prakash P Sinha says
    
    April 15, 2020 at 12:26 am
    
    Yes Abdullah, This is a valid point. However Vishal has mentioned that on upcoming modules, he is planning to add.
    
    I requested to Abdullah to please explain the same in this comments as well, so that any of the new visitor can understand after reading this comments.
    
    Reply

somnath says

March 14, 2020 at 10:57 am

x=''
y=''
p ='SoMnaTh'
for i in range(len(p)):         #upper case letter
        if p[i].upper()==p[i]:
                x =x+p[i]
for i in range(len(p)):        #lower case letter
        if p[i].lower()==p[i]:
                y=y+p[i]                
z = x+y                               #concatenate

Seetharam says

February 4, 2020 at 11:56 am
Question:1
Given a string of odd length greater 7,return a string made of the middle three chars of a given String
Answer:
```
a=input("Enter the String:")
b=0
n=0
for i in a:
	b=b+1
if(b%2==1):
	n=b//2
	print((a[n-1:n+2]))
```
Reply

El Mehdi Qaos says

January 25, 2020 at 5:41 am

solution 9:

def sum_average(s: str):
    l = []
    num = 0
    i = 0
    while i < len(s):
        if s[i].isdigit():
            num = int(s[i])
            j = i + 1
            while j < len(s) and s[j].isdigit():
                num = num * 10 + int(s[j])
                j += 1
                i += 1
            l.append(num)
        i += 1
    return [sum(l), sum(l)/len(l)]
 
result = sum_average('English = 78 Science = 83 Math = 68 History = 65')
print('sum', result[0], 'average', result[1])

Zlatko says

November 23, 2019 at 8:46 pm

Solution for Q1:

word = input("word: ")

if len(word)>=7 and len(word)%2 != 0:
    n = (len(word)-3)//2
    newWord1 = word[n:]
    middleThree = newWord1[:3]
    print("original word is: ", word)
    print("middle three chars are: ",middleThree)
    
else:
    print("word have less then 7 letters")

Waqar says

August 3, 2019 at 4:43 pm

string = input("Please enter your string: ")
my_string = len(string)+1
if len(string) % 2 ==0:
    our_string = len(string)
else:
    our_string = my_string
upper = (int(our_string) / 2)-2
lower = ((int(our_string)/2)* (-1))+1
upper = int(upper)
lower = int(lower)
print(string[upper:lower])

Uday Kiran Reddy Pateel says

August 1, 2019 at 6:30 pm

Solution for Q9:

def sumofnum(input):
    x=input.split()
    sum=0
    count=0

    for eachvalue in input:
            if eachvalue.isnumeric():
                sum+=int(eachvalue)
                count+=1
            else:
                print(eachvalue)

    average = (sum/count)
    print("Sum of integers:", sum)
    print("Count of Integers:", count)
    print("Average of integers:", average)


sumofnum("56xyz1234")

Barry Ford says

July 4, 2019 at 4:12 am

Bill’s solution to Question 4 is very elegant. Being a beginner, I really don’t understand it, but thanks for posting. I’ll keep studying.

Reply

Barry Ford says

July 4, 2019 at 3:58 am

Question 4 – I asked for user input. I think my solution is more straightforward than the one provided.

def printLowerfirst(s1):
    lowers = ''
    uppers = ''
    for char in my_word:
        if char.islower():
            lowers += char
        else:
            uppers += char
    return lowers + uppers

word = input("Enter a word: ")
print("Your word:", word)
print("Your word with lowercase letters first:", printLowerfirst(word))

Vitaliy says

June 20, 2019 at 8:12 pm

Q10

str1 = input("Input word : ")
 print({c : str1.count(c) for c in str1})

Tarun Verma says

May 29, 2019 at 2:24 pm
Question 6: Please don’t give complicated solutions to easy problems, when we can solve these by easy methods
```
s1 = "Pynative"
s2 = "Website"
L1 = list(s1)
L2 = list(s2)
mix = ""
for n in range(len(L1)):
  mix = mix + L1[0] + L2[-1]
  del(L1[0])
  del(L2[-1])
print (mix)
```
Reply
- Vignesh says
  
  July 13, 2019 at 9:08 pm
  
  Your answer is right if both the string lengths are equal. Else it is coming as list index out of range.
  
  Reply
- sonali says
  
  July 31, 2020 at 2:34 pm
  
  bro please check your code, it has errors
  
  Reply
- Abdullah says
  
  January 3, 2023 at 1:35 pm
  
  🤔🤔
  
  Reply

Bill Hardwick says

May 17, 2019 at 3:36 pm

Q10 – I offer my slightly different solution:

import pprint
def char_frequency(txt):
  char_counts = {}
  for i in range(len(txt)):
    char_counts.setdefault(txt[i], 0)
    char_counts[txt[i]] += 1
  return char_counts
pprint.pprint(char_frequency('abbcccddddeeeeeeeeeee'))

{'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 11}

Bill Hardwick says

May 17, 2019 at 3:18 pm

Q9 A very minor point, but in both the example and solution you use the word Percentage instead of Average. My solution without using regex (which mostly confuses me) is as follows:

def sum_and_average(string):
  lst = string.split()
  sum_int = 0
  count_int = 0
  for token in lst:
    if token.isdigit():
      count_int += 1
      sum_int += int(token)
  return (sum_int, sum_int/count_int)  
  
marks = 'English = 78 Science = 83 Math = 68 History = 65'
result = sum_and_average(marks)
print(f'Total marks earned = {result[0]}')
print(f'Average mark  = {result[1]}')

Total marks earned = 294
Average mark = 73.5

Bill Hardwick says

May 16, 2019 at 7:37 pm

Q7 I know it’s not what you were looking for, but my essentially one-liner is

def is_balanced(a, b):
  return set(a).issubset(set(b))
s1 = 'rose'
s2 = 'My arm is sore'
print(f'Are "{s1}" and "{s2}" balanced?  {is_balanced(s1, s2)}')
s2 = 'My arm is sort of limp'
print(f'Are "{s1}" and "{s2}" balanced?  {is_balanced(s1, s2)}')

Are “rose” and “My arm is sore” balanced? True
Are “rose” and “My arm is sort of limp” balanced? False

Bill Hardwick says

May 16, 2019 at 7:34 pm

Q6 again. I offer an alternative solution:

def mix_string(x, y):
  in_range = min(len(x), len(y))
  end = -1
  result = []
  for i in range(in_range):
    result.append(x[i])
    result.append(y[end])
    end -= 1
  if len(x) > len(y):
    result.append(x[abs(end) - 1:])
  elif len(y) > len(x):
    result.append(y[:end + 1])
  return ''.join(result)

Tarun Verma says

May 29, 2019 at 2:25 pm

s1 = "Pynative"
s2 = "Website"
mix = ""
L1 = list(s1)
L2 = list(s2)
for n in range(len(L1)):
  mix = mix + L1[0] + L2[-1]
  del(L1[0])
  del(L2[-1])
print (mix)

Bill Hardwick says

May 16, 2019 at 6:34 pm

It took me longer to understand what was required for Q6 than to come up with the solution! May I suggest a rewording of the question along the following lines:

Given two strings, s1 and s2, create a “mix string” such that the new string is made up of the first character of s1 followed by the last char of s2, then the second character of s1 followed by the penultimate character of s2, and so on. Any left-over characters from the longer string go at the end of the result.

Reply
- Vishal says
  
  May 16, 2019 at 10:22 pm
  
  Hey, Bill will reword the question 6. Thank you for all your comments and suggestion. Yes, there can be multiple alternative solutions for each question. I kept it simple because it is beginners exercises.
  
  Reply
Bill Hardwick says

May 16, 2019 at 6:07 pm
The example shown for Q4 is incorrect: only some of the lowercase characters appear before the uppercase ones.
My alternative solution to the one given is:
```
txt = 'PyNaTive'
lst = list(txt)
lst.sort(reverse=True)
txt2 = ''.join(lst)
print(txt2)
```
yvieaTPN
Reply
- Barry Ford says
  
  July 4, 2019 at 4:31 am
  
  I just figured out what you did here. You used the ASCII values. I’m a beginner and the only reason I figured it out was that I kept wondering what was sorted in reverse. To confirm, I ran the program having it print ord() of each character. I could also have simply looked them up.
  Pretty neat solution.
  
  Reply
- Nvm says
  
  October 31, 2023 at 7:07 pm
  
  It does not match the expected output.
  
  Reply
Bill Hardwick says

May 16, 2019 at 5:54 pm
Having interpreted the ambiguous wording of Q3 as meaning the first middle and last characters of s1 then the same for s2, my solution is:
```
def get_middle(x):
  return int((len(x) + 1) /2 - 1)
s1 = 'playlet'
s2 = 'heaps of iron'
s3 = s1[0] + s1[get_middle(s1)] + s1[-1] + s2[0] + s2[get_middle(s2)] + s2[-1]
print(s1)
print(s2)
print(s3)
```
Output:

playlet
heaps of iron
python
Reply

Python String Exercise with Solutions

Table of contents

Exercise 1A: Create a string made of the first, middle and last character

Exercise 1B: Create a string made of the middle three characters

Exercise 2: Append new string in the middle of a given string

Exercise 3: Create a new string made of the first, middle, and last characters of each input string

Exercise 4: Arrange string characters such that lowercase letters should come first

Exercise 5: Count all letters, digits, and special symbols from a given string

Exercise 6: Create a mixed String using the following rules

Exercise 7: String characters balance Test

Exercise 8: Find all occurrences of a substring in a given string by ignoring the case

Exercise 9: Calculate the sum and average of the digits present in a string

Exercise 10: Write a program to count occurrences of all characters within a string

Exercise 11: Reverse a given string

Exercise 12: Find the last position of a given substring

Exercise 13: Split a string on hyphens

Exercise 14: Remove empty strings from a list of strings

Exercise 15: Remove special symbols / punctuation from a string

Exercise 16: Removal all characters from a string except integers

Exercise 17: Find words with both alphabets and numbers

Exercise 18: Replace each special symbol with # in the following string

About Vishal

Related Tutorial Topics:

Python Exercises and Quizzes

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Table of contents

Exercise 1A: Create a string made of the first, middle and last character

Exercise 1B: Create a string made of the middle three characters

Exercise 2: Append new string in the middle of a given string

Exercise 3: Create a new string made of the first, middle, and last characters of each input string

Exercise 4: Arrange string characters such that lowercase letters should come first

Exercise 5: Count all letters, digits, and special symbols from a given string

Exercise 6: Create a mixed String using the following rules

Exercise 7: String characters balance Test

Exercise 8: Find all occurrences of a substring in a given string by ignoring the case

Exercise 9: Calculate the sum and average of the digits present in a string

Exercise 10: Write a program to count occurrences of all characters within a string

Exercise 11: Reverse a given string

Exercise 12: Find the last position of a given substring

Exercise 13: Split a string on hyphens

Exercise 14: Remove empty strings from a list of strings

Exercise 15: Remove special symbols / punctuation from a string

Exercise 16: Removal all characters from a string except integers

Exercise 17: Find words with both alphabets and numbers

Exercise 18: Replace each special symbol with # in the following string

About Vishal

Related Tutorial Topics:

Python Exercises and Quizzes

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