Python Data Structure Exercise for Beginners

Updated on: May 22, 2025 | 119 Comments

This data structures exercise is designed for beginners to understand and practice fundamental data structures in Python. You’ll practice Python List, Set, Dictionary, and Tuple questions.

Data structures are crucial for organizing and holding data. To perform any programming task effectively in Python, a strong grasp of data structures is essential.

Solve this exercise to develop a solid understanding of basic data structures in Python.

Also, Solve:

This Exercise includes the followings: –

This exercise includes 10 questions, each with a provided solution.
The questions cover various methods for manipulating lists, dictionaries, sets, and tuples.
Use an Online Code Editor to solve the exercises.

Exercise 1: List Creation using two lists

Write a code to create a new list using odd-indexed elements from the first list and even-indexed elements from the second

Given two lists, l1 and l2, write a program to create a third list l3 by picking an odd-index element from the list l1 and even index elements from the list l2.

Given:

l1 = [3, 6, 9, 12, 15, 18, 21]
l2 = [4, 8, 12, 16, 20, 24, 28]Code language: Python (python)

Expected Output:

Element at odd-index positions from list one
[6, 12, 18]
Element at even-index positions from list two
[4, 12, 20, 28]

Printing Final third list
[6, 12, 18, 4, 12, 20, 28]

Show Hint

Use list slicing

Show Solution

To access a range of items in a list, use the slicing operator :. With this operator, you can specify where to start the slicing, end, and specify the step.

For example, the expression list1[ start : stop : step] returns the portion of the list from index start to index stop, at a step size step.

for 1st list: Start from the 1st index with step value 2 so it will pick elements present at index 1, 3, 5, and so on
for 2nd list: Start from the 0th index with step value 2 so it will pick elements present at index 0, 2, 4, and so on

list1 = [3, 6, 9, 12, 15, 18, 21]
list2 = [4, 8, 12, 16, 20, 24, 28]
res = list()

odd_elements = list1[1::2]
print("Element at odd-index positions from list one")
print(odd_elements)

even_elements = list2[0::2]
print("Element at even-index positions from list two")
print(even_elements)

print("Printing Final third list")
res.extend(odd_elements)
res.extend(even_elements)
print(res)Code language: Python (python)

Exercise 2: Remove and add item in a list

Write a program to remove the item present at index 4 and add it to the 2nd position and at the end of the list.

Given:

list1 = [54, 44, 27, 79, 91, 41]Code language: Python (python)

Expected Output:

List After removing element at index 4  [34, 54, 67, 89, 43, 94]
List after Adding element at index 2  [34, 54, 11, 67, 89, 43, 94]
List after Adding element at last  [34, 54, 11, 67, 89, 43, 94, 11]

Show Hint

Use the list methods, pop(), insert() and append()

Show Solution

pop(index): Removes and returns the item at the given index from the list.
insert(index, item): Add the item at the specified position(index) in the list
append(item): Add item at the end of the list.

sample_list = [34, 54, 67, 89, 11, 43, 94]

print("Original list ", sample_list)
element = sample_list.pop(4)
print("List After removing element at index 4 ", sample_list)

sample_list.insert(2, element)
print("List after Adding element at index 2 ", sample_list)

sample_list.append(element)
print("List after Adding element at last ", sample_list)Code language: Python (python)

Exercise 3: Slice list into 3 equal chunks and reverse each chunk

Given:

sample_list = [11, 45, 8, 23, 14, 12, 78, 45, 89]Code language: Python (python)

Expected Outcome:

Chunk  1 [11, 45, 8]
After reversing it  [8, 45, 11]
Chunk  2 [23, 14, 12]
After reversing it  [12, 14, 23]
Chunk  3 [78, 45, 89]
After reversing it  [89, 45, 78]

Show Hint

Divide the length of a list by 3 to get the each chunk size
Run loop three times and use the slice() function to get the chunk and reverse it

Show Solution

Get the length of a list using a len() function
Divide the length by 3 to get the chunk size
Run loop three times
In each iteration, get a chunk using a slice(start, end, step) function and reverse it using the reversed() function
In each iteration, start and end value will change

sample_list = [11, 45, 8, 23, 14, 12, 78, 45, 89]
print("Original list ", sample_list)

length = len(sample_list)
chunk_size = int(length / 3)
start = 0
end = chunk_size

# run loop 3 times
for i in range(3):
    # get indexes
    indexes = slice(start, end)
    
    # get chunk
    list_chunk = sample_list[indexes]
    print("Chunk ", i, list_chunk)
    
    # reverse chunk
    print("After reversing it ", list(reversed(list_chunk)))

    start = end
    end += chunk_sizeCode language: Python (python)

Exercise 4: Count the occurrence of each element from a list

Write a program to iterate a given list and count the occurrence of each element and create a dictionary to show the count of each element.

Given:

sample_list = [11, 45, 8, 11, 23, 45, 23, 45, 89]Code language: Python (python)

Expected Output:

Printing count of each item   {11: 2, 45: 3, 8: 1, 23: 2, 89: 1}

Show Solution

sample_list = [11, 45, 8, 11, 23, 45, 23, 45, 89]
print("Original list ", sample_list)

count_dict = dict()
for item in sample_list:
    if item in count_dict:
        count_dict[item] += 1
    else:
        count_dict[item] = 1

print("Printing count of each item  ", count_dict)Code language: Python (python)

Exercise 5: Paired Elements from Two Lists as a Set

Write a code to create a Python set such that it shows the element from both lists in a pair.

Given:

first_list = [2, 3, 4, 5, 6, 7, 8]
second_list = [4, 9, 16, 25, 36, 49, 64]Code language: Python (python)

Expected Output:

Result is  {(6, 36), (8, 64), (4, 16), (5, 25), (3, 9), (7, 49), (2, 4)}

Show Hint

Use the zip() function. This function takes two or more iterables (like list, dict, string), aggregates them in a tuple, and returns it.

Show Solution

first_list = [2, 3, 4, 5, 6, 7, 8]
print("First List ", first_list)

second_list = [4, 9, 16, 25, 36, 49, 64]
print("Second List ", second_list)

result = zip(first_list, second_list)
result_set = set(result)
print(result_set)Code language: Python (python)

Exercise 6: Set Intersection and Removal

Write a code to find the intersection (common) of two sets and remove those elements from the first set.

See: Python Set

Given:

first_set = {23, 42, 65, 57, 78, 83, 29}
second_set = {57, 83, 29, 67, 73, 43, 48}Code language: Python (python)

Expected Output:

Intersection is  {57, 83, 29}
First Set after removing common element  {65, 42, 78, 23}

Show Hint

Use the intersection() and remove() method of a set

Show Solution

Get the common items using the first_set.intersection(second_set)
Next, iterate common items using a for loop
In each iteration, use the remove() method of on first set and pass the current item to it.

first_set = {23, 42, 65, 57, 78, 83, 29}
second_set = {57, 83, 29, 67, 73, 43, 48}

print("First Set ", first_set)
print("Second Set ", second_set)

intersection = first_set.intersection(second_set)
print("Intersection is ", intersection)
for item in intersection:
    first_set.remove(item)

print("First Set after removing common element ", first_set)Code language: Python (python)

Exercise 7: Subset or Superset of another set

Write a code to checks if one set is a subset or superset of another set. If found, delete all elements from that set.

Given:

first_set = {27, 43, 34}
second_set = {34, 93, 22, 27, 43, 53, 48}Code language: Python (python)

Expected Output:

First set is subset of second set - True
Second set is subset of First set -  False

First set is Super set of second set -  False
Second set is Super set of First set -  True

First Set  set()
Second Set  {67, 73, 43, 48, 83, 57, 29}

Show Hint

Use the below methods of a set class

issubset()
issuperset()
clear()

Show Solution

first_set = {57, 83, 29}
second_set = {57, 83, 29, 67, 73, 43, 48}

print("First Set ", first_set)
print("Second Set ", second_set)

print("First set is subset of second set -", first_set.issubset(second_set))
print("Second set is subset of First set - ", second_set.issubset(first_set))

print("First set is Super set of second set - ", first_set.issuperset(second_set))
print("Second set is Super set of First set - ", second_set.issuperset(first_set))

if first_set.issubset(second_set):
    first_set.clear()

if second_set.issubset(first_set):
    second_set.clear()

print("First Set ", first_set)
print("Second Set ", second_set)Code language: Python (python)

Write a program to iterate a given list and check if a given element exists as a key’s value in a dictionary. If not, delete it from the list

Given:

roll_number = [47, 64, 69, 37, 76, 83, 95, 97]
sample_dict = {'Jhon':47, 'Emma':69, 'Kelly':76, 'Jason':97}Code language: Python (python)

Expected Outcome:

After removing unwanted elements from list [47, 69, 76, 97]

Show Solution

roll_number = [47, 64, 69, 37, 76, 83, 95, 97]
sample_dict = {'Jhon': 47, 'Emma': 69, 'Kelly': 76, 'Jason': 97}

print("List:", roll_number)
print("Dictionary:", sample_dict)

# create new list
roll_number[:] = [item for item in roll_number if item in sample_dict.values()]
print("after removing unwanted elements from list:", roll_number)
Code language: Python (python)

Exercise 9: Extract Unique Dictionary Values to List

Write a code to get all values from the dictionary and add them to a list but don’t add duplicates

Given:

speed = {'jan': 47, 'feb': 52, 'march': 47, 'April': 44, 'May': 52, 'June': 53, 'july': 54, 'Aug': 44, 'Sept': 54}Code language: Python (python)

Expected Outcome:

[47, 52, 44, 53, 54]

Show Solution

speed = {'jan': 47, 'feb': 52, 'march': 47, 'April': 44, 'May': 52, 'June': 53,
         'july': 54, 'Aug': 44, 'Sept': 54}

print("Dictionary's values - ", speed.values())

speed_list = list()

# iterate dict values
for val in speed.values():
    # check if value not present in a list
    if val not in speed_list:
        speed_list.append(val)
print("unique list", speed_list)Code language: Python (python)

Exercise 10: remove duplicates from a list

Write a code to remove duplicates from a list and create a tuple and find the minimum and maximum number

Given:

sample_list = [87, 45, 41, 65, 94, 41, 99, 94]Code language: Python (python)

Expected Outcome:

unique items [87, 45, 41, 65, 99]
tuple (87, 45, 41, 65, 99)
min: 41
max: 99

Show Solution

sample_list = [87, 52, 44, 53, 54, 87, 52, 53]

print("Original list", sample_list)

sample_list = list(set(sample_list))
print("unique list", sample_list)

t = tuple(sample_list)
print("tuple ", t)

print("Minimum number is: ", min(t))
print("Maximum number is: ", max(t))Code language: Python (python)

Comments

Stef says

December 26, 2024 at 11:08 pm

Great site!
Perhaps there’s a mistake in Ex 2:
list1 = [54, 44, 27, 79, 91, 41]
should be:
sample_list = [34, 54, 67, 89, 11, 43, 94]

Reply
vilakshan kaushik says

May 7, 2024 at 10:45 pm

Q3: My answer to solve this question is

lst = [11, 45, 8, 23, 14, 12, 78, 45, 89]

start = 0
end = len(lst)

for i in range(start,end,3):
print(list(reversed(lst[i:i+3])))

Reply
akri_ma says

January 13, 2024 at 8:02 pm

Hi, there is an error in Exercise 2:
Given list is [54, 44, 27, 79, 91, 41], but in answer is [34, 54, 67, 89, 11, 43, 94].
Change it for the future.

The website is really good to start!

Reply
fateme says

November 15, 2023 at 10:11 am

Q8. Alternative answer:

roll_number = [47, 64, 69, 37, 76, 83, 95, 97]
sample_dict = {‘Jhon’:47, ‘Emma’:69, ‘Kelly’:76, ‘Jason’:97}

b = []
for key,value in sample_dict.items():
if value in roll_number:
b.append(value)
print(“after removing unwanted elements from list:”, b)

Reply
fateme says

November 15, 2023 at 10:05 am

Q8. Alternative answer:

roll_number = [47, 64, 69, 37, 76, 83, 95, 97]
sample_dict = {‘Jhon’:47, ‘Emma’:69, ‘Kelly’:76, ‘Jason’:97}

b = []
for key,value in sample_dict.items():
if value in roll_number:
b.append(value)
print(“after removing unwanted elements from list:”, b)

Reply
fateme says

November 14, 2023 at 12:35 pm

Q4. Short answer:

sample_list = [11, 45, 8, 11, 23, 45, 23, 45, 89]

dict_list = {}
for i in sample_list:
dict_list[i] = sample_list.count(i)
print(dict_list)

Reply
Nvm says

November 11, 2023 at 7:05 pm

Exercise 10:
Expected outcome is incorrect.

sample_list = [87, 45, 41, 65, 94, 41, 99, 94]

unique items [87, 45, 41, 65, 99]

41 should not be in the unique items list

Reply
Muhammad Usman says

October 26, 2023 at 10:55 am

EXERCISE : 4

sample_list = [11, 45, 8, 11, 23, 45, 23, 45, 89]
data = dict()
for i in range(len(sample_list)):
data.update({sample_list[i]:sample_list.count(sample_list[i])})
print(“Printing count of each item : “,data)

Reply
Anon says

April 14, 2023 at 2:45 pm

Just wanted to say that this is a great list of quizzes for beginners to learn and brush up on Python skills.
Big thanks and keep up the good work.

Reply
Mahesh Reddy says

March 7, 2023 at 11:58 am

exercise:10

sample_list = [87, 45, 41, 65, 94, 41, 99, 94]
l=list() for i in sample_list: if i not in l: l.append(i) print("unique items",l) tup=tuple(l) print("tuple" ,tup)
print("min: ",min(l),"\nmax: ",max(l))

Reply
Mahesh Reddy says

March 7, 2023 at 1:07 am

Exercise 8:

some change to the existing code.I Hope it’s helpful to you..

roll_number = [47, 64, 69, 37, 76, 83, 95, 97] sample_dict = {'Jhon':47, 'Emma':69, 'Kelly':76, 'Jason':97}
lst=list() for item in roll_number: if item in sample_dict.values(): lst.append(item) print("After removing unwanted elements from list",lst)

Reply
Mahesh Reddy says

March 6, 2023 at 10:46 pm

Exercise 3:
I Hope this answer is helpful..
#given

sample_list = [11, 45, 8, 23, 14, 12, 78, 45, 89]

#solution

c1=sample_list[0:3] print("chunk 1",c1) #reverse c1 c1.reverse() print("After reversing it",c1 )
c2=sample_list[3:6] print("chunk 2",c2) #reverse c2 c2.reverse() print("After reversing it ",c2)
c3=sample_list[6:] print("chunk 3",c3) #reverse c3 c3.reverse() print("After reversing it",c3)

Reply
- Joonatan says
  
  April 19, 2025 at 6:16 am
  
  sample_list = [11, 45, 8, 23, 14, 12, 78, 45, 89]
  l = len(sample_list) // 3
  for i in range(0, 3):
  chunk = sample_list[i*l:(i+1)*l]
  print(f”Chunk {i+1} {chunk}”)
  print(f”After reversing it {chunk[::-1]}”)
  
  Reply
Adrian says

February 3, 2023 at 8:38 pm

In Exercise 2 is bug, there are two different lists presented in input and output.

Reply
Miri says

January 26, 2023 at 1:29 am

My solution for Exercise 9 (one-liner):

print(list(set(speed.values())))

Reply
- Peri says
  
  August 24, 2023 at 12:31 pm
  
  Yup. similar idea:
  my_list = list(dict.fromkeys(speed.values()))
  
  Reply
Shubham Navghare says

December 21, 2022 at 1:49 am

For Exercise 8

roll_number = [47, 64, 69, 37, 76, 83, 95, 97]
sample_dict = {'Jhon':47, 'Emma':69, 'Kelly':76, 'Jason':97} result =[]
for i in roll_number: if i == i in sample_dict.values(): result.append(i) print(result)

Reply
Vishal Soni says

December 13, 2022 at 12:46 pm

Question 9 :
speed = {'jan': 47, 'feb': 52, 'march': 47, 'April': 44, 'May': 52, 'June': 53, 'july': 54, 'Aug': 44, 'Sept': 54}
lst = [set(speed.values())] print(lst)
Question 10 : sample_list = [87, 45, 41, 65, 94, 41, 99, 94] tup = (set(sample_list)) print(max(tup)) print(min(tup))

Reply
Anton D says

December 12, 2022 at 12:20 am

speed = {'jan': 47, 'feb': 52, 'march': 47, 'April': 44, 'May': 52, 'June': 53, 'july': 54, 'Aug': 44, 'Sept': 54} my_set = set(speed.values()) print(list(my_set))

Does this work for exercise 9?

Reply
- Monishree RmeshBabu says
  
  June 28, 2023 at 1:07 pm
  
  yes it does
  
  Reply
Khan says

November 2, 2022 at 7:07 am

For exercise 8
roll_number = [47, 64, 69, 37, 76, 83, 95, 97] sample_dict = {'Jhon':47, 'Emma':69, 'Kelly':76, 'Jason':97}
result = list() for i in roll_number: if i in sample_dict.values(): result.append(i) else: result print(result)

Reply
arek says

October 21, 2022 at 6:43 pm

In Exercise 10, the output ought to be: sample_tuple = (87, 45, 65, 99), because number 41 is duplicated in the original list.

Reply

luke says

August 3, 2022 at 2:39 pm

exercise1
l1 = [3, 6, 9, 12, 15, 18, 21]
l2 = [4, 8, 12, 16, 20, 24, 28]
l3 = []
for item in range(1,len(l1)-1+1,2):
    l3.append(l1[item])
for item2 in range(0,len(l2)-1+1,2):
    l3.append(l2[item2])
print(l3)
silution2
l1 = [3, 6, 9, 12, 15, 18, 21]
l2 = [4, 8, 12, 16, 20, 24, 28]
list = [l1[1::2],l2[0::2]]

print(list)

swat says

June 22, 2022 at 1:44 am

speed = {'jan': 47, 'feb': 52, 'march': 47, 'April': 44, 'May': 52, 'June': 53,'july': 54, 'Aug': 44, 'Sept': 54}
new=set()
for i in speed.values():
    new.add(i)
new_1=list(new)
print(new_1)

swat says

June 22, 2022 at 1:45 am

for exercise 9

Reply

pavan says

September 23, 2022 at 11:31 am

exercise 9

speed = {'jan': 47, 'feb': 52, 'march': 47, 'April': 44, 'May': 52, 'June': 53, 'july': 54, 'Aug': 44, 'Sept': 54}
list_a=[]
for i in speed.values():
    if(i not in list_a):
        list_a.append(i)
print(list_a)

Okbazghi says

January 30, 2025 at 12:22 am

Excellent idea!

Reply

hiom says

May 8, 2022 at 4:33 am

Cane same one explain to me exercise N4

Azia says

May 23, 2022 at 9:43 am

Hello
I will propose to you 3 other solutions for #4. Maybe they will help you understand the author’s solution


#option 1 - use a set to get distinct elements and then loop on the set

sample_list = [11, 45, 8, 11, 23, 45, 23, 45, 89]
sample_set = set(sample_list)

dict = {}
# print(type(dict))

for item in sample_set: 
    dict[item] = sample_list.count(item)
    
print('Printing count of each item ', dict)


-------------------------------------------
#option 2 - Use a loop

sample_list = [11, 45, 8, 11, 23, 45, 23, 45, 89] 

count_dict = {}
# print(type(dict))

for item in sample_list:
    if item in count_dict.keys():
        pass
    else:
        count_dict[item] = sample_list.count(x)
    
print('Printing count of each item ', count_dict)

---------------------------------------

#option 3  - use dictionary comprehension

sample_list = [11, 45, 8, 11, 23, 45, 23, 45, 89] 
 
dict = { k: sample_list.count(k) for k in sample_list}
print('Printing count of each item ', dict)

naresh irigineni says

October 20, 2022 at 8:57 am

Hi, Azia this is Naresh I checked your program you don take X which is not defined in the program that’s why the program throws the error.
line count_dict[item] = sample_list.count(x) "remove the x insert the item
thanks

sample_list = [11, 45, 8, 11, 23, 45, 23, 45, 89]
count_dict = {} # print(type(dict)) for item in sample_list: if item in count_dict.keys(): pass else: count_dict[item] = sample_list.count(x)
print('Printing count of each item ', count_dict)

Reply

Swats says

June 21, 2022 at 6:03 am

sample_list = [11, 45, 8, 11, 23, 45, 23, 45, 89]
my_dic={}
for i in sample_list:
    count=sample_list.count(i)
    my_dic[i]=count
print(my_dic)

Alvin says

May 5, 2022 at 6:45 pm

exercise 4

s1 = [11, 45, 8, 23, 14, 12, 78, 45, 89]
lengthChunk=int(len(s1)/3)
temp=[]
for x in range(0,9,lengthChunk):
    temp=s1[x:x+lengthChunk]
    temp.reverse()
    print(temp)

Alvin says

May 5, 2022 at 6:52 pm
the solution above is for exercise 3 sorry.My alternative for exercise 4 would be the following
```
s1 = [11, 45, 8, 11, 23, 45, 23, 45, 89]
for x in set(s1):
    print(str(x)+":"+str(s1.count(x)))
```
Reply

Kai says

April 25, 2022 at 7:23 pm

Hi there! I am a newbie in Python, so I thought I would share my method of solving Question 7 as my way is a little different from what you guys did. However, the answer is still the same.

first_set = {27, 43, 34}
second_set = {34, 93, 22, 27, 43, 53, 48}

a = first_set.issubset(second_set)
b = second_set.issubset(first_set)

if a == True:
    print("First set is subset of second set - True")
else:
    print("First set is subset of second set - False")
if b == True:
    print("Second set is subset of First set - True")
else:
    print("Second set is subset of First set - False")

c = first_set.issuperset(second_set)
d = second_set.issuperset(first_set)

if c == True:
    print("First set is Super set of second set - True")
else:
    print("First set is Super set of second set - False")
    
if d == True:
    print("Second set is Super set of First set - True")
else:
    print("Second set is Super set of First set - False")

if a or c == True:
    first_set.clear()
    
print("First Set", first_set)
print("Second Set", second_set)

sonali says

April 6, 2022 at 6:09 pm

exercise 1

l1=[10,20,30,40,50]
l2=[10,40,50,60,70,90]
print(l1[0::2])
print(l2[1::2])

Rishikesh Shah says

April 4, 2022 at 10:11 pm

Exercise 9:

import collections
speed = {'jan': 47, 'feb': 52, 'march': 47, 'April': 44, 'May': 52, 'June': 53, 'july': 54, 'Aug': 44, 'Sept': 54}
dict_val = speed.values()
new_list = [item for item, count in collections.Counter(dict_val).items() if count > 1 or count >= 1]
print(new_list)

Rishikesh Shah says

April 4, 2022 at 8:07 pm

Exercise 7:

first_set = {27, 43, 34}
second_set = {34, 93, 22, 27, 43, 53, 48}
print("First set is subset of second set - ", first_set.issubset(second_set))
print("Second set is subset of First set - ", second_set.issubset(first_set))
print("First set is super of second set - ", first_set.issuperset(second_set))
print("Second set is super of First set - ", second_set.issuperset(first_set))
res_first_set = first_set.difference_update((second_set))
print("First Set", first_set)
print("Second Set", second_set)

Rishikesh Shah says

March 23, 2022 at 12:39 pm
To Exercise 2:
Hi Vishal, thanks for this awesome tutorial. It helped me a lot.
I found a mistake in question Nr. 2
The expected output according to the question should be:
```
List After removing element at index 4  [54, 44, 27, 79, 41]
List after Adding element at index 2 [54, 44, 91, 27, 79, 41]
List after Adding element at last  [54, 44, 91, 27, 79, 41, 91]
```
Reply
- harshul Patel says
  
  November 3, 2022 at 5:28 pm
  
  yes, you are right bro.. 🙂
  
  Reply

Aditya Patel says

March 3, 2022 at 12:45 pm

Q1 –
new_list = l1[1:len(l1):2] + l2[0:len(l2):2]

Rishikesh Shah says

March 23, 2022 at 12:01 pm

Ex- 1

l1 = [3, 6, 9, 12, 15, 18, 21]
l2 = [4, 8, 12, 16, 20, 24, 28]
odd_list = []
even_list = []
new_lst = []
counter = 0
for i in range(len(l1)):
    if i % 2 != 0:
        odd_list.append(l1[i])
print("Element at odd-index position from list one\n", odd_list) 
for j in range(len(l2)):
    if j % 2 == 0:
        even_list.append(l2[j])
print("Element at even-index positions from list two\n", even_list)  
new_lst = [*odd_list, *even_list]  
print("Printing Final third list\n", new_lst)

zxenon_ says

February 18, 2022 at 2:32 am

For exercise 6

# We can use set methods “intersection()” and “difference()”

first_set = {23, 42, 65, 57, 78, 83, 29}
second_set = {57, 83, 29, 67, 73, 43, 48}

print(f"Intersection is {first_set.intersection(second_set)}")
print(f"First set after removing common elements {first_set.difference(first_set.intersection(second_set))}")

RT says

January 18, 2022 at 4:48 pm

Exercise 5 solution:

first_list = [2, 3, 4, 5, 6, 7, 8]
second_list = [4, 9, 16, 25, 36, 49, 64]
third_list = set()

for i, e in enumerate(first_list):
	pair = (e, second_list[i])
	third_list.add(pair)
print(third_list)

suman says

March 18, 2022 at 11:37 pm

for i in zip(first_list,second_list):
    print(i)

Maria says

April 20, 2022 at 2:40 pm


def list_pairs():
    first_list = [2, 3, 4, 5, 6, 7, 8]
    second_list = [4, 9, 16, 25, 36, 49, 64]
    my_set = [(x, y) for x in first_list for y in second_list if x ** 2 == y]
    print('Result is:', set(my_set))

list_pairs()

Danilo says

December 13, 2021 at 1:21 pm

Question 8


roll_number = [47, 64, 69, 37, 76, 83, 95, 97]
sample_dict = {'John':47, 'Emma':69, 'Kelly':76, 'Jason':97}

new_list = []
for i in roll_number:
    if i in sample_dict.values():
        new_list.append(i)

print("after removing unwanted elements from list:", new_list)

Danilo says

December 10, 2021 at 12:49 pm

Question 1


l1 = [3, 6, 9, 12, 15, 18, 21]
l2 = [4, 8, 12, 16, 20, 24, 28]

l1a = [l1[i] for i in range(0, len(l1), 2)] # Odd indexes
l2a = [l2[j] for j in range(1, len(l2), 2)] # Even indexes

l3 = l1a + l2a # Mix Odd and Even indexes

# Print
print("Element at odd-index position from list one", "\n", l1a)
print("Element at even-index position from list two", "\n", l2a)
print("Printing final third list", "\n", l3)

sama71 says

October 26, 2021 at 1:14 pm

sample = [11, 45, 8, 11, 23, 45, 23, 45, 89]
dic = {}
key_list = []
value_list = []
for n in sample:
    key_list.append(n)
    c = sample.count(n)
    value_list.append(c)
    
# first method
    for key in key_list:
        for value in value_list:
            dic[key] = value
print(dic)             


# second method   
print ({key_list[i] : value_list[i] for i in range(len(key_list))})
         

# third method
print(dict(zip(key_list, value_list)))

mar says

December 21, 2021 at 6:41 pm

sample_list = [11, 45, 8, 11, 23, 45, 23, 45, 89]

my_dict = {}
for item in sample_list:
    
    x = sample_list.count(item)
    
    my_dict[item] = x

print('Printing count of each item', my_dict)

Shiv says

September 9, 2021 at 1:46 pm


list=[11, 45, 8, 11, 23, 45, 23, 45, 89]
dict = {}
for i in range(0,len(list)):
    dict[list[i]] = list.count(list[i])
print(dict)

Mazhar says

September 4, 2021 at 10:32 am

Exercise 1:

def check(l1,l2):
    third_list = []

    for num in range(1,len(l1),2):
        third_list.append(l1[num])

    for num in range(0,len(l2),2):
        third_list.append(l2[num])

    print(third_list) 

listOne = [3, 6, 9, 12, 15, 18, 21]
listTwo = [4, 8, 12, 16, 20, 24, 28]
check(listOne,listTwo)

Mark Kustelski says

August 11, 2021 at 5:01 pm

list1 = [11,45,8,23,14,12,78,45,89]
list2=list1[:3:]
list3=list1[3:6:]
list4=list1[6::]
print(list2[::-1], list3[::-1], list4[::-1])

Andy Lee Parker says

June 28, 2021 at 11:39 pm

Here’s my solution for #1.

listOne = [3, 6, 9, 12, 15, 18, 21]
listTwo = [4, 8, 12, 16, 20, 24, 28]
listThree = listOne[1::2] + listTwo[0::2]
print(listThree)

Narender says

May 31, 2021 at 9:39 pm

Can you check the Example 2 Question and Answer

Reply

Hobby Programer says

May 25, 2021 at 1:40 am

Hello
I do not understand

listOne = [3, 6, 9, 12, 15, 18, 21]
listTwo = [4, 8, 12, 16, 20, 24, 28]
listThree = list()

oddElements = listOne[1::2]
print("Element at odd-index positions from list one")
print(oddElements)

EvenElement = listTwo[0::2]
print("Element at even-index positions from list two")
print(EvenElement)

print("Printing Final third list")
listThree.extend(oddElements)
listThree.extend(EvenElement)
print(listThree)

What I did was:

listOne = [3, 6, 9, 12, 15, 18, 21]
listTwo = [4, 8, 12, 16, 20, 24, 28]
odd=[]
even=[]
listThree=[]
for i in listOne:
    if i % 2 == 1:
        odd.append(i)
for i in listTwo:
    if i % 2 == 0:
        even.append(i)
for i in odd:
        listThree.append(i)
for i in even:
        listThree.append(i)
print("Element at odd-index positions from list one")
print(odd)
print("Element at even-index positions from list two")
print(even)
print("Printing Final third list")
print(listThree)

and I got it wrong. What is :: used for?

frk_th says

May 28, 2021 at 8:53 pm
```
listOne = [3, 6, 9, 12, 15, 18, 21]
```
[x::y] means a slice of the listOne
where the x is the index to start from and y is the step size
So, for example;
listOne[0:3] ##By the way you can write this as listOne[::3]
I will start from index 0, which is ‘3’
Then I will use the 3rd index from the starting index; which is index 3 = ’12’
So, my result will be;
```
[0, 12, 21]
```
Reply

Raju says

June 9, 2021 at 2:45 pm

lst1 = [3, 6, 9, 12, 15, 18, 21]
lst2= [4, 8, 12, 16, 20, 24, 28]
odd_index=[]
even_index=[]
combo_index=[]
for i in range (len(lst1)):
    if i%2==1:
        odd_index.append(lst1[i])
        combo_index.append(lst1[i])

print("Elements of odd index list:",odd_index)
for j in range (len(lst2)):
    if j%2==0:
        even_index.append(lst2[j])
        combo_index.append(lst2[j])
print("Elements of even index list:",even_index)
print("Final third list:",combo_index)

Chandini Maheshwari says

June 14, 2021 at 7:33 pm

list1=[10,33,77,64,88,63]
list2=[20,60,77,51,90,11]
list3=[]
print("Original List1 is",list1)
print("Original List2 is",list2)
print("Odd List is")
for i in list1:
    if i%2!=0:
        print(i,end=" ")
        list3.append(i)
        print(" ")
print("Even List is")
for i in list2:
    if i%2==0:
        print(i,end=" ")
        list3.append(i)
        print(" ")
print("Merge List1 and List2 is")
print(list3,end=" ")
print(" ")

VIKRAM says

July 5, 2021 at 9:44 pm

you have to check indexes that they are odd or even, instead you are checking for elements odd or even

Reply

happy says

May 8, 2021 at 5:42 pm

can i get solution for this???

You are required to design the data structure to display the individual player stats for cricket players. A player may have represented more than one team and may have played in more than one format such as Test, ODI and T20.

Problem Statement
Create a list of different data fields and use appropriate Python data types to represent each of them.

Reply

Pawel says

March 25, 2021 at 1:38 pm

Hi,
Exercise 10

sampleList = [87, 45, 41, 65, 94, 41, 99, 94]
unique items [87, 45, 41, 65, 99] missing 94 :)

I thought that you want to remove all element what was duplicated, but 41 is in unique item

Narmin says

August 8, 2021 at 12:22 am

from random import*
n=int(input("say:"))
A=[randint(0,100) for i in range(n)]
print(A)
B=[]
for i in range(n):
	for j in range(n):
		if A[i]==A[j] and A[i] not in B:
			B+=[A[i]]
print(B)
maxx=A[0]
minn=A[0]
for i in range(1,n):
	if A[i]>maxx:
			maxx=A[i]
	if A[i]<minn:
		minn=A[i]
print(maxx)
print(minn)

Narmin says

August 8, 2021 at 12:23 am

Exercise 10

Reply

Marco says

March 12, 2021 at 9:31 pm

An alternative way to solve question # 8. I do not know why, but it works! PS: Thank you very much for this website, it’s helping me tremendously!!! =)


rollNumber = [47, 64, 69, 37, 76, 83, 95, 97]
sampleDict ={'Jhon':47, 'Emma':69, 'Kelly':76, 'Jason':97}
for i in rollNumber[:]:
    if i not in sampleDict.values():
        rollNumber.remove(i)
print(rollNumber)

Sat says

February 27, 2021 at 7:53 pm
Q9:
```
print(list(set(speed.values())))
```
Reply

Sat says

February 27, 2021 at 4:31 pm

Q3:

sampleList = [11, 45, 8, 23, 14, 12, 78, 45, 89]
chunk  = 3
for i in range(0, len(sampleList ), chunk):
    print(list1[i:i+chunk][::-1])

someboi says

February 11, 2021 at 11:24 pm


 def something(list):
    index = int(len(data) / 3)
    i = 0
    while i < index:
        chunk = data[index*i:index*(i+1)]
        chunk = chunk[::-1]
        print(chunk)
        i = i + 1

data = [11, 45, 8, 23, 14, 12, 78, 45, 89]
something(data)

Naruto Uzumaki says

September 30, 2021 at 2:33 pm

sampleList = [11, 45, 8, 23, 14, 12, 78, 45, 89]
l=len(sampleList)//3
c=0
d=[]
for j in range(1,l+1):
    
    for i in sampleList[c:c+3]:
        
        d.append(i)
        
    print(d[::-1])
    c=c+3
    d=[]

Suman kumar Shrestha says

February 7, 2021 at 5:04 am

Can anyone tell me what’s wrong in this code?
it prints after removing unwanted elements from list [47, 69, 76, 95, 97]
Why is 95 still in the list


rollNumber = [47, 64, 69, 37, 76, 83, 95, 97]
sampleDict ={'Jhon':47, 'Emma':69, 'Kelly':76, 'Jason':97}

for i in rollNumber:
    if i not in sampleDict.values():
        rollNumber.remove(i)

print("after removing unwanted elements from list ", rollNumber)

haddad says

August 20, 2021 at 7:20 pm

rollNumber = [47, 64, 69, 37, 76, 83, 95, 97]
sampleDict ={'Jhon':47, 'Emma':69, 'Kelly':76, 'Jason':97}

for i in rollNumber[::-1]:
    if i not in sampleDict.values():
        rollNumber.remove(i)
print("after removing unwanted elements from list ", rollNumber)

Danilo says

December 13, 2021 at 4:43 pm


rollNumber = [47, 64, 69, 37, 76, 83, 95, 97]
sampleDict ={'Jhon':47, 'Emma':69, 'Kelly':76, 'Jason':97}

for i in rollNumber[0:9]: # or rollNumber[:]  <<<< apparently you need to define the index
    if i not in sampleDict.values():
        rollNumber.remove(i)

print("after removing unwanted elements from list ", rollNumber)

Rajkumar says

January 12, 2021 at 5:43 pm

A solution to question 9

speed ={'jan':47,'feb':52,'march':47,'April':44,'May':52,'June':53, 'july':54, 'Aug':44, 'Sept':54}

new_list = []

for key, val in speed.items():
    if val not in new_list:
        new_list.append(val)
print(new_list)

Raghu says

November 6, 2020 at 7:20 pm
#Given a dictionary get all values from the dictionary and add it in a list but don’t add duplicates
```
speed ={'jan':47, 'feb':52, 'march':47, 'April':44, 'May':52, 'June':53, 'july':54, 'Aug':44, 'Sept':54}
list2[:] = [item for item in speed.values()]
print(list(set(list2)))
```
Reply
- nikhil says
  
  December 3, 2020 at 9:37 pm
  
  yupp comprehensive lists great
  
  Reply

CaraDoCoidgo says

September 14, 2020 at 1:57 am

Q.4:

lst = [11, 45, 8, 11, 23, 45, 23, 45, 89]
dc = {}
for x in lst:
    dc[x] = lst.count(x)
print(dc)

DMS KARUNARATNE says

September 8, 2020 at 9:46 am

Exercise Question 4

from collections import Counter
orilist=[11, 45, 8, 11, 23, 45, 23, 45, 89]
z=Counter(orilist)
print(dict(z))

Galbatrollix says

August 25, 2020 at 1:54 am
Q5:

You can do this by using set comprehension with zip() function:
```
def set_of_pairs(list1,list2):
    return {(x,y) for x,y in zip(list1,list2)}
```
Reply

sheena says

August 18, 2020 at 11:42 am

my solution for Q4

 Original_list = [11, 45, 8, 11, 23, 45, 23, 45, 89]
dict_new=dict()
for i in Original_list:
    c=Original_list.count(i)
    dict_new[i]=c
print(dict_new)

Shivam says

August 6, 2020 at 8:07 pm

Hi Vishal,
Thanks for making pynative.com. its really helpful. Given exercise are grt to make some basic concepts.
only request i have is please add some more questions in exercise.

Reply
- Vishal says
  
  August 7, 2020 at 11:13 am
  
  Hey Shivam, Yes I will add more such examples for practice
  
  Reply

Dalvir says

July 31, 2020 at 10:54 am

solution to Q3:


sampleList = [11, 45, 8, 23, 14, 12, 78, 45, 89]
chunk1 = sampleList[:3]
print(chunk1[::-1])
chunk2 = sampleList[3:6]
print(chunk2[::-1])
chunk3 = sampleList[6:9]
print(chunk3[::-1])

Charan says

October 18, 2020 at 8:30 pm

perfect and simple solution. Well done

Reply

Raghu Varier says

November 6, 2020 at 7:18 pm

#Given a list slice it into a 3 equal chunks and reverse each list

list1 = input().split()
l = len(list1)
for i in range(0,3):
  sublist = []
  for j in range(i*l//3,(i+1)*l//3):
    sublist.append(list1[j])
  sublist.reverse()
  print(sublist)

Try this for any list length

Raghu says

November 6, 2020 at 7:19 pm

#Given a list slice it into a 3 equal chunks and reverse each list

list1 = input().split()
l = len(list1)
for i in range(0,3):
  sublist = []
  for j in range(i*l//3,(i+1)*l//3):
    sublist.append(list1[j])
  sublist.reverse()
  print(sublist)

Saurabh says

July 13, 2020 at 3:26 am
Q4
A simple approach for Q.4
Please tell if you like:-))
```
l=[11, 45, 8, 11, 23, 45, 23, 45, 89]
#creating empty Dictionary
Count_Dict={}
for i in l:
    count=0
    for j in l:
        if i==j:
            count+=1
        Count_Dict[i]=count
print(Count_Dict)
```
Hope you liked it:-))
Reply
- maya says
  
  November 15, 2023 at 12:27 am
  
  count_dic =dict()
  for i in sample_list:
  count_dic[i] = sample_list.count(i)
  
  print(count_dic)
  
  Reply
bostan Khan says

June 14, 2020 at 5:22 pm

thanks a lot for writing this article. its very helpful for me the God give you a long live

Reply

Narendra says

June 11, 2020 at 3:16 pm

question 3


List=[11, 45, 8, 23, 14, 12, 78, 45, 89]
s = len(List)//2
chunk1=List[:3]
chunk2=List[3:s+2]
chunk3=List[s+2:]
print('chunck 1',chunk1)
chunk1.reverse()
print('After reversed :',chunk1)
print('chunk 2',chunk2)
chunk2.reverse()
print('After reversed :',chunk2)
print('chunk 3:',chunk3)
chunk3.reverse()
print('After reversed :',chunk3)

Olli says

June 17, 2020 at 3:19 pm


list1 = [11, 45, 8, 23, 14, 12, 78, 45, 89]

chunk_size = int(len(list1)/3)

chunk1 = list1[:chunk_size]
chunk2 = list1[chunk_size:chunk_size*2]
chunk3 = list1[chunk_size*2:]

print("Original list", list1)
print("Chunk 1", chunk1)
print("After reversing it", chunk1[::-1])
print("Chunk 2", chunk2)
print("After reversing it", chunk2[::-1])
print("Chunk 3", chunk3)
print("After reversing it", chunk3[::-1])

Ramneek Singh says

May 10, 2020 at 4:04 pm

speed ={'jan':47, 'feb':52, 'march':47, 'April':44, 'May':52, 'June':53, 'july':54, 'Aug':44, 'Sept':54}
set1=set()
for values in speed.values():
  set1.add(values)
print(list(set1))

Ramneek Singh says

May 10, 2020 at 1:15 am
Hi Vishal

What am I doing wrong here? The for loop is ignoring 95 from the list and I can’t find out why
```
rollNumber  = [47, 64, 69, 37, 76, 83, 95, 97]
sampleDict ={'Jhon':47, 'Emma':69, 'Kelly':76, 'Jason':97}
for i in rollNumber:
  if i not in sampleDict.values():
    rollNumber.remove(i)
print(f'after removing unwanted elements from list {rollNumber}')
```
Reply
- Pieter says
  
  August 12, 2020 at 1:01 am
  
  I have the same issue when doing it the way you do it Vishal.
  But when using the exact code in she given solution it works.
  
  Reply

Ramneek Singh says

May 9, 2020 at 9:21 pm

#My code on Q4

OPTION1
list1=[11, 45, 8, 23, 14, 12, 78, 45, 89]
limit=int(len(list1)/3)
chunk1=[]
for i in range(limit):
  chunk1.append(list1.pop(0))
print(f'Chunk 1 {chunk1}')
chunk1.reverse()
print(f'After reversing it {chunk1}')

chunk2=[]
for i in range(limit):
  chunk2.append(list1.pop(0))
print(f'Chunk 2 {chunk2}')
chunk2.reverse()
print(f'After reversing it {chunk2}')

chunk3=[]
for i in range(limit):
  chunk3.append(list1.pop(0))
print(f'Chunk 3 {chunk3}')
chunk3.reverse()
print(f'After reversing it {chunk3}')

#OPTION2 (Using Function)
def chunkandreverse(list1,limit):
  counter=0
  while list1:
    counter+=1
    chunk=[]
    for i in range(limit):
      chunk.append(list1.pop(0))
    print(f'Chunk {counter} {chunk}')
    chunk.reverse()
    print(f'After reversing it {chunk}')
    chunkandreverse(list1,limit)

list1=[11, 45, 8, 23, 14, 12, 78, 45, 89]
limit=int(len(list1)/3)
chunkandreverse(list1,limit)

Ramneek Singh says

May 9, 2020 at 9:25 pm

Apologies for copying twice. I missed the tags as suggested for posting code.
The code I have posted is for Q3

Reply

Ramneek Singh says

May 9, 2020 at 10:39 pm

”’
Fixed some defects. Please ignore previous or may be you can compare and see what was fixed. The previous was failing if we added more elements in list. The non function one was simply ignoring elements which were not exact set of 3. Try several possible inputs as below
list1=[11, 45, 8, 23, 14, 12, 78, 45, 89]
list1=[11, 45, 8, 23, 14, 12, 78, 45, 89, 90]
list1=[11, 45, 8, 23, 14, 12, 78, 45, 89, 90, 54]
list1=[11, 45, 8, 23, 14, 12, 78, 45, 89, 90, 54, 12]
”’

#OPTION1 Without Function
list1=[11, 45, 8, 23, 14, 12, 78, 45,89,44,33,11,90]
limit=int(len(list1)/3)
chunk1=[]
for i in range(limit):
  chunk1.append(list1.pop(0))
print(f'Chunk 1 {chunk1}')
chunk1.reverse()
print(f'After reversing it {chunk1}')

chunk2=[]
for i in range(limit):
  chunk2.append(list1.pop(0))
print(f'Chunk 2 {chunk2}')
chunk2.reverse()
print(f'After reversing it {chunk2}')

chunk3=[]
for i in range(limit):
  chunk3.append(list1.pop(0))
if list1:
  chunk3.extend(list1)
print(f'Chunk 3 {chunk3}')
chunk3.reverse()
print(f'After reversing it {chunk3}')


#OPTION2 (Using Function)
def chunkandreverse(chunks,list1,limit,counter):

  counter+=1
  chunk=[]

  for i in range(limit):
    chunk.append(list1.pop(0))
  
  if chunks == 1:
    chunk.extend(list1)
    
  print(f'Chunk {counter} {chunk}')
  chunk.reverse()
  print(f'After reversing it {chunk}')
    
  chunks-=1
  
  if chunks > 0:   
    chunkandreverse(chunks,list1,limit,counter)
  
    

list1=[11, 45, 8, 23, 14, 12, 78, 45,89,44,33,11,90]
counter=0
chunks=3
limit=int(len(list1)/chunks)
chunkandreverse(chunks,list1,limit,0)

Nupur says

May 9, 2020 at 4:25 am

solution for Q6:

set1 = {65, 42, 78, 83, 23, 57, 29}
set2 = {67, 73, 43, 48, 83, 57, 29}
set1.difference_update(set1 & set2)
print(set1)

ershad says

May 5, 2020 at 1:33 am

I can’t see any reason for if else condition you used in solution 3 !
could you explain please?

Reply

Nikhil K says

May 1, 2020 at 4:03 pm

Q3:

sample_list = [11, 45, 8, 23, 14, 12, 78, 45, 89]
chunk = len(sample_list)//3
for i in range(0, len(sample_list), chunk):
        ls = sample_list[i:i+chunk]
        if i == 0:
            print("Chunk 1 ", ls)
            print("After reversing it ", ls[::-1])
        elif i == 3:
            print("Chunk 2", ls)
            print("After reversing it ", ls[::-1])
        else:
            print("Chunk 3", ls)
            print("After reversing it ", ls[::-1])

Ali says

March 23, 2020 at 1:23 pm

Question 3:

orgList = [11, 45, 8, 23, 14, 12, 78, 45, 89]
l1 = orgList[0:3]
l1.reverse()
print(l1)
l2 = orgList[3:6]
l2.reverse()
print(l2)
l3 = orgList[6:9]
l3.reverse()
print(l3)

Maria says

April 20, 2022 at 2:12 pm

def slice_chunks():
    sample_list = [11, 45, 8, 23, 14, 12, 78, 45, 89]
    chunk1 = slice(3)
    chunk2 = slice(3, 6, 1)
    chunk3 = slice(6, 9, 1)
    print('Chunk  1:', sample_list[chunk1])
    reversed1 = list(reversed(sample_list[chunk1]))
    print('After reversing it:', reversed1)

    print('Chunk  2:', sample_list[chunk2])
    reversed2 = list(reversed(sample_list[chunk2]))
    print('After reversing it:', reversed2)

    print('Chunk  3:', sample_list[chunk3])
    reversed3 = list(reversed(sample_list[chunk3]))
    print('After reversing it:', reversed3)

slice_chunks()

Lakhtibi Mourad says

February 14, 2020 at 10:51 am

Q6:
happy to share this,,42 yrs old beginner ,,love python

sa={65, 42, 78, 83, 23, 57, 29}
ba={67, 73, 43, 48, 83, 57, 29}
inter_set={x for x in sa for y in ba if x==y}
sa_1=[x for x in sa if x not in inter_set]

print("First set after removing elemnt",set(sa_1))
print(inter_set)

Collins Oyugi says

December 26, 2019 at 9:10 am

sampleList = [87, 45, 41, 65, 94, 41, 99, 94]
my_tuple = ()
for i in sampleList:
    if i not in my_tuple:
        my_tuple += (i,)
print(my_tuple)
print("The Max Value is" , max(my_tuple))
print("The Minimum Value is ",min(my_tuple))

Giorgos says

November 11, 2019 at 3:37 pm

Q10
the expected outcome is wrong and leads to wrong code to solve it compared to the solution.
Given your solution the output should be:

Original list [87, 45, 41, 65, 94, 41, 99, 94]
unique list [65, 99, 41, 45, 87, 94]
tuple (65, 99, 41, 45, 87, 94)
Minimum number is: 41
Maximum number is: 99

94 should be included, as set doesn’t remove both instances of duplicate numbers.

Reply

Vitaliy says

June 21, 2019 at 6:15 pm

Q9 as variant

speed ={'jan':47, 'feb':52, 'march':47, 'April':44, 'May':52, 'June':53, 'july':54, 'Aug':44, 'Sept':54}
print(list(set(speed.values())))

Vitaliy says

June 21, 2019 at 12:35 pm
Q6
```
firstSet = {23, 42, 65, 57, 78, 83, 29}
secondSet = {57, 83, 29, 67, 73, 43, 48}

print(firstSet - secondSet)
```
Reply
- Saurabh says
  
  July 13, 2020 at 3:32 am
  
  Can you really tell from where have you learned python, Your solutions are literally amazing
  
  Reply

Vitaliy says

June 21, 2019 at 9:57 am

s_List = [11, 45, 8, 11, 23, 45, 23, 45, 89]
print({i: s_List.count(i) for i in s_List})

Vitaliy says

June 20, 2019 at 9:54 pm

sampleList = [11, 45, 8, 23, 14, 12, 78, 45, 89]
c_size = len(sampleList) // 3
print("", sampleList[c_size - 1::-1], "\n",sampleList[(c_size * 2) - 1:c_size - 1:-1],
          "\n",sampleList[:(c_size * 2) - 1:-1])

Vitaliy says

June 20, 2019 at 9:25 pm
Q2
```
List = [34, 54, 67, 89, 11, 43, 94]
List[2], List[3], List[4] = List[4], List[2], List[3]
List.append(List[2])
print(List)
```
Reply
- amine says
  
  August 30, 2019 at 10:16 pm
  
  u really did it with a smart way . Thanks for sharing ur way . can u tell me where did u practiced python?
  
  Reply
Rohan says

June 15, 2019 at 6:41 pm

please solve this problem
Scenario
You are required to design the data structure to display the individual player stats for cricket players. A player may have represented more than one team and may have played in more than one format such as Test, ODI and T20.
Problem Statement
Create a list of different data fields and use appropriate Python data types to represent each of them

Reply

zubair khalid says

June 5, 2019 at 6:10 pm

sampleList = [1, 2, 3, 4, 5, 6, 7, 8, 9]
a = sampleList[:3]
b = sampleList[3:6]
c = sampleList[6:9]
a.reverse()
b.reverse()
c.reverse()
print(a)
print(b)
print(c)

Bill Hardwick says

May 18, 2019 at 3:14 pm
Q9 – My one-line solution:
```
list_ = list(set(speed.values()))
```
Reply
Bill Hardwick says

May 17, 2019 at 7:33 pm

Q7 – Slight typo in the Expected Outcome which shows “firstSet = {}” (curly braces), which I must admit is what I was anticipating.
Having read about sets but not having worked with them, I was puzzled as to why my solution was showing “firstSet = set()” (set prefix and round brackets) instead. It was only when I checked against your solution that I discovered we had done essentially the same and obtained the same result! Please amend the expected outcome line to show the correct result and possibly spare others the several minutes of head-scratching I experienced.
I now realise that set() is how you would declare an empty set rather than as {} to differentiate it from an empty dictionary, hence that result. So at least the head-scratching resulted in a little more learning.

Reply
Bill Hardwick says

May 17, 2019 at 7:05 pm

Q6 – Or slightly more succinctly in one line:

setA = setA.difference(setA.intersection(setB))

Reply

Bill Hardwick says

May 17, 2019 at 6:17 pm

Q4. Purely a matter of personal taste, but I prefer the dict’s setdefault method to testing whether the dict currently contains the term. my solution is therefore:

hopkins = ["Not", "I'll",  "not", "carrion", "comfort", "Despair",  "not", "feast", "on", "thee"]
dic = {}
for w in hopkins:
  dic.setdefault(w.lower(), 0)
  dic[w.lower()] += 1
print(dic)

{'not': 3, "i'll": 1, 'carrion': 1, 'comfort': 1, 'despair': 1, 'feast': 1, 'on': 1, 'thee': 1}

(And I hope these ‘pre’ tags do what I expect !?!)

Bill Hardwick says

May 17, 2019 at 5:56 pm

Q3 My slightly different approach which pads the original list with None values if necessary to make it exactly divisible into equal thirds:

def thirds_reversed(lst):
  rem = len(lst)%3
  if rem > 0:
    for i in range(3 - rem):
      lst.append(None)

  slice = int(len(lst)/3)
  thirds = [] 
  for i in range(3):
    third = lst[i*slice:(i+1)*slice]
    third.reverse()
    thirds.append(third)
  return(thirds)
print(thirds_reversed([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]))

[[4, 3, 2, 1], [8, 7, 6, 5], [None, 11, 10, 9]]

Varun says

February 15, 2020 at 9:38 pm

For Q3, If this helps someone –

sample_list = [11, 45, 8, 23, 14, 12, 78, 45, 89]
chunk = len(sample_list)//3

for i in range(0, len(sample_list), chunk):
        ls = sample_list[i:i+chunk]
        print(ls[::-1])

Output –

[8, 45, 11]
[12, 14, 23]
[89, 45, 78]

Bill Hardwick says

May 17, 2019 at 4:22 pm

Q2 Minor error in the wording. You ask to remove the ‘element at index 4’ – given the example and the solution, this should either read ‘the fourth item in the list’ or ‘the element at index 3’. With the current wording, you should be throwing around 91, not 79.

By the way, a belated thank you for your work on this site. It is providing very useful memory jogging and consolidation for what I have learned to date. I look forward to the numpy and subsequent exercises, which will be mostly breaking new ground for me.

Reply

Python Data Structure Exercise for Beginners

Exercise 1: List Creation using two lists

Exercise 2: Remove and add item in a list

Exercise 3: Slice list into 3 equal chunks and reverse each chunk

Exercise 4: Count the occurrence of each element from a list

Exercise 5: Paired Elements from Two Lists as a Set

Exercise 6: Set Intersection and Removal

Exercise 7: Subset or Superset of another set

Exercise 8: Filter List Against Dictionary Values

Exercise 9: Extract Unique Dictionary Values to List

Exercise 10: remove duplicates from a list

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Exercise 1: List Creation using two lists

Exercise 2: Remove and add item in a list

Exercise 3: Slice list into 3 equal chunks and reverse each chunk

Exercise 4: Count the occurrence of each element from a list

Exercise 5: Paired Elements from Two Lists as a Set

Exercise 6: Set Intersection and Removal

Exercise 7: Subset or Superset of another set

Exercise 8: Filter List Against Dictionary Values

Exercise 9: Extract Unique Dictionary Values to List

Exercise 10: remove duplicates from a list

About Vishal

Related Tutorial Topics:

Python Exercises and Quizzes

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