I have a list of integers that I would like to convert to one number like:
numList = [1, 2, 3]
num = magic(numList)
print num, type(num)
>>> 123, <type 'int'>
What is the best way to implement the magic function?
EDIT
I did find this, but it seems like there has to be a better way.
# Over-explaining a bit:
def magic(numList): # [1,2,3]
s = map(str, numList) # ['1','2','3']
s = ''.join(s) # '123'
s = int(s) # 123
return s
# How I'd probably write it:
def magic(numList):
s = ''.join(map(str, numList))
return int(s)
# As a one-liner
num = int(''.join(map(str,numList)))
# Functionally:
s = reduce(lambda x,y: x+str(y), numList, '')
num = int(s)
# Using some oft-forgotten built-ins:
s = filter(str.isdigit, repr(numList))
num = int(s)
''.join(map(str, numList)) ? Also, for your "Cleverly" option, you need to int() the result.
reduce(lambda n,d:10*n+d,.. is even faster than 'int(''.join(map(..`Two solutions:
>>> nums = [1, 2, 3]
>>> magic = lambda nums: int(''.join(str(i) for i in nums)) # Generator exp.
>>> magic(nums)
123
>>> magic = lambda nums: sum(digit * 10 ** (len(nums) - 1 - i) # Summation
... for i, digit in enumerate(nums))
>>> magic(nums)
123
The map-oriented solution actually comes out ahead on my box -- you definitely should not use sum for things that might be large numbers:
import collections
import random
import timeit
import matplotlib.pyplot as pyplot
MICROSECONDS_PER_SECOND = 1E6
FUNS = []
def test_fun(fun):
FUNS.append(fun)
return fun
@test_fun
def with_map(nums):
return int(''.join(map(str, nums)))
@test_fun
def with_interpolation(nums):
return int(''.join('%d' % num for num in nums))
@test_fun
def with_genexp(nums):
return int(''.join(str(num) for num in nums))
@test_fun
def with_sum(nums):
return sum(digit * 10 ** (len(nums) - 1 - i)
for i, digit in enumerate(nums))
@test_fun
def with_reduce(nums):
return int(reduce(lambda x, y: x + str(y), nums, ''))
@test_fun
def with_builtins(nums):
return int(filter(str.isdigit, repr(nums)))
@test_fun
def with_accumulator(nums):
tot = 0
for num in nums:
tot *= 10
tot += num
return tot
def time_test(digit_count, test_count=10000):
"""
:return: Map from func name to (normalized) microseconds per pass.
"""
print 'Digit count:', digit_count
nums = [random.randrange(1, 10) for i in xrange(digit_count)]
stmt = 'to_int(%r)' % nums
result_by_method = {}
for fun in FUNS:
setup = 'from %s import %s as to_int' % (__name__, fun.func_name)
t = timeit.Timer(stmt, setup)
per_pass = t.timeit(number=test_count) / test_count
per_pass *= MICROSECONDS_PER_SECOND
print '%20s: %.2f usec/pass' % (fun.func_name, per_pass)
result_by_method[fun.func_name] = per_pass
return result_by_method
if __name__ == '__main__':
pass_times_by_method = collections.defaultdict(list)
assert_results = [fun([1, 2, 3]) for fun in FUNS]
assert all(result == 123 for result in assert_results)
digit_counts = range(1, 100, 2)
for digit_count in digit_counts:
for method, result in time_test(digit_count).iteritems():
pass_times_by_method[method].append(result)
for method, pass_times in pass_times_by_method.iteritems():
pyplot.plot(digit_counts, pass_times, label=method)
pyplot.legend(loc='upper left')
pyplot.xlabel('Number of Digits')
pyplot.ylabel('Microseconds')
pyplot.show()
digit_count. See stackoverflow.com/questions/489999/…
with_accumulator looks better on my machine M1 + python3def magic(number):
return int(''.join(str(i) for i in number))
[ ] and do the str'ing as a generation expression. int(''.join(str(i) for i in number)) - it's.. two bytes quicker!Just for completeness, here's a variant that uses print() (works on Python 2.6-3.x):
from __future__ import print_function
try: from cStringIO import StringIO
except ImportError:
from io import StringIO
def to_int(nums, _s = StringIO()):
print(*nums, sep='', end='', file=_s)
s = _s.getvalue()
_s.truncate(0)
return int(s)
I've measured performance of @cdleary's functions. The results are slightly different.
Each function tested with the input list generated by:
def randrange1_10(digit_count): # same as @cdleary
return [random.randrange(1, 10) for i in xrange(digit_count)]
You may supply your own function via --sequence-creator=yourmodule.yourfunction command-line argument (see below).
The fastest functions for a given number of integers in a list (len(nums) == digit_count) are:
len(nums) in 1..30
def _accumulator(nums):
tot = 0
for num in nums:
tot *= 10
tot += num
return tot
len(nums) in 30..1000
def _map(nums):
return int(''.join(map(str, nums)))
def _imap(nums):
return int(''.join(imap(str, nums)))
|------------------------------+-------------------|
| Fitting polynom | Function |
|------------------------------+-------------------|
| 1.00 log2(N) + 1.25e-015 | N |
| 2.00 log2(N) + 5.31e-018 | N*N |
| 1.19 log2(N) + 1.116 | N*log2(N) |
| 1.37 log2(N) + 2.232 | N*log2(N)*log2(N) |
|------------------------------+-------------------|
| 1.21 log2(N) + 0.063 | _interpolation |
| 1.24 log2(N) - 0.610 | _genexp |
| 1.25 log2(N) - 0.968 | _imap |
| 1.30 log2(N) - 1.917 | _map |
To plot the first figure download cdleary.py and make-figures.py and run (numpy and matplotlib must be installed to plot):
$ python cdleary.py
Or
$ python make-figures.py --sort-function=cdleary._map \
> --sort-function=cdleary._imap \
> --sort-function=cdleary._interpolation \
> --sort-function=cdleary._genexp --sort-function=cdleary._sum \
> --sort-function=cdleary._reduce --sort-function=cdleary._builtins \
> --sort-function=cdleary._accumulator \
> --sequence-creator=cdleary.randrange1_10 --maxn=1000
def magic(numbers):
return int(''.join([ "%d"%x for x in numbers]))
pseudo-code:
int magic(list nums)
{
int tot = 0
while (!nums.isEmpty())
{
int digit = nums.takeFirst()
tot *= 10
tot += digit
}
return tot
}
A one-liner without needing to cast to and from str
def magic(num):
return sum(e * 10**i for i, e in enumerate(num[::-1]))
if the list contains only integer:
reduce(lambda x,y: x*10+y, list)
This method works in 2.x as long as each element in the list is only a single digit. But you shouldn't actually use this. It's horrible.
>>> magic = lambda l:int(`l`[1::3])
>>> magic([3,1,3,3,7])
31337
Using a generator expression:
def magic(numbers):
digits = ''.join(str(n) for n in numbers)
return int(digits)
This seems pretty clean, to me.
def magic( aList, base=10 ):
n= 0
for d in aList:
n = base*n + d
return n
reduce(lambda x, y: base*x + y, aList, 0) is even cleaner ;)
reduce() often creates additional problems when used carelessly. I strongly encourage your to post your own answer, since yours is so different from this one.I found some examples are not compatible with python 3 I test one from @Triptych
s = filter(str.isdigit, repr(numList))
num = int(s)
in python 3 it's gonna give error
TypeError: int() argument must be a string, a bytes-like object or a number, not 'filter'
i think the more simple and compatible way would be
def magic(num_list):
return int("".join(map(str, num_list)))
This may be helpful
def digits_to_number(digits):
return reduce(lambda x,y : x+y, map(str,digits))
print digits_to_number([1,2,3,4,5])
If you happen to be using numpy (with import numpy as np):
In [24]: x
Out[24]: array([1, 2, 3, 4, 5])
In [25]: np.dot(x, 10**np.arange(len(x)-1, -1, -1))
Out[25]: 12345
I think this seems to be the simplest solution with no need for any fcn
res = int(''.join(numList))
NOTE: This implementation is in Python 3
I found this thread while trying to convert a list to the real value of the underlying int in terms of a C-style pointer, but none of the other answers appear to work for this case. I think the following solution works as intended and could be useful to others even though it doesn't necessarily answer the original question.
def listToInt(x, reverseBytes=False):
if reverseBytes:
x = x[::-1]
return reduce(lambda x,y: x*256+y, x)
listToInt([1, 249]) == 505
listToInt([249, 1], True) == 505
The primary value of this is to emulate the behavior of casting a byte array to another data type, e.g. uint16, which Python can't seem to do natively, on either a big or little endian system.
Too late for the party here is the approch without converting to string
x=0
k=len(a)-1
for i in a:
x+=(10**k)*i
k-=1
Hope this one-line formula can be useful too:
s=sum([numlist[i] * 10**(len(numlist)-i-1) for i in range(len(numlist))])
A pair of non-strs based approaches.
Assumed that each term in numList is a natural number smaller than 10.
With reduce:
from functools import reduce
numList = [1, 2, 3, 4]
tot = reduce(lambda i, j: i+j[1]*10**(len(numList)-j[0]-1), enumerate(numList), 0)
Used zip/map to couple coefficients with the powers of the base.
The zip-way:
numList = [1, 2, 3]
tot = sum(coef*10**i for coef, i in zip(numList, range(len(numList)-1, -1, -1)))
print(tot)
The map-way:
numList = [1, 2, 3]
base_pows = map(int(10).__pow__, range(len(numList)-1, -1, -1))
tot = sum(map(int.__mul__, numList, base_pows))
print(tot)
int(''.join([str(n) for n in numList]))