Converting integer to digit list

What is the quickest and cleanest way to convert an integer into a list?

For example, change 132 into [1,3,2] and 23 into [2,3]. I have a variable which is an int, and I want to be able to compare the individual digits so I thought making it into a list would be best, since I can just do int(number[0]), int(number[1]) to easily convert the list element back into int for digit operations.

Solution 1:

Convert the integer to string first, and then use map to apply int on it:

>>> num = 132
>>> map(int, str(num))    #note, This will return a map object in python 3.
[1, 3, 2]

or using a list comprehension:

>>> [int(x) for x in str(num)]
[1, 3, 2]

Solution 2:

There are already great methods already mentioned on this page, however it does seem a little obscure as to which to use. So I have added some mesurements so you can more easily decide for yourself:

A large number has been used (for overhead) 1111111111111122222222222222222333333333333333333333

Using map(int, str(num)):

import timeit

def method():
    num = 1111111111111122222222222222222333333333333333333333
    return map(int, str(num))

print(timeit.timeit("method()", setup="from __main__ import method", number=10000)

Output: 0.018631496999999997

Using list comprehension:

import timeit

def method():
    num = 1111111111111122222222222222222333333333333333333333
    return [int(x) for x in str(num)]

print(timeit.timeit("method()", setup="from __main__ import method", number=10000))

Output: 0.28403817900000006

Code taken from this answer

The results show that the first method involving inbuilt methods is much faster than list comprehension.

The "mathematical way":

import timeit

def method():
    q = 1111111111111122222222222222222333333333333333333333
    ret = []
    while q != 0:
        q, r = divmod(q, 10) # Divide by 10, see the remainder
        ret.insert(0, r) # The remainder is the first to the right digit
    return ret

print(timeit.timeit("method()", setup="from __main__ import method", number=10000))

Output: 0.38133582499999996

Code taken from this answer

The list(str(123)) method (does not provide the right output):

import timeit

def method():
    return list(str(1111111111111122222222222222222333333333333333333333))
    
print(timeit.timeit("method()", setup="from __main__ import method", number=10000))

Output: 0.028560138000000013

Code taken from this answer

The answer by Duberly González Molinari:

import timeit

def method():
    n = 1111111111111122222222222222222333333333333333333333
    l = []
    while n != 0:
        l = [n % 10] + l
        n = n // 10
    return l

print(timeit.timeit("method()", setup="from __main__ import method", number=10000))

Output: 0.37039988200000007

Code taken from this answer

Remarks:

In all cases the map(int, str(num)) is the fastest method (and is therefore probably the best method to use). List comprehension is the second fastest (but the method using map(int, str(num)) is probably the most desirable of the two.

Those that reinvent the wheel are interesting but are probably not so desirable in real use.

Solution 3:

The shortest and best way is already answered, but the first thing I thought of was the mathematical way, so here it is:

def intlist(n):
    q = n
    ret = []
    while q != 0:
        q, r = divmod(q, 10) # Divide by 10, see the remainder
        ret.insert(0, r) # The remainder is the first to the right digit
    return ret

print intlist(3)
print '-'
print intlist(10)
print '--'
print intlist(137)

It's just another interesting approach, you definitely don't have to use such a thing in practical use cases.

Solution 4:

n = int(raw_input("n= "))

def int_to_list(n):
    l = []
    while n != 0:
        l = [n % 10] + l
        n = n // 10
    return l

print int_to_list(n)