How to split an integer into an array of digits?

python

My integer input is suppose 12345, I want to split and put it into an array as 1, 2, 3, 4, 5.
How will I be able to do it?

>>> [int(i) for i in str(12345)]

[1, 2, 3, 4, 5]

return array as string

>>> list(str(12345))
['1', '2', '3', '4', '5']

return array as integer

>>> map(int,str(12345))
[1, 2, 3, 4, 5]

[int(i) for i in str(number)]

or, if do not want to use a list comprehension or you want to use a base different from 10

from __future__ import division # for compatibility of // between Python 2 and 3
def digits(number, base=10):
    assert number >= 0
    if number == 0:
        return [0]
    l = []
    while number > 0:
        l.append(number % base)
        number = number // base
    return l

I'd rather not turn an integer into a string, so here's the function I use for this:

def digitize(n, base=10):
    if n == 0:
        yield 0
    while n:
        n, d = divmod(n, base)
        yield d

Examples:

tuple(digitize(123456789)) == (9, 8, 7, 6, 5, 4, 3, 2, 1)
tuple(digitize(0b1101110, 2)) == (0, 1, 1, 1, 0, 1, 1)
tuple(digitize(0x123456789ABCDEF, 16)) == (15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1)

As you can see, this will yield digits from right to left. If you'd like the digits from left to right, you'll need to create a sequence out of it, then reverse it:

reversed(tuple(digitize(x)))

You can also use this function for base conversion as you split the integer. The following example splits a hexadecimal number into binary nibbles as tuples:

import itertools as it
tuple(it.zip_longest(*[digitize(0x123456789ABCDEF, 2)]*4, fillvalue=0)) == ((1, 1, 1, 1), (0, 1, 1, 1), (1, 0, 1, 1), (0, 0, 1, 1), (1, 1, 0, 1), (0, 1, 0, 1), (1, 0, 0, 1), (0, 0, 0, 1), (1, 1, 1, 0), (0, 1, 1, 0), (1, 0, 1, 0), (0, 0, 1, 0), (1, 1, 0, 0), (0, 1, 0, 0), (1, 0, 0, 0))

Note that this method doesn't handle decimals, but could be adapted to.

How to split an integer into an array of digits?

Related

Recent Posts