如何将一个整数分割成一列数字？

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]

Strings are just as iterable as arrays, so just convert it to string:

str(12345)

>>> [int(i) for i in 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

like @nd says but using the built-in function of int to convert to a different base

>>> [ int(i,16) for i in '0123456789ABCDEF' ]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]


>>> [int(i,2) for i in "100 010 110 111".split()]
[4, 2, 6, 7]

I don't know what is the final objective but take a look also inside the decimal module of python for doing stuff like

>>> Decimal('3.1415926535') + Decimal('2.7182818285')
Decimal('5.85987')

Splitting a single number to it's digits (as answered by all):

>>> [int(i) for i in str(12345)]
[1, 2, 3, 4, 5]

But, to get digits from a list of numbers:

>>> [int(d) for d in ''.join(str(x) for x in [12, 34, 5])]
[1, 2, 3, 4, 5]

So like to know, if we can do the above, more efficiently.

While list(map(int, str(x))) is the Pythonic approach, you can formulate logic to derive digits without any type conversion:

from math import log10


def digitize(x):
n = int(log10(x))
for i in range(n, -1, -1):
factor = 10**i
k = x // factor
yield k
x -= k * factor


res = list(digitize(5243))


[5, 2, 4, 3]

One benefit of a generator is you can feed seamlessly to set, tuple, next, etc, without any additional logic.

Using join and split methods of strings:

>>> a=12345
>>> list(map(int,' '.join(str(a)).split()))
[1, 2, 3, 4, 5]
>>> [int(i) for i in ' '.join(str(a)).split()]
[1, 2, 3, 4, 5]
>>>

Here we also use map or a list comprehension to get a list.

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.

Another solution that does not involve converting to/from strings:

from math import log10


def decompose(n):
if n == 0:
return [0]
b = int(log10(n)) + 1
return [(n // (10 ** i)) % 10 for i in reversed(range(b))]

You can simply do:

list(str(number))


e.g.


list(str(12345))
lst = ["1","2","3","4","5"]

to convert the list of strings to numbers you can do a simple list comprehension:

[int(num) for num in lst]

Simply turn it into a string, split, and turn it back into an array integer
nums = []
c = 12345
for i in str(c):
l = i.split()[0]
nums.append(l)
np.array(nums)

lol....crude but still works