Divide number into digits and save them in list (array) using python

Question

I want to divide number into digits and save them in list (or array) in python. So firstly I should create list like

dig = [0 for i in range(10)]

and then

i = 0
while num > 9:
    dig[i] = num % 10
    i += 1
    num /= 10
dig[i] = num

But I don't really like just creating list for 10 spaces, is it possible to get length of number without repeating loop

i = 0
num2 = num
while num2 > 9:
    num2 /= 10
    i += 1
i += 1

and then repeat first part of code? Or just make as I made in first place? I don't know exact length of number but it won't be very long

So any advices? Maybe you know better way to divide numbers into digits, or maybe something else.

Answer 1

Since you're just adding the digits from smallest to greatest in order, just use an empty list:

dig = []
i = 0
while num > 9:
    dig.append(num % 10)
    i += 1
    num /= 10
dig.append(num)

Alternatively, just do

dig = list(int(d) for d in str(num))

Which will turn i.e. 123 into '123' then turn each digit back into a number and put them into a list, resulting in [1, 2, 3].

If you want it in the same order as your version, use

dig = reversed(int(d) for d in str(num))

If you really just want to get the length of a number, it's easiest to do len(str(num)) which turns it into a string then gets the length.

Answer 2

Assuming num is a positive integer, the number of digits is equal to int(math.log10(num) + 1).

But better than that is to just use the append method of list -- then you won't have to know the length beforehand.

Even easier (though the list will be in reverse order):

dig = [int(x) for x in str(num)]

or

dig = map(int, str(num))

Answer 3

Although map is often said to be unpythonic, this task is definitely for it:

>>> d = 123456
>>> map(int, str(d))
[1, 2, 3, 4, 5, 6]