Splitting the digits of an integer into groups of 1, 2, 3 etc. digits

Question

I want to write a program to split an integer's digits into groups of n digits, assuming that the number of digits is divisible by n.

For example, say I had the integer 123456789, and n=3, this would produce the list:

[123, 456, 789]

Or if the number was 12345678, and n=2, I would want the list:

[12, 34, 56, 78]

So the order of the digits remains the same. It is okay if the numbers in the list are strings, as this is easy to change.

EDIT: I apologise it seems that this question has already been asked. I shall look there for answers. Thank you to those who answered.

There are many ways to do that. Which one did you try and what went wrong? Show us. — user, Apr 08 '15 at 11:23
Please show us what have you tried so far. Stack Overflow is not free code-writing service. — Łukasz Rogalski, Apr 08 '15 at 11:26
There are dozens of questions similar to this already; see e.g. http://stackoverflow.com/questions/312443/how-do-you-split-a-list-into-evenly-sized-chunks-in-python — jonrsharpe, Apr 08 '15 at 11:29
Are you sure you want to leave off the last `9` in your second example? — Jamie Cockburn, Apr 08 '15 at 11:32

itzMEonTV · Answer 1 · 2015-04-08T11:40:52.133

2

I think you can create a generator function

def split_by_n( seq, n ):
    """A generator to divide a sequence into chunks of n units."""
    seq = str(seq)
    while seq:
        yield int(seq[:n])
        seq = seq[n:]

>>>list(split_by_n(1234567890,2))
[12, 34, 56, 78, 90]

edited Apr 08 '15 at 11:40

answered Apr 08 '15 at 11:28

itzMEonTV

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Splitting the digits of an integer into groups of 1, 2, 3 etc. digits

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