take a list of numbers (1 element) and break it into chunks with python?

Question

I would like to take a var containing numbers and create a list of 2 digit numbers.

For instance:

x = 123456

I want to create a list of 2 digit chunks

y = [12,34,56]

I can't seem to figure this one out.

possible duplicate of [How do you split a list into evenly sized chunks in Python?](http://stackoverflow.com/questions/312443/how-do-you-split-a-list-into-evenly-sized-chunks-in-python) — Martijn Pieters, Jan 02 '14 at 19:28
How do you want to handle the case of `x = 120034`? The only way to have `00` is to use a string. — DSM, Jan 02 '14 at 19:33
Also, how do you want to handle `x = 12345`? Or `x = 0`? Or `x = -123456`? — abarnert, Jan 02 '14 at 19:36
`n = 2; s = str(x); result = map(int, (s[i:i+n] for i in range(0, len(s), n)))` — Joel Cornett, Jan 02 '14 at 19:46

Hyperboreus · Answer 1 · 2014-01-02T19:38:01.800

3

Use modulo and floor division.

def chunks(n):
    if n < 0: raise Exception ("Don't")
    while n:
        yield n % 100
        n //= 100

a = [c for c in chunks (123456)][::-1]
print(a)

Also PS: For input 12345 the output is [1, 23, 45].

And PPS: Is this for FFT multiplication?

edited Jan 02 '14 at 19:38

answered Jan 02 '14 at 19:32

Hyperboreus

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Probably should add an explanatory note that `chunks` yields in reverse order, necessitating the reversing slice (`[::-1]`). Also, it might be clearer to do `reversed([c for c in chunks (123456)])`. – Steven Rumbalski Jan 02 '14 at 19:54
@StevenRumbalski Very correct. The main point was to avoid costly string operations. The only possible application of splitting a number into its coefficients of a given base I see is FFT, and then hopefully the order doesn't matter as long as you stick to one and the same. – Hyperboreus Jan 02 '14 at 20:00
Thanks, this works for a string but how bout an int? – MacR6 Jan 02 '14 at 20:41
1

@user2270470 This doesn't work for a string... It takes an `int` (in my example `123456`). – Hyperboreus Jan 02 '14 at 20:42

kostbash · Answer 2 · 2014-01-02T19:58:27.243

0

If x is string:

x = '1234563'

a = [x[i * 2 : (i + 1) * 2] for i in range(len(x) // 2)]

If x is int:

x = 1234563

l = len(str(x))
a = [(x // (10 ** (i - 2))) % 100 for i in range(l, 2, -2)]

edited Jan 02 '14 at 19:58

answered Jan 02 '14 at 19:45

kostbash

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score 0 · Answer 3 · answered Jan 02 '14 at 19:45

0

>>> x = 123456
>>> [int(str(x)[i:i+2]) for i in range(0, len((str(x)), 2)]
[12, 34, 56]

answered Jan 02 '14 at 19:45

ndpu

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score 0 · Answer 4 · answered Jan 02 '14 at 19:54

def trunk(numbers, chunkSize):
    new_list = []
    nums = str(numbers)
    for x in xrange(0, len(nums), chunkSize):
        new_list.append(int(nums[x:chunkSize+x]))
    return new_list

>>> x = 123456
>>> trunk(x, 2)
[12, 34, 56]
>>> x = 12345
>>> trunk(x, 2)
[12, 34, 5]

take a list of numbers (1 element) and break it into chunks with python?

4 Answers4