How to split an array into chunks of a given length in python?

Question

What is the fastest and shortest method to turn this:

ids = [1, 2, 3, 4, 5, 6, 7, 8, 9]

for example into this:

ids = [[1, 2], [3, 4], [5, 6], [7, 8], [9]]

by giving the input 2 as the fixed length.

Of course there are some easy ways to make this but none of them occur to me as fast or pretty.

Because you did not google? It is the first result for `split an array into chunks of a given length in python stackoverflow`. — lllrnr101, Apr 03 '21 at 16:15

Spiros Gkogkas · Accepted Answer · 2021-04-03T17:13:22.770

14

Why don't you try out a list comprehension?

Example:

[ids[i:i+2] for i in range(0,len(ids),2)]

Output:

[[1, 2], [3, 4], [5, 6], [7, 8], [9]]

edited Apr 03 '21 at 17:13

answered Apr 03 '21 at 16:12

Spiros Gkogkas

432
5
11

1

Yeah just found it thanks to leoOrions comment too. stackoverflow.com/a/312464/4983469 But Thanks! – StrangeGirlMurph Apr 03 '21 at 16:14
2

range should also include a step I think. (If you posted this as answer and not direction) – lllrnr101 Apr 03 '21 at 16:17
@lllrnr101 I thought that was obvious but alright :) – Spiros Gkogkas Apr 03 '21 at 16:41
Your output and code do not match. That is all I wanted to say. Because range() is still not using step. So the sub-lists still overlap which were not overlapping in OPs question. But since it is accepted I think it solves his problem. – lllrnr101 Apr 03 '21 at 16:58
@lllrnr101 You are right I fixed it, I was a bit in a rush so my apologies. – Spiros Gkogkas Apr 03 '21 at 17:14

score 0 · Answer 2 · answered Apr 03 '21 at 16:17

bad but effective:

ids = [1, 2, 3, 4, 5, 6, 7, 8, 9]
new_ids = []


function = lambda x, y: [x, y]

for i in range(0, len(ids), 2):
    try:
        new_ids.append(function(ids[i], ids[i + 1]))
    except IndexError:
        new_ids.append(ids[i])
ids = new_ids
print(ids)

How to split an array into chunks of a given length in python?

2 Answers2

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