Slice list in loop by step

Question

looking for a nice way to solve slice problem.

list_a = [1,2,3,4,5,6,7,8,9,10,11]
step = 5

print -> [1,2,3,4,5]
print -> [6,7,8,9,10]
print -> [11]

What exactly are you trying to do here? What is your goal and what have you tried? — rangeseeker, Oct 31 '21 at 03:05
So you want to divide a list into small ones based on a size? — Pedro Maia, Oct 31 '21 at 03:09
Does this answer your question? [How do you split a list into evenly sized chunks?](https://stackoverflow.com/questions/312443/how-do-you-split-a-list-into-evenly-sized-chunks) — Pedro Maia, Oct 31 '21 at 03:13

score 3 · Accepted Answer · edited Oct 31 '21 at 11:48

3

This is what you want to do:

output = [list_a[i:i + step] for i in range(0, len(list_a), step)]

When you pass:

list_a = ['1','2','3','4','5','6','7','8','9','10']

You will get like this:

output = [['1', '2', '3'], ['4', '5', '6'], ['7', '8', '9'], ['10']]

edited Oct 31 '21 at 11:48

Python learner

answered Oct 31 '21 at 03:10

rangeseeker

random · Answer 2 · 2021-10-31T03:10:32.467

1

Just use slice notation: list_a[:step]. With try and except: try: print(list_a[:step]) except: print('Out of Range')

edited Oct 31 '21 at 03:10

answered Oct 31 '21 at 03:06

random

1

Remember the first `:` is for the slice, the second is step. So `list_a[slice_start:slice_end:step]` – rangeseeker Oct 31 '21 at 03:08
How avoid -> out of range? – Denis Oct 31 '21 at 03:08
1

Use try and accept. – random Oct 31 '21 at 03:09
It not rally nice way. – Denis Oct 31 '21 at 03:09

2 Answers2