Iterating over a list based on indices in another list

Question

I have a list of strings - foo and another list of integers- bar which keeps the track of important indices in foo.

For example:

foo = [{}.format(i) for i in range(1, 11)] # not necessarily of this format
bar = [0, 3, 5]

I would like to create a recipe for creating a list of lists, each list obtained by splitting foo based on indices in bar.

Expected output for the above example:

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

For achieving this, I have created the following function which works fine:

result = []
for index, value in enumerate(b):
    if index == len(b) - 1:
        result.append(a[value:])
    elif index == 0 and value != 0:
        result.append(a[0: value])
    else:
        result.append(a[value: b[index + 1]])

However, I find this code highly Non-Pythonic, thanks to my C-Java background.
I would like to know a better solution to this problem (maybe we can use itertools somehow).

Typical slicing problem. Maybe this helps: http://stackoverflow.com/questions/509211/explain-pythons-slice-notation — Jan Zeiseweis, Jan 23 '17 at 12:49
@JanZeiseweis Thank you for the reference but it was of not much use for this particular question. — Kshitij Saraogi, Jan 23 '17 at 12:54
@schwobaseggl Thanks for noticing. I will flag it for duplication — Kshitij Saraogi, Jan 23 '17 at 12:59

score 2 · Answer 1 · answered Jan 23 '17 at 12:52

2

You could do as follows:

In [3]: [foo[a:b] for a, b in zip(bar, bar[1:]+[None])]
Out[3]: [['1', '2', '3'], ['4', '5'], ['6', '7', '8', '9', '10']]

answered Jan 23 '17 at 12:52

J. P. Petersen

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If `bar` contains an index equal to the length of `foo` this will generates an empty list! try `bar=[0, 3, 5, 10]` – Mazdak Jan 23 '17 at 12:58
A index the length of `foo` would be also be outside the `foo` list? – J. P. Petersen Jan 23 '17 at 13:02

score 1 · Answer 2 · answered Jan 23 '17 at 12:51

1

Here is one way using a list comprehension:

In [107]: bar = bar + [len(foo)] if bar[-1] < len(foo) else bar


In [110]: [foo[i:j] for i, j in zip(bar, bar[1:])]
Out[110]: [['1', '2', '3'], ['4', '5'], ['6', '7', '8', '9', '10']]

answered Jan 23 '17 at 12:51

Mazdak

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Iterating over a list based on indices in another list

2 Answers2