Simplest way to return an array that is nested in multiple arrays

Question

I have a situation where I have data that sometimes can be nested in multiple array layers.

Some times the data can be nested like:

[ [ 'green', 'blue', 'red' ] ]

Other times

[[[ ['green', 'blue', 'red' ] ]]]

I want to extract the array and return it, what would be the most pythonic way of doing this?

Are there any other elements in the parent lists? or are you simply looking to strip the excess nesting? — Martijn Pieters, Nov 28 '12 at 16:34
See http://stackoverflow.com/questions/952914/making-a-flat-list-out-of-list-of-lists-in-python — Praveen Gollakota, Nov 28 '12 at 16:35
Stripping the "nesting" is easy - but I'd look more to not having it in the first place ie, if unpredictable results - then there's a flaw somewhere that needs addressing... — Jon Clements, Nov 28 '12 at 16:36
@JonClements I agree with you on your approach of not having bad data in the first place but for the time being I need to strip the nesting. — Nirma, Nov 28 '12 at 16:41

Steven Rumbalski · Answer 1 · 2012-11-28T17:26:08.920

3

def get_nested_list(a):
    if len(a) == 1 and isinstance(a[0], list):
        return get_nested_list(a[0])
    return a

Examples:

>>> get_nested_list([[[ ['green', 'blue', 'red' ] ]]])
['green', 'blue', 'red']
>>> get_nested_list([[[[1, 2],[3]]]])
[[1, 2], [3]]

edited Nov 28 '12 at 17:26

answered Nov 28 '12 at 16:40

Steven Rumbalski

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This is useful only if a single element is nested. – asheeshr Nov 28 '12 at 16:44
@AshRj: Now handles multi-element lists with some sanity. – Steven Rumbalski Nov 28 '12 at 16:59
I am assuming the first test is to check for empty lists ? But there already is a check len(a)==1 which will handle empty lists .. – asheeshr Nov 28 '12 at 17:02
The len(a)==1 should be before the isinstance check for it to handle empty lists – asheeshr Nov 28 '12 at 17:08

score 3 · Accepted Answer · answered Nov 28 '12 at 16:58

3

Numpy is your best friend as always :

>>> import numpy as np
>>> a = [[[ ['green', 'blue', 'red' ] ]]]
>>> print np.squeeze(a)
['green' 'blue' 'red']

The numpy function squeeze() remove all the dimensions that are 1 in your array.

answered Nov 28 '12 at 16:58

Thomas Leonard

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Don't agree with `your best friend as always` - but +1 – Jon Clements Nov 28 '12 at 18:04
:) why? Numpy has ever let you down in the past? – Thomas Leonard Nov 30 '12 at 16:08

Cameron Sparr · Answer 3 · 2012-11-28T16:51:56.680

0

recursive solution:

def strip(a):
    if len(a)==1:
        if isinstance(a[0], list):
            a=a[0]
            return strip(a)
        else:
            return a
    return a

edited Nov 28 '12 at 16:51

answered Nov 28 '12 at 16:42

Cameron Sparr

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@AshRj, what would be the expected output in that case though? – Cameron Sparr Nov 28 '12 at 16:46
Depending on state of the program, you could want to return either the first list or the second list. – asheeshr Nov 28 '12 at 16:48
`strip([['a']]) -> RuntimeError: maximum recursion depth exceeded` – Steven Rumbalski Nov 28 '12 at 16:49
true about the single element lists, I added an instance check – Cameron Sparr Nov 28 '12 at 16:52
@COpython The OP hasnt **specified** that there will only be one element :) But ,anyways, point taken. – asheeshr Nov 28 '12 at 16:53
Also note the OP has a nested list of strings, not numbers (and that matters). – martineau Nov 28 '12 at 16:58

Simplest way to return an array that is nested in multiple arrays

3 Answers3