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
– Martijn Pieters, Commented Nov 28, 2012 at 16:34
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
– Jon Clements, Commented Nov 28, 2012 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
– Nirma, Commented Nov 28, 2012 at 16:41

Thomas Leonard · Accepted Answer · 2012-11-28 16:58:02Z

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, 2012 at 16:58

Thomas Leonard

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2 Comments

Jon Clements Over a year ago

Don't agree with your best friend as always - but +1

Thomas Leonard Over a year ago

:) why? Numpy has ever let you down in the past?

Steven Rumbalski · Accepted Answer · 2012-11-28 17:26:08Z

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, 2012 at 17:26

answered Nov 28, 2012 at 16:40

Steven Rumbalski

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4 Comments

asheeshr Over a year ago

This is useful only if a single element is nested.

Steven Rumbalski Over a year ago

@AshRj: Now handles multi-element lists with some sanity.

asheeshr Over a year ago

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 Over a year ago

The len(a)==1 should be before the isinstance check for it to handle empty lists

Cameron Sparr · Accepted Answer · 2012-11-28 16:51:56Z

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, 2012 at 16:51

answered Nov 28, 2012 at 16:42

Cameron Sparr

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6 Comments

Cameron Sparr Over a year ago

@AshRj, what would be the expected output in that case though?

asheeshr Over a year ago

Depending on state of the program, you could want to return either the first list or the second list.

Steven Rumbalski Over a year ago

strip([['a']]) -> RuntimeError: maximum recursion depth exceeded

Cameron Sparr Over a year ago

true about the single element lists, I added an instance check

asheeshr Over a year ago

@COpython The OP hasnt specified that there will only be one element :) But ,anyways, point taken.

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