Iterating through lists of different lengths which are inside a list

Question

I have the following list:

a = [[1, [0], [0], [1, [0]]], [1, [0], [0], [1, [0]]], [1, [0], [0]]]

and I would like to take all the integers and make a string out of them:

b = '1001010010100'

Is there any way I can do this? Thank you in advance!

I think the more general question here is how to iterate over n-dimensional lists. Very good question... — Oisin, May 20 '16 at 15:58

score 22 · Accepted Answer · answered May 20 '16 at 16:03

22

Here is a rebellious approach:

a = [[1, [0], [0], [1, [0]]], [1, [0], [0], [1, [0]]], [1, [0], [0]]]
b = ''.join(c for c in str(a) if c.isdigit())

answered May 20 '16 at 16:03

hilberts_drinking_problem

11,322
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This was pretty smart. – Vedang Mehta May 20 '16 at 16:07
13

...and a +1 for ingenuity... I wouldn't really recommend this approach though... – Jared Goguen May 20 '16 at 16:10
2

@JaredGoguen why though? – Sabito stands with Ukraine Oct 23 '20 at 06:36
@Sabito錆兎 it's a very clever but brittle solution because it solves only this particular case. what if we only want to return digits that are n-levels deep? iterating through `a` is more "pythonic" because it maintains the structure of `a` while the operation `str(a)` destroys the structure of `a`. still I felt compelled to give `+1` for this solution :P – Derek O Jan 01 '22 at 07:18

Jared Goguen · Answer 2 · 2016-05-20T16:07:04.953

5

You can write a function that recursively iterates through your nested listed and tries to convert each element to an iterator.

def recurse_iter(it):
    if isinstance(it, basestring):
        yield it
    else:
        try:
            for element in iter(it):
                for subelement in recurse_iter(element):
                    yield subelement
        except TypeError:
            yield it

This hideous function will produce a list of the strings and non-iterable members in an object.

a = [[1, [0], [0], [1, [0]]], [1, [0], [0], [1, [0]]], [1, [0], [0]]]
list(recurse_iter(a))
# [1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0]

Converting this to a string is straight forward enough.

''.join(map(str, recurse_iter(a)))

edited May 20 '16 at 16:07

answered May 20 '16 at 15:59

Jared Goguen

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While it works for this case, a general solutions should probably check if an element is a string so that it doesn't start yielding individual characters. – Brian Schlenker May 20 '16 at 16:02
@BrianSchlenker That case will cause an infinite recursion. I'll update in a second. – Jared Goguen May 20 '16 at 16:03
I like this answer since it gives you the option to [conditionally] modify all items in the list, no matter how deep they are. – Oisin May 20 '16 at 16:16

score 1 · Answer 3 · answered May 20 '16 at 15:59

1

Code -

def solve(x):                                                        
    if isinstance(x, int):
        return str(x)
    return ''.join(solve(y) for y in x)

print(solve(a))

Output -

1001010010100

answered May 20 '16 at 15:59

Vedang Mehta

2,214
9
22

score 1 · Answer 4 · answered May 20 '16 at 16:00

1

You are looking for a flatten function:

def flatten_str(x):
    if isinstance(x, list):
        return "".join(flatten_str(a) for a in x)
    else:
        return str(x)

answered May 20 '16 at 16:00

malbarbo

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Iterating through lists of different lengths which are inside a list

4 Answers4