How to flatten a hetrogenous list of list into a single list in python?

Question

I have a list of objects where objects can be lists or scalars. I want an flattened list with only scalars. Eg:

L = [35,53,[525,6743],64,63,[743,754,757]]
outputList = [35,53,525,6743,64,63,743,754,757]

P.S. The answers in this question does not work for heterogeneous lists. Flattening a shallow list in Python

this works if pop() returns a only one scalar at a time http://stackoverflow.com/a/10546929/1321404 you can modify this code with if len(returned_pop_element)>1 then call recursive function again for returned_pop_element(list). — Dmitry Zagorulkin, May 17 '12 at 08:23
Also see: http://stackoverflow.com/questions/5828123/nested-list-and-count/5828872 — sateesh, May 17 '12 at 08:26

score 10 · Accepted Answer · answered May 17 '12 at 08:36

10

Here is a relatively simple recursive version which will flatten any depth of list

l = [35,53,[525,6743],64,63,[743,754,757]]

def flatten(xs):
    result = []
    if isinstance(xs, (list, tuple)):
        for x in xs:
            result.extend(flatten(x))
    else:
        result.append(xs)
    return result

print flatten(l)

answered May 17 '12 at 08:36

Nick Craig-Wood

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I think I can make it `isinstance(xs,collections.Iterable) and not isinstance(xs,str)` so that it includes `set` and other possible iterables as well. – balki May 18 '12 at 06:50
`and not isinstance(xs,basestring)` for python < 3, but yes good idea – Nick Craig-Wood May 18 '12 at 07:00

nye17 · Answer 2 · 2012-05-17T08:49:58.120

5

it could be done neatly in one line using numpy

import numpy as np
np.hstack(l)

you end up with an ndarray

array([  35,   53,  525, 6743,   64,   63,  743,  754,  757])

edited May 17 '12 at 08:49

answered May 17 '12 at 08:33

nye17

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jamylak · Answer 3 · 2012-05-17T09:06:13.920

5

>>> data = [35,53,[525,6743],64,63,[743,754,757]]
>>> def flatten(L):
        for item in L:
            if isinstance(item,list):
                for subitem in item:
                    yield subitem
            else:
                yield item


>>> list(flatten(data))
[35, 53, 525, 6743, 64, 63, 743, 754, 757]

Here is a one-liner version for code-golf purposes (it doesn't look good :D )

>>> [y for x in data for y in (x if isinstance(x,list) else [x])]
[35, 53, 525, 6743, 64, 63, 743, 754, 757]

edited May 17 '12 at 09:06

answered May 17 '12 at 08:40

jamylak

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The first version breaks strings into characters, I don't think that is desirable. – Janne Karila May 17 '12 at 09:00
@JanneKarila It doesn't say there will be strings. – jamylak May 17 '12 at 09:00
If you use `hasattr(item, '__iter__')` you can avoid the string problem without limiting the range of iterables. – Joel Cornett May 17 '12 at 09:15
@JoelCornett The question only mentions the use of `list` so I will just use `isinstance` as the other answers have also done. – jamylak May 17 '12 at 09:17

score 3 · Answer 4 · edited May 17 '12 at 09:18

3

l = [35,53,[525,6743],64,63,[743,754,757]]
outputList = []

for i in l:
    if isinstance(i, list):
        outputList.extend(i)
    else:
        outputList.append(i)

edited May 17 '12 at 09:18

jamylak

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answered May 17 '12 at 08:22

Vikas

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@jamylak, thanks for editing, but i like two spaces for indentation :-( – Vikas May 17 '12 at 09:22
@jamylak, thanks for the ref. I always thought that the recommendation is to use spaces. Did not know it was 4 spaces. – Vikas May 17 '12 at 09:29
Yeah but of course they are filled in by the editor when you press tab anyway. It is usually set to 4 or 3 by default. Also for some reason i cannot write @Vikas at the beginning of my comments, I don't know why it is not letting me... – jamylak May 17 '12 at 09:31

score 1 · Answer 5 · edited May 23 '17 at 12:32

1

Here's a oneliner, based on the question you've mentioned:

list(itertools.chain(*((sl if isinstance(sl, list) else [sl]) for sl in l)))

UPDATE: And a fully iterator-based version:

from itertools import imap, chain
list(chain.from_iterable(imap(lambda x: x if isinstance(x, list) else [x], l)))

edited May 23 '17 at 12:32

Community

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answered May 17 '12 at 08:24

ubik

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It's a oneliner, it's not meant to be pretty. – ubik May 17 '12 at 08:45
1

Well in that case i think i have a smaller one liner, i will post it. – jamylak May 17 '12 at 08:46
1

Additionally, `sum((i if isinstance(i, list) else [i] for i in L), [])` – Joel Cornett May 17 '12 at 09:23
@JoelCornett +1 That is what I thought of as well but i didn't like it since it has to construct a new list each iteration. – jamylak May 17 '12 at 09:26

score 1 · Answer 6 · edited May 17 '12 at 09:25

1

outputList = []
for e in l:
    if type(e) == list:
        outputList += e
    else:
        outputList.append(e)

>>> outputList
[35, 53, 525, 6743, 64, 63, 743, 754, 757]

edited May 17 '12 at 09:25

jamylak

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answered May 17 '12 at 08:24

Maria Zverina

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score 0 · Answer 7 · answered May 17 '12 at 08:28

0

def nchain(iterable):
    for elem in iterable:
        if type(elem) is list:
            for elem2 in elem:
                yield elem2
        else:
            yield elem

answered May 17 '12 at 08:28

Christian Witts

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Josiah · Answer 8 · 2012-05-17T09:04:35.820

0

Recursive function that will allow for infinite tree depth:

def flatten(l):
    if isinstance(l,(list,tuple)):
        if len(l):
            return flatten(l[0]) + flatten(l[1:])
        return []
    else:
        return [l]

>>>flatten([35,53,[525,[1,2],6743],64,63,[743,754,757]])
[35, 53, 525, 1, 2, 6743, 64, 63, 743, 754, 757]

I tried to avoid isinstance so as to allow for generic types, but old version would infinite loop on strings. Now it flattens strings correctly (Not by characters now, but as if it's pretending a string is a scalar).

edited May 17 '12 at 09:04

answered May 17 '12 at 08:39

Josiah

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I would not expect strings to be flattened (broken to single characters). – Janne Karila May 17 '12 at 08:57
Technically strings are iterables, which is why I included it. It does seem kind of odd when I look at it more closely. – Josiah May 17 '12 at 09:01

score 0 · Answer 9 · answered May 17 '12 at 13:50

0

>>> L = [35,53,[525,6743],64,63,[743,754,757]]
>>> K = []
>>> [K.extend([i]) if type(i) == int else K.extend(i) for i in L ]
[None, None, None, None, None, None]
>>> K
[35, 53, 525, 6743, 64, 63, 743, 754, 757]

answered May 17 '12 at 13:50

fraxel

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score 0 · Answer 10 · answered Feb 26 '15 at 21:00

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This solution is only for your specific situation (scalars within lists) and assumes the scalars are integer. It is a terrible solution but it is incredibly short.

outputlist = map(int,",".split(str(L).replace("[","").replace("]","")))

answered Feb 26 '15 at 21:00

Tuomas Laakkonen

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score 0 · Answer 11 · answered Feb 10 '17 at 06:13

The answer is quite simple. Take advantage of recursion.

def flatten(nst_lst, final_list):

    for val in nst_lst:
        if isinstance(val, list):
            flatten(val, final_list)
        else:
            final_list.append(val)
    return final_list

#Sample usage
fl_list = []
lst_to_flatten = [["this",["a",["thing"],"a"],"is"],["a","easy"]]

print(flatten(lst_to_flatten, fl_list))

How to flatten a hetrogenous list of list into a single list in python?

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