flatten list of list through list comprehension

Question

I am trying to flatten a list using list comprehension in python. My list is somewhat like

[[1, 2, 3], [4, 5, 6], 7, 8]

just for printing then individual item in this list of list I wrote this function:

def flat(listoflist):
    for item in listoflist:
        if type(item) != list:
            print(item)
        else:
            for num in item:
                print(num)

interactive output:

>>> flat(list1)
1
2
3
4
5
6
7
8

Then I used the same logic to flatten my list through list comprehension I am getting the following error

list2 = [item if type(item) != list else num for num in item for item in list1]

Which gives me the following error:

Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: 'int' object is not iterable

How can I flatten this type of list-of-list using using list comprehension ?

Does this answer your question? [How do I make a flat list out of a list of lists?](https://stackoverflow.com/questions/952914/how-do-i-make-a-flat-list-out-of-a-list-of-lists) — questionto42, Jul 08 '22 at 18:38
in pseudocode, `flatten(xs) = [ y | for x from xs , for y from ( if (type x == list) then flatten(x) else [x] ) ]`. now just make it valid Python syntax. — Will Ness, Apr 13 '23 at 20:22

score 28 · Answer 1 · answered Jan 02 '14 at 09:33

28

No-one has given the usual answer:

def flat(l):
  return [y for x in l for y in x]

There are dupes of this question floating around StackOverflow.

answered Jan 02 '14 at 09:33

GreenAsJade

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6

Your code does not work for the list the OP provided because his list includes non-iterable objects, namely integers. Your code does however work if every item in the list is an iterable object. See my comment below. – JDG May 26 '15 at 12:55
1

Thanks for pointing that out - I'll actually need to find a moment to mess around with that and come to grips... – GreenAsJade May 26 '15 at 23:36
1

Double `for` is interesting. The above problem can also be solved using: `[z for y in [x if isinstance(x, list) else [x] for x in [1, [2]]] for z in y]`. – Nishant Mar 19 '22 at 10:39

Ashwini Chaudhary · Accepted Answer · 2013-06-27T09:21:04.177

13

>>> from collections import Iterable
>>> from itertools import chain

One-liner:

>>> list(chain.from_iterable(item if isinstance(item,Iterable) and
                    not isinstance(item, basestring) else [item] for item in lis))
[1, 2, 3, 4, 5, 6, 7, 8]

A readable version:

>>> def func(x):                                         #use `str` in py3.x 
...     if isinstance(x, Iterable) and not isinstance(x, basestring): 
...         return x
...     return [x]
... 
>>> list(chain.from_iterable(func(x) for x in lis))
[1, 2, 3, 4, 5, 6, 7, 8]
#works for strings as well
>>> lis = [[1, 2, 3], [4, 5, 6], 7, 8, "foobar"]
>>> list(chain.from_iterable(func(x) for x in lis))                                                                
[1, 2, 3, 4, 5, 6, 7, 8, 'foobar']

Using nested list comprehension:(Going to be slow compared to itertools.chain):

>>> [ele for item in (func(x) for x in lis) for ele in item]
[1, 2, 3, 4, 5, 6, 7, 8, 'foobar']

edited Jun 27 '13 at 09:21

answered Jun 27 '13 at 09:04

Ashwini Chaudhary

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2

This is great for the asker using 2.x, but it's worth a note that 3.x users should use `str` instead of `basestring`. – Gareth Latty Jun 27 '13 at 09:11
@Ashwini What should be done if my elements are of type dict instead of integer? – Anurag Sharma Jun 28 '13 at 09:19
@AnuragSharma small example? – Ashwini Chaudhary Jun 28 '13 at 09:21
@AshwiniChaudhary for example ---> [[{'ascdd': 'skj'}, {'fd': 'srsr'}], {'dsf': 'ds', 'ds': 'dsffd'}] – Anurag Sharma Jun 28 '13 at 09:23
@AnuragSharma what is the expected output for this case? – Ashwini Chaudhary Jun 28 '13 at 09:26
@AshwiniChaudhary expected output ---> [{'ascdd': 'skj'}, {'fd': 'srsr'},{'dsf': 'ds', 'ds': 'dsffd'}] or simply a list of dictionary – Anurag Sharma Jun 28 '13 at 09:27

TerryA · Answer 3 · 2013-06-27T09:18:41.627

3

You're trying to iterate through a number, which you can't do (hence the error).

If you're using python 2.7:

>>> from compiler.ast import flatten
>>> flatten(l)
[1, 2, 3, 4, 5, 6, 7, 8]

But do note that the module is now deprecated, and no longer exists in Python 3

edited Jun 27 '13 at 09:18

answered Jun 27 '13 at 09:03

TerryA

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1

Type checking lists is a bad idea. It makes this much less flexible. – Gareth Latty Jun 27 '13 at 09:13
1

if `l` is `[[1, 2, 3], [4, 5, 6], 7, 888]`, the first solution yields `[1, 2, 3, 4, 5, 6, 7, 8, 8, 8]`. – falsetru Jun 27 '13 at 09:14
Removed the first bit, but it's a shame that my other solution won't work in python 3 – TerryA Jun 27 '13 at 09:18

score 3 · Answer 4 · answered Jun 27 '13 at 09:18

An alternative solution using a generator:

import collections

def flatten(iterable):
    for item in iterable:
        if isinstance(item, collections.Iterable) and not isinstance(item, str):  # `basestring` < 3.x
            yield from item  # `for subitem in item: yield item` < 3.3
        else:
            yield item

>>> list(flatten([[1, 2, 3], [4, 5, 6], 7, 8]))
[1, 2, 3, 4, 5, 6, 7, 8]

Alessandro Anderson · Answer 5 · 2018-08-25T09:53:14.390

2

def nnl(nl):    # non nested list

    nn = []

    for x in nl:
        if type(x) == type(5):
            nn.append(x)

    if type(x) == type([]):
        n = nnl(x)

        for y in n:
            nn.append(y)
    return nn

print (nnl([[9, 4, 5], [3, 8,[5]], 6]))  # output: [9, 4, 5, 3, 8, 5, 6]

edited Aug 25 '18 at 09:53

answered Aug 24 '18 at 13:55

Alessandro Anderson

21
2

Probably an explanation would be helpful as well. – scopchanov Aug 24 '18 at 14:17
This solution uses **recursion.** For every element in the nested list you check if the element is an integer or a list. If an integer, we append that integer in the list nn. If the element in the nested list is a list, call a recursive function, you will do to the sublist what you did to the initial nested list. Then append the elements that are not a list again in nn. The output of nn will be a non nested list. Happy coding everybody! – Alessandro Anderson Aug 25 '18 at 09:46
Great! Now please [edit your answer](https://stackoverflow.com/posts/52006022/edit) adding this information. – scopchanov Aug 25 '18 at 09:49
1

Done - Namasté. – Alessandro Anderson Aug 25 '18 at 09:54

score 0 · Answer 6 · answered Apr 16 '23 at 09:07

Here's a more efficient and fast solution using precision and generators that will also work with highly nested lists.

Implementation

def _flatten_list(l: list):
    for item in l:
        if isinstance(item, list):
            yield from flatten_list(item)
        else:
            yield item
            
def flatten_list(l: list):
    return list(_flatten_list(l))
        
example = [1, 2, [3, 4, [5, 6], 7], 8, [9, [10]]]
print(flatten_list(example))

Output

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

flatten list of list through list comprehension

6 Answers6

Implementation

Output

Linked

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