Flatten an irregular (arbitrarily nested) list of lists


Yes, I know this subject has been covered before:

but as far as I know, all solutions, except for one, fail on a list like [[[1, 2, 3], [4, 5]], 6], where the desired output is [1, 2, 3, 4, 5, 6] (or perhaps even better, an iterator).

The only solution I saw that works for an arbitrary nesting is found in this question:

def flatten(x):
    result = []
    for el in x:
        if hasattr(el, "__iter__") and not isinstance(el, basestring):
    return result

Is this the best approach? Did I overlook something? Any problems?

Asked By: telliott99



My solution:

import collections

def flatten(x):
    if isinstance(x, collections.Iterable):
        return [a for i in x for a in flatten(i)]
        return [x]

A little more concise, but pretty much the same.

Answered By: Josh Lee

Using generator functions can make your example easier to read and improve performance.

Python 2

Using the Iterable ABC added in 2.6:

from collections import Iterable

def flatten(xs):
    for x in xs:
        if isinstance(x, Iterable) and not isinstance(x, basestring):
            for item in flatten(x):
                yield item
            yield x

Python 3

In Python 3, basestring is no more, but the tuple (str, bytes) gives the same effect. Also, the yield from operator returns an item from a generator one at a time.

from collections.abc import Iterable

def flatten(xs):
    for x in xs:
        if isinstance(x, Iterable) and not isinstance(x, (str, bytes)):
            yield from flatten(x)
            yield x
Answered By: Cristian

This version of flatten avoids python’s recursion limit (and thus works with arbitrarily deep, nested iterables). It is a generator which can handle strings and arbitrary iterables (even infinite ones).

import itertools as IT
import collections

def flatten(iterable, ltypes=collections.Iterable):
    remainder = iter(iterable)
    while True:
        first = next(remainder)
        if isinstance(first, ltypes) and not isinstance(first, (str, bytes)):
            remainder = IT.chain(first, remainder)
            yield first

Here are some examples demonstrating its use:

# [1, 1, 1, 1, 1, 1, 1, 1, 1, 1]

                                       'foo bar'.split(),
# [2, 2, 2, 10, 20, 30, 'foo', 'bar', 1, 1]

# [1, 2, 3, 4]

seq = ([[chr(i),chr(i-32)] for i in range(ord('a'), ord('z')+1)] + list(range(0,9)))
# ['a', 'A', 'b', 'B', 'c', 'C', 'd', 'D', 'e', 'E', 'f', 'F', 'g', 'G', 'h', 'H',
# 'i', 'I', 'j', 'J', 'k', 'K', 'l', 'L', 'm', 'M', 'n', 'N', 'o', 'O', 'p', 'P',
# 'q', 'Q', 'r', 'R', 's', 'S', 't', 'T', 'u', 'U', 'v', 'V', 'w', 'W', 'x', 'X',
# 'y', 'Y', 'z', 'Z', 0, 1, 2, 3, 4, 5, 6, 7, 8]

Although flatten can handle infinite generators, it can not handle infinite nesting:

def infinitely_nested():
    while True:
        yield IT.chain(infinitely_nested(), IT.repeat(1))

print(list(IT.islice(flatten(infinitely_nested()), 10)))
# hangs
Answered By: unutbu

Generator version of @unutbu’s non-recursive solution, as requested by @Andrew in a comment:

def genflat(l, ltypes=collections.Sequence):
    l = list(l)
    i = 0
    while i < len(l):
        while isinstance(l[i], ltypes):
            if not l[i]:
                i -= 1
                l[i:i + 1] = l[i]
        yield l[i]
        i += 1

Slightly simplified version of this generator:

def genflat(l, ltypes=collections.Sequence):
    l = list(l)
    while l:
        while l and isinstance(l[0], ltypes):
            l[0:1] = l[0]
        if l: yield l.pop(0)
Answered By: Alex Martelli

If you like recursion, this might be a solution of interest to you:

def f(E):
    if E==[]: 
        return []
    elif type(E) != list: 
        return [E]
        a = f(E[0])
        b = f(E[1:])
        return a

I actually adapted this from some practice Scheme code that I had written a while back.


Answered By: inspectorG4dget

I’m new to python and come from a lisp background. This is what I came up with (check out the var names for lulz):

def flatten(lst):
    if lst:
        if isinstance(car,(list,tuple)):
            if cdr: return flatten(car) + flatten(cdr)
            return flatten(car)
        if cdr: return [car] + flatten(cdr)
        return [car]

Seems to work. Test:



[1, 2, 3, 4, 5, 6, 7, 8, 1, 2]
Answered By: Michael Puckett
def flatten(xs):
    res = []
    def loop(ys):
        for i in ys:
            if isinstance(i, list):
    return res
Answered By: kev

I prefer simple answers. No generators. No recursion or recursion limits. Just iteration:

def flatten(TheList):
    listIsNested = True

    while listIsNested:                 #outer loop
        keepChecking = False
        Temp = []

        for element in TheList:         #inner loop
            if isinstance(element,list):
                keepChecking = True

        listIsNested = keepChecking     #determine if outer loop exits
        TheList = Temp[:]

    return TheList

This works with two lists: an inner for loop and an outer while loop.

The inner for loop iterates through the list. If it finds a list element, it (1) uses list.extend() to flatten that part one level of nesting and (2) switches keepChecking to True. keepchecking is used to control the outer while loop. If the outer loop gets set to true, it triggers the inner loop for another pass.

Those passes keep happening until no more nested lists are found. When a pass finally occurs where none are found, keepChecking never gets tripped to true, which means listIsNested stays false and the outer while loop exits.

The flattened list is then returned.



[1, 2, 3, 4, 100, 200, 300, 1000, 2000, 3000]

Answered By: clay

Here’s another answer that is even more interesting…

import re

def Flatten(TheList):
    a = str(TheList)
    b,_Anon = re.subn(r'[[,]]', ' ', a)
    c = b.split()
    d = [int(x) for x in c]


Basically, it converts the nested list to a string, uses a regex to strip out the nested syntax, and then converts the result back to a (flattened) list.

Answered By: clay

Although an elegant and very pythonic answer has been selected I would present my solution just for the review:

def flat(l):
    ret = []
    for i in l:
        if isinstance(i, list) or isinstance(i, tuple):
    return ret

Please tell how good or bad this code is?

Answered By: Xolve

Here is my functional version of recursive flatten which handles both tuples and lists, and lets you throw in any mix of positional arguments. Returns a generator which produces the entire sequence in order, arg by arg:

flatten = lambda *n: (e for a in n
    for e in (flatten(*a) if isinstance(a, (tuple, list)) else (a,)))


l1 = ['a', ['b', ('c', 'd')]]
l2 = [0, 1, (2, 3), [[4, 5, (6, 7, (8,), [9]), 10]], (11,)]
print list(flatten(l1, -2, -1, l2))
['a', 'b', 'c', 'd', -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
Answered By: samplebias

I don’t see anything like this posted around here and just got here from a closed question on the same subject, but why not just do something like this(if you know the type of the list you want to split):

>>> a = [1, 2, 3, 5, 10, [1, 25, 11, [1, 0]]]    
>>> g = str(a).replace('[', '').replace(']', '')    
>>> b = [int(x) for x in g.split(',') if x.strip()]

You would need to know the type of the elements but I think this can be generalised and in terms of speed I think it would be faster.

Answered By: Bogdan

Generator using recursion and duck typing (updated for Python 3):

def flatten(L):
    for item in L:
            yield from flatten(item)
        except TypeError:
            yield item

list(flatten([[[1, 2, 3], [4, 5]], 6]))
>>>[1, 2, 3, 4, 5, 6]
Answered By: dansalmo
L2 = [o for k in [[j] if not isinstance(j,list) else j for j in [k for i in [[m] if not 
isinstance(m,list) else m for m in L] for k in i]] for o in k]
Answered By: fotocoder

It was fun trying to create a function that could flatten irregular list in Python, but of course that is what Python is for (to make programming fun). The following generator works fairly well with some caveats:

def flatten(iterable):
        for item in iterable:
            yield from flatten(item)
    except TypeError:
        yield iterable

It will flatten datatypes that you might want left alone (like bytearray, bytes, and str objects). Also, the code relies on the fact that requesting an iterator from a non-iterable raises a TypeError.

>>> L = [[[1, 2, 3], [4, 5]], 6]
>>> def flatten(iterable):
        for item in iterable:
            yield from flatten(item)
    except TypeError:
        yield iterable

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


I disagree with the previous implementation. The problem is that you should not be able to flatten something that is not an iterable. It is confusing and gives the wrong impression of the argument.

>>> list(flatten(123))

The following generator is almost the same as the first but does not have the problem of trying to flatten a non-iterable object. It fails as one would expect when an inappropriate argument is given to it.

def flatten(iterable):
    for item in iterable:
            yield from flatten(item)
        except TypeError:
            yield item

Testing the generator works fine with the list that was provided. However, the new code will raise a TypeError when a non-iterable object is given to it. Example are shown below of the new behavior.

>>> L = [[[1, 2, 3], [4, 5]], 6]
>>> list(flatten(L))
[1, 2, 3, 4, 5, 6]
>>> list(flatten(123))
Traceback (most recent call last):
  File "<pyshell#32>", line 1, in <module>
  File "<pyshell#27>", line 2, in flatten
    for item in iterable:
TypeError: 'int' object is not iterable
Answered By: Noctis Skytower

Here’s the compiler.ast.flatten implementation in 2.7.5:

def flatten(seq):
    l = []
    for elt in seq:
        t = type(elt)
        if t is tuple or t is list:
            for elt2 in flatten(elt):
    return l

There are better, faster methods (If you’ve reached here, you have seen them already)

Also note:

Deprecated since version 2.6: The compiler package has been removed in Python 3.

Answered By: pradyunsg

Here’s a simple function that flattens lists of arbitrary depth. No recursion, to avoid stack overflow.

from copy import deepcopy

def flatten_list(nested_list):
    """Flatten an arbitrarily nested list, without recursion (to avoid
    stack overflows). Returns a new list, the original list is unchanged.

    >> list(flatten_list([1, 2, 3, [4], [], [[[[[[[[[5]]]]]]]]]]))
    [1, 2, 3, 4, 5]
    >> list(flatten_list([[1, 2], 3]))
    [1, 2, 3]

    nested_list = deepcopy(nested_list)

    while nested_list:
        sublist = nested_list.pop(0)

        if isinstance(sublist, list):
            nested_list = sublist + nested_list
            yield sublist
Answered By: Wilfred Hughes

totally hacky but I think it would work (depending on your data_type)

flat_list = ast.literal_eval("[%s]"%re.sub("[[]]","",str(the_list)))
Answered By: Joran Beasley

Here is another py2 approach, Im not sure if its the fastest or the most elegant nor safest …

from collections import Iterable
from itertools import imap, repeat, chain

def flat(seqs, ignore=(int, long, float, basestring)):
    return repeat(seqs, 1) if any(imap(isinstance, repeat(seqs), ignore)) or not isinstance(seqs, Iterable) else chain.from_iterable(imap(flat, seqs))

It can ignore any specific (or derived) type you would like, it returns an iterator, so you can convert it to any specific container such as list, tuple, dict or simply consume it in order to reduce memory footprint, for better or worse it can handle initial non-iterable objects such as int …

Note most of the heavy lifting is done in C, since as far as I know thats how itertools are implemented, so while it is recursive, AFAIK it isn’t bounded by python recursion depth since the function calls are happening in C, though this doesn’t mean you are bounded by memory, specially in OS X where its stack size has a hard limit as of today (OS X Mavericks) …

there is a slightly faster approach, but less portable method, only use it if you can assume that the base elements of the input can be explicitly determined otherwise, you’ll get an infinite recursion, and OS X with its limited stack size, will throw a segmentation fault fairly quickly …

def flat(seqs, ignore={int, long, float, str, unicode}):
    return repeat(seqs, 1) if type(seqs) in ignore or not isinstance(seqs, Iterable) else chain.from_iterable(imap(flat, seqs))

here we are using sets to check for the type so it takes O(1) vs O(number of types) to check whether or not an element should be ignored, though of course any value with derived type of the stated ignored types will fail, this is why its using str, unicode so use it with caution …


import random

def test_flat(test_size=2000):
    def increase_depth(value, depth=1):
        for func in xrange(depth):
            value = repeat(value, 1)
        return value

    def random_sub_chaining(nested_values):
        for values in nested_values:
            yield chain((values,), chain.from_iterable(imap(next, repeat(nested_values, random.randint(1, 10)))))

    expected_values = zip(xrange(test_size), imap(str, xrange(test_size)))
    nested_values = random_sub_chaining((increase_depth(value, depth) for depth, value in enumerate(expected_values)))
    assert not any(imap(cmp, chain.from_iterable(expected_values), flat(chain(((),), nested_values, ((),)))))

>>> test_flat()
>>> list(flat([[[1, 2, 3], [4, 5]], 6]))
[1, 2, 3, 4, 5, 6]

$ uname -a
Darwin Samys-MacBook-Pro.local 13.3.0 Darwin Kernel Version 13.3.0: Tue Jun  3 21:27:35 PDT 2014; root:xnu-2422.110.17~1/RELEASE_X86_64 x86_64
$ python --version
Python 2.7.5
Answered By: Samy Vilar

I used recursive to solve nested list with any depth

def combine_nlist(nlist,init=0,combiner=lambda x,y: x+y):
    apply function: combiner to a nested list element by element(treated as flatten list)
    for each_item in nlist:
        if isinstance(each_item,list):
            current_value =combine_nlist(each_item,current_value,combiner)
            current_value = combiner(current_value,each_item)
    return current_value

So after i define function combine_nlist, it is easy to use this function do flatting. Or you can combine it into one function. I like my solution because it can be applied to any nested list.

def flatten_nlist(nlist):
    return combine_nlist(nlist,[],lambda x,y:x+[y])


In [379]: flatten_nlist([1,2,3,[4,5],[6],[[[7],8],9],10])
Out[379]: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Answered By: Oldyoung

Without using any library:

def flat(l):
    def _flat(l, r):    
        if type(l) is not list:
            for i in l:
                r = r + flat(i)
        return r
    return _flat(l, [])

# example
test = [[1], [[2]], [3], [['a','b','c'] , [['z','x','y']], ['d','f','g']], 4]    
print flat(test) # prints [1, 2, 3, 'a', 'b', 'c', 'z', 'x', 'y', 'd', 'f', 'g', 4]
Answered By: Nir Alfasi

Using itertools.chain:

import itertools
from collections import Iterable

def list_flatten(lst):
    flat_lst = []
    for item in itertools.chain(lst):
        if isinstance(item, Iterable):
            item = list_flatten(item)
    return flat_lst

Or without chaining:

def flatten(q, final):
    if not q:
    if isinstance(q, list):
        if not isinstance(q[0], list):
            flatten(q[0], final)
        flatten(q[1:], final)
Answered By: Saksham Varma

Shamelessly taken from my own answer to another question.

This function

  • Does not use isinstance, because it’s evil and breaks duck typing.
  • Uses reduce recursively. There has to be an answer using reduce.
  • Works with arbitrary nested-lists whose elements are either nested-lists, or non-nested lists of atoms, or atoms (subjected to recursion limit).
  • Does not LBYL.
  • But not with nested-lists that contain strings as atoms.

Code below:

def flattener(left, right):
        res = reduce(flattener, right, left)
    except TypeError:
        res = left
    return res

def flatten(seq):
    return reduce(flattener, seq, [])

>>> nested_list = [0, [1], [[[[2]]]],
                   [3, [], [4, 5]],
                   [6, [7, 8],
                    9, [[[]], 10,
                   11, [], [],
>>> flatten(nested_list)
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
Answered By: Cong Ma

I’m surprised no one has thought of this. Damn recursion I don’t get the recursive answers that the advanced people here made. anyway here is my attempt on this. caveat is it’s very specific to the OP’s use case

import re

L = [[[1, 2, 3], [4, 5]], 6]
flattened_list = re.sub("[[]]", "", str(L)).replace(" ", "").split(",")
new_list = list(map(int, flattened_list))


[1, 2, 3, 4, 5, 6]
Answered By: Zion

The easiest way is to use the morph library using pip install morph.

The code is:

import morph

list = [[[1, 2, 3], [4, 5]], 6]
flattened_list = morph.flatten(list)  # returns [1, 2, 3, 4, 5, 6]
Answered By: YPCrumble

We can also use the ‘type’ function of python. When iterating the list we check if the item is a list or not. If not we ‘append’ it else we ‘extend’ it. Here is a sample code –

for i in l:
    if type(i) is list:
print x


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

For more info on append() and extend() check this website :

Answered By: noobcoder

I am a dumb guy so I’ll give a “dumb” solution. All that recursion hurts my brain.

flattened_list = []
nested_list = [[[1, 2, 3], [4, 5]], 6]

def flatten(nested_list, container):
    for item in nested_list:
        if isintance(item, list):
            flatten(item, container)

>>> flatten(nested_list, flattened_list)
>>> flattened_list
[1, 2, 3, 4, 5, 6]

I get that it’s using a side effect but well that’s to the best of my comprehension of recursion can go

Answered By: halcyonjuly7

I didn’t go through all the already available answers here, but here is a one liner I came up with, borrowing from lisp’s way of first and rest list processing

def flatten(l): return flatten(l[0]) + (flatten(l[1:]) if len(l) > 1 else []) if type(l) is list else [l]

here is one simple and one not-so-simple case –

>>> flatten([1,[2,3],4])
[1, 2, 3, 4]

>>> flatten([1, [2, 3], 4, [5, [6, {'name': 'some_name', 'age':30}, 7]], [8, 9, [10, [11, [12, [13, {'some', 'set'}, 14, [15, 'some_string'], 16], 17, 18], 19], 20], 21, 22, [23, 24], 25], 26, 27, 28, 29, 30])
[1, 2, 3, 4, 5, 6, {'age': 30, 'name': 'some_name'}, 7, 8, 9, 10, 11, 12, 13, set(['set', 'some']), 14, 15, 'some_string', 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30]
Answered By: Shreyas

I am aware that there are already many awesome answers but i wanted to add an answer that uses the functional programming method of solving the question. In this answer i make use of double recursion :

def flatten_list(seq):
    if not seq:
        return []
    elif isinstance(seq[0],list):
        return (flatten_list(seq[0])+flatten_list(seq[1:]))
        return [seq[0]]+flatten_list(seq[1:])



[1, 2, 3, 4, 5, 6, 7]
Answered By: Leo wahyd

This will flatten a list or dictionary (or list of lists or dictionaries of dictionaries etc). It assumes that the values are strings and it creates a string that concatenates each item with a separator argument. If you wanted you could use the separator to split the result into a list object afterward. It uses recursion if the next value is a list or a string. Use the key argument to tell whether you want the keys or the values (set key to false) from the dictionary object.

def flatten_obj(n_obj, key=True, my_sep=''):
    my_string = ''
    if type(n_obj) == list:
        for val in n_obj:
            my_sep_setter = my_sep if my_string != '' else ''
            if type(val) == list or type(val) == dict:
                my_string += my_sep_setter + flatten_obj(val, key, my_sep)
                my_string += my_sep_setter + val
    elif type(n_obj) == dict:
        for k, v in n_obj.items():
            my_sep_setter = my_sep if my_string != '' else ''
            d_val = k if key else v
            if type(v) == list or type(v) == dict:
                my_string += my_sep_setter + flatten_obj(v, key, my_sep)
                my_string += my_sep_setter + d_val
    elif type(n_obj) == str:
        my_sep_setter = my_sep if my_string != '' else ''
        my_string += my_sep_setter + n_obj
        return my_string
    return my_string

print(flatten_obj(['just', 'a', ['test', 'to', 'try'], 'right', 'now', ['or', 'later', 'today'],
                [{'dictionary_test': 'test'}, {'dictionary_test_two': 'later_today'}, 'my power is 9000']], my_sep=', ')


just, a, test, to, try, right, now, or, later, today, dictionary_test, dictionary_test_two, my power is 9000
Answered By: Matt Farguson


from collections import Iterable

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

def flatten(thing):
    result = []

    if isinstance(thing, Iterable):
        for item in thing:

    return result

flat = flatten(L)
Answered By: Fuji Komalan

I’m not sure if this is necessarily quicker or more effective, but this is what I do:

def flatten(lst):
    return eval('[' + str(lst).replace('[', '').replace(']', '') + ']')

L = [[[1, 2, 3], [4, 5]], 6]

The flatten function here turns the list into a string, takes out all of the square brackets, attaches square brackets back onto the ends, and turns it back into a list.

Although, if you knew you would have square brackets in your list in strings, like [[1, 2], "[3, 4] and [5]"], you would have to do something else.

Answered By: diligar

Iterative solution with Python 3

This solution may work with all objects except str and bytes.

from collections import Iterable
from collections import Iterator

def flat_iter(obj):
    stack = [obj]
    while stack:
        element = stack.pop()
        if element and isinstance(element, Iterator):
            except StopIteration:
        elif isinstance(element, Iterable) and not isinstance(element, (str, bytes)):
            yield element

tree_list = [[(1,2,3),(4,5,6, (7,8, 'next element is 5')), (5,6), [[[3,4,5],'foo1'],'foo2'],'foo3']]

not_iterable = 10

it1 = flat_iter(tree_list)
it2 = flat_iter(not_iterable)


[1, 2, 3, 4, 5, 6, 7, 8, ‘next element is 5’, 5, 6, 3, 4, 5, ‘foo1’, ‘foo2’, ‘foo3’]


Answered By: DeaD_EyE

You could use deepflatten from the 3rd party package iteration_utilities:

>>> from iteration_utilities import deepflatten
>>> L = [[[1, 2, 3], [4, 5]], 6]
>>> list(deepflatten(L))
[1, 2, 3, 4, 5, 6]

>>> list(deepflatten(L, types=list))  # only flatten "inner" lists
[1, 2, 3, 4, 5, 6]

It’s an iterator so you need to iterate it (for example by wrapping it with list or using it in a loop). Internally it uses an iterative approach instead of an recursive approach and it’s written as C extension so it can be faster than pure python approaches:

>>> %timeit list(deepflatten(L))
12.6 µs ± 298 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
>>> %timeit list(deepflatten(L, types=list))
8.7 µs ± 139 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

>>> %timeit list(flatten(L))   # Cristian - Python 3.x approach from https://stackoverflow.com/a/2158532/5393381
86.4 µs ± 4.42 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

>>> %timeit list(flatten(L))   # Josh Lee - https://stackoverflow.com/a/2158522/5393381
107 µs ± 2.99 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

>>> %timeit list(genflat(L, list))  # Alex Martelli - https://stackoverflow.com/a/2159079/5393381
23.1 µs ± 710 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

I’m the author of the iteration_utilities library.

Answered By: MSeifert

Simple Function Without using instances

L = [[[1, 2, 3], [4, 5]], 6]
l1 = []
def FlattenList(List1):
    for i in range(len(List1)):
        if type(List1[i]) == type([]):
    return l1

[1, 2, 3, 4, 5, 6]
Answered By: Shivpe_R

From my previous answer, this function flattens most cases I can think of. I believe this works down to python 2.3.

def flatten(item, keepcls=(), keepobj=()):
    if not hasattr(item, '__iter__') or isinstance(item, keepcls) or item in keepobj:
        yield item
        for i in item:
            for j in flatten(i, keepcls, keepobj + (item,)):
                yield j

Circular lists

>>> list(flatten([1, 2, [...], 3]))
[1, 2, [1, 2, [...], 3], 3]

Depth first lists

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

Nested repeated lists:

>>> list(flatten([[1,2],[1,[1,2]],[1,2]]))
[1, 2, 1, 1, 2, 1, 2]

Lists with dicts (or other objects to not flatten)

>>> list(flatten([1,2, {'a':1, 'b':2}, 'text'], keepcls=(dict, str)))
[1, 2, {'a': 1, 'b': 2}, 'text']

Any iterables

>>> list(flatten((x for x in [1,2, set([3,(4,5),6])])))
[1, 2, 4, 5, 3, 6]

You may want to keep some default classes in keepcls to make calling
the function more terse.

Answered By: wihlke

When trying to answer such a question you really need to give the limitations of the code you propose as a solution. If it was only about performances I wouldn’t mind too much, but most of the codes proposed as solution (including the accepted answer) fail to flatten any list that has a depth greater than 1000.

When I say most of the codes I mean all codes that use any form of recursion (or call a standard library function that is recursive). All these codes fail because for every of the recursive call made, the (call) stack grow by one unit, and the (default) python call stack has a size of 1000.

If you’re not too familiar with the call stack, then maybe the following will help (otherwise you can just scroll to the Implementation).

Call stack size and recursive programming (dungeon analogy)

Finding the treasure and exit

Imagine you enter a huge dungeon with numbered rooms, looking for a treasure. You don’t know the place but you have some indications on how to find the treasure. Each indication is a riddle (difficulty varies, but you can’t predict how hard they will be). You decide to think a little bit about a strategy to save time, you make two observations:

  1. It’s hard (long) to find the treasure as you’ll have to solve (potentially hard) riddles to get there.
  2. Once the treasure found, returning to the entrance may be easy, you just have to use the same path in the other direction (though this needs a bit of memory to recall your path).

When entering the dungeon, you notice a small notebook here. You decide to use it to write down every room you exit after solving a riddle (when entering a new room), this way you’ll be able to return back to the entrance. That’s a genius idea, you won’t even spend a cent implementing your strategy.

You enter the dungeon, solving with great success the first 1001 riddles, but here comes something you hadn’t planed, you have no space left in the notebook you borrowed. You decide to abandon your quest as you prefer not having the treasure than being lost forever inside the dungeon (that looks smart indeed).

Executing a recursive program

Basically, it’s the exact same thing as finding the treasure. The dungeon is the computer’s memory, your goal now is not to find a treasure but to compute some function (find f(x) for a given x). The indications simply are sub-routines that will help you solving f(x). Your strategy is the same as the call stack strategy, the notebook is the stack, the rooms are the functions’ return addresses:

x = ["over here", "am", "I"]
y = sorted(x) # You're about to enter a room named `sorted`, note down the current room address here so you can return back: 0x4004f4 (that room address looks weird)
# Seems like you went back from your quest using the return address 0x4004f4
# Let's see what you've collected 
print(' '.join(y))

The problem you encountered in the dungeon will be the same here, the call stack has a finite size (here 1000) and therefore, if you enter too many functions without returning back then you’ll fill the call stack and have an error that look like “Dear adventurer, I’m very sorry but your notebook is full”: RecursionError: maximum recursion depth exceeded. Note that you don’t need recursion to fill the call stack, but it’s very unlikely that a non-recursive program call 1000 functions without ever returning. It’s important to also understand that once you returned from a function, the call stack is freed from the address used (hence the name “stack”, return address are pushed in before entering a function and pulled out when returning). In the special case of a simple recursion (a function f that call itself once — over and over –) you will enter f over and over until the computation is finished (until the treasure is found) and return from f until you go back to the place where you called f in the first place. The call stack will never be freed from anything until the end where it will be freed from all return addresses one after the other.

How to avoid this issue?

That’s actually pretty simple: “don’t use recursion if you don’t know how deep it can go”. That’s not always true as in some cases, Tail Call recursion can be Optimized (TCO). But in python, this is not the case, and even “well written” recursive function will not optimize stack use. There is an interesting post from Guido about this question: Tail Recursion Elimination.

There is a technique that you can use to make any recursive function iterative, this technique we could call bring your own notebook. For example, in our particular case we simply are exploring a list, entering a room is equivalent to entering a sublist, the question you should ask yourself is how can I get back from a list to its parent list? The answer is not that complex, repeat the following until the stack is empty:

  1. push the current list address and index in a stack when entering a new sublist (note that a list address+index is also an address, therefore we just use the exact same technique used by the call stack);
  2. every time an item is found, yield it (or add them in a list);
  3. once a list is fully explored, go back to the parent list using the stack return address (and index).

Also note that this is equivalent to a DFS in a tree where some nodes are sublists A = [1, 2] and some are simple items: 0, 1, 2, 3, 4 (for L = [0, [1,2], 3, 4]). The tree looks like this:

           |     |     |     |
           0   --A--   3     4
               |   |
               1   2

The DFS traversal pre-order is: L, 0, A, 1, 2, 3, 4. Remember, in order to implement an iterative DFS you also “need” a stack. The implementation I proposed before result in having the following states (for the stack and the flat_list):

init.:  stack=[(L, 0)]
**0**:  stack=[(L, 0)],         flat_list=[0]
**A**:  stack=[(L, 1), (A, 0)], flat_list=[0]
**1**:  stack=[(L, 1), (A, 0)], flat_list=[0, 1]
**2**:  stack=[(L, 1), (A, 1)], flat_list=[0, 1, 2]
**3**:  stack=[(L, 2)],         flat_list=[0, 1, 2, 3]
**3**:  stack=[(L, 3)],         flat_list=[0, 1, 2, 3, 4]
return: stack=[],               flat_list=[0, 1, 2, 3, 4]

In this example, the stack maximum size is 2, because the input list (and therefore the tree) have depth 2.


For the implementation, in python you can simplify a little bit by using iterators instead of simple lists. References to the (sub)iterators will be used to store sublists return addresses (instead of having both the list address and the index). This is not a big difference but I feel this is more readable (and also a bit faster):

def flatten(iterable):
    return list(items_from(iterable))

def items_from(iterable):
    cursor_stack = [iter(iterable)]
    while cursor_stack:
        sub_iterable = cursor_stack[-1]
            item = next(sub_iterable)
        except StopIteration:   # post-order
        if is_list_like(item):  # pre-order
        elif item is not None:
            yield item          # in-order

def is_list_like(item):
    return isinstance(item, list)

Also, notice that in is_list_like I have isinstance(item, list), which could be changed to handle more input types, here I just wanted to have the simplest version where (iterable) is just a list. But you could also do that:

def is_list_like(item):
        return not isinstance(item, str)  # strings are not lists (hmm...) 
    except TypeError:
        return False

This considers strings as “simple items” and therefore flatten_iter([["test", "a"], "b]) will return ["test", "a", "b"] and not ["t", "e", "s", "t", "a", "b"]. Remark that in that case, iter(item) is called twice on each item, let’s pretend it’s an exercise for the reader to make this cleaner.

Testing and remarks on other implementations

In the end, remember that you can’t print a infinitely nested list L using print(L) because internally it will use recursive calls to __repr__ (RecursionError: maximum recursion depth exceeded while getting the repr of an object). For the same reason, solutions to flatten involving str will fail with the same error message.

If you need to test your solution, you can use this function to generate a simple nested list:

def build_deep_list(depth):
    """Returns a list of the form $l_{depth} = [depth-1, l_{depth-1}]$
    with $depth > 1$ and $l_0 = [0]$.
    sub_list = [0]
    for d in range(1, depth):
        sub_list = [d, sub_list]
    return sub_list

Which gives: build_deep_list(5) >>> [4, [3, [2, [1, [0]]]]].

Answered By: cglacet

This is a simple implement of flatten on python2

flatten=lambda l: reduce(lambda x,y:x+y,map(flatten,l),[]) if isinstance(l,list) else [l]

print flatten(test)

#output [1, 2, 3, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14]
Answered By: Statham

No recursion or nested loops. A few lines. Well formatted and easy to read:

def flatten_deep(arr: list):
    """ Flattens arbitrarily-nested list `arr` into single-dimensional. """

    while arr:
        if isinstance(arr[0], list):  # Checks whether first element is a list
            arr = arr[0] + arr[1:]  # If so, flattens that first element one level
            yield arr.pop(0)  # Otherwise yield as part of the flat array


From my own code at https://github.com/jorgeorpinel/flatten_nested_lists/blob/master/flatten.py

Just use a funcy library:
pip install funcy

import funcy

funcy.flatten([[[[1, 1], 1], 2], 3]) # returns generator
funcy.lflatten([[[[1, 1], 1], 2], 3]) # returns list
Answered By: ADR
def nested_list(depth):
    l = [depth]
    for i in range(depth-1, 0, -1):
        l = [i, l]
    return l


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

def Flatten(ul):
    fl = []
    for i in ul:
        if type(i) is list:
            fl += Flatten(i)
            fl += [i]
    return fl


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


l = nested_list(100)


import collections

def flatten(l):
    for el in l:
        if isinstance(el, collections.Iterable) and not isinstance(el, (str, bytes)):
            yield from flatten(el)
            yield el
%%timeit -n 1000

320 µs ± 14.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%%timeit -n 1000

60 µs ± 10.2 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

list(flatten(l)) == Flatten(l)


Answered By: Alexey Golyshev

The think that following would probably work in python 3:

def get_flat_iter(xparent):
        r = xparent
        if hasattr(xx, '__iter__'):
            iparent = iter(xparent)
            if iparent != xparent:
                r = map(a, xparent)
    return r

irregular_list = [1, [2, [3, 4]]]
flat_list = list(irregular_list)
print(flat_list) # [1, 2, 3, 4]
Answered By: Toothpick Anemone
def flatten(item) -> list:
    if not isinstance(item, list): return item
    return reduce(lambda x, y: x + [y] if not isinstance(y, list) else x + [*flatten(y)], item, [])

Two-line reduce function.

Answered By: Union find

I modified the code of the accepted answer and added a keyword max_depth to only flatten up to a specified depth. max_depth=0 means, the list stays as it is. Maybe somebody can use it:

def flatten(l, __depth=0, max_depth=100):

    for el in l:

        if isinstance(el, collections.Iterable) and not isinstance(el, (str, bytes)):

            __depth += 1
            if __depth <= max_depth:
                yield from flatten(el, __depth=__depth, max_depth=max_depth)
                yield el
            __depth -= 1


            yield el

Some examples:

# A
l = []
depth = 5
for i in range(depth):
    el = i
    for j in range(i):
        el = [el]
# [0, [1], [[2]], [[[3]]], [[[[4]]]]]

for i in range(depth):
    print(list(flatten_gen(l, max_depth=i)))
# [0, [1], [[2]], [[[3]]], [[[[4]]]]]
# [0,  1,   [2],   [[3]],   [[[4]]]]
# [0,  1,    2,     [3],     [[4]]]
# [0,  1,    2,      3,       [4]]
# [0,  1,    2,      3,        4]

# B
l = [[1, 2], [3, 4, [5, 6, [7, [8, [9]]], 10], 12, [13]], 14, [15]]

for i in range(6):
    print(list(flatten_gen(l, max_depth=i)))
# [[1, 2], [3, 4, [5, 6, [7, [8, [9]]], 10], 12, [13]], 14, [15]]
# [ 1, 2,   3, 4, [5, 6, [7, [8, [9]]], 10], 12, [13],  14,  15]
# [ 1, 2,   3, 4,  5, 6, [7, [8, [9]]], 10,  12,  13,   14,  15]
# [ 1, 2,   3, 4,  5, 6,  7, [8, [9]],  10,  12,  13,   14,  15]
# [ 1, 2,   3, 4,  5, 6,  7,  8, [9],   10,  12,  13,   14,  15]
# [ 1, 2,   3, 4,  5, 6,  7,  8,  9,    10,  12,  13,   14,  15]
Answered By: scleronomic

Most of the answers are using a loop to go through the items. Here I have a variant that is using an EAFP way to do things: try to get an iterator on your input, if it succeeds run your function first on the first element, next on the remainder of this iterator. If you can’t get an iterator, or if it’s a string or a bytes object: yield the element.

Thanks to the suggestion from A. Kareem, who found out that my code was very slow, due to the fact that the recursion took too long for string and byte objects, here is an improved version of my code.

def flatten(x, it = None):
        if type(x) in (str, bytes):
            yield x
            if not it:
                it = iter(x)
            yield from flatten(next(it))
        if type(x) not in (str, bytes):
            yield from flatten(x, it)
    except StopIteration:
    except Exception:
        yield x

oldlist = [1,[[[["test",3]]]],((4,5,6)),[ bytes("test", encoding="utf-8"),7,[8,9]]]
newlist = [ x for x in flatten(oldlist) ]
# [1, 'test', 3, 4, 5, 6, b'test', 7, 8, 9]
Answered By: Marko

Pandas has a function that does this. It returns an iterator as you mentioned.

In [1]: import pandas
In [2]: pandas.core.common.flatten([[[1, 2, 3], [4, 5]], 6])
Out[2]: <generator object flatten at 0x7f12ade66200>
In [3]: list(pandas.core.common.flatten([[[1, 2, 3], [4, 5]], 6]))
Out[3]: [1, 2, 3, 4, 5, 6]
Answered By: Mark Harrison

This solution is based on the python’s iteration-utilities library and its function deepflatten

from iteration_utilities import deepflatten
Answered By: MefAldemisov

I’ve tried solving it without using any library. Simply using two nested functions does the job.

def first(list_to_flatten):
    a = []

    def second(list_to_flatten):
        for i in list_to_flatten:
            if type(i) is not list:
                list_to_flatten = i

    return a

list_to_flatten = [1, 2, [3, 4, [5, 6, [7, 8, [9, 10]]]]]
a = first(list_to_flatten)

>>> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Answered By: archit jain

A much more efficient version of this answer: https://stackoverflow.com/a/20495215/8887313

If you have control over the creation of the list and are willing to mutate it, then it is much more efficient to use a deque (instead of pop(0) and list contatenation).

import collections

def flatten_and_consume(nested_deque: collections.deque):
    while nested_deque:
        elt = nested_deque.popleft()

        elt_is_sublist = isinstance(elt, collections.deque)
        if elt_is_sublist:
            yield elt
Answered By: ATOMP

This is how I did it with recursion:

def flatten(x):
    if not any(isinstance(e, list) for e in x):
        return x
    while type(x[-1]) == int:
        x = [x[-1]] + [x[:-1]]
    return flatten(x = x + x.pop(-1))

Or even:

def flatten(x):
    if not any(isinstance(e, list) for e in x):
        return x
    return flatten(x = x + x.pop([isinstance(e, list) for e in x].index(1)))
Answered By: arara

No frills. Only chills.

recursive_list_of_lists = [1,2,3,[1,2,[[3,4,[5]],7,0,1,10],100,[101,[101,[[101]],2]],0]]

k = []

def flatten(subl):
    for i in subl:
        if type(i) != type([1]):



[1, 2, 3, 1, 2, 3, 4, 5, 7, 0, 1, 10, 100, 101, 101, 101, 2, 0]

Answered By: Rex