Reputation: 1000
Just started toying around with Python so please bear with me :)
Assume the following list which contains nested lists:
[[[[[1, 3, 4, 5]], [1, 3, 8]], [[1, 7, 8]]], [[[6, 7, 8]]], [9]]
In a different representation:
[
[
[
[
[1, 3, 4, 5]
],
[1, 3, 8]
],
[
[1, 7, 8]
]
],
[
[
[6, 7, 8]
]
],
[9]
]
How would you go about extracting those inner lists so that a result with the following form would be returned:
[[1, 3, 4, 5], [1, 3, 8], [1, 7, 8], [6, 7, 8], [9]]
Many thanks!
EDIT (Thanks @falsetru):
Empty inner-list or mixed type lists will never be part of the input.
Upvotes: 23
Views: 1967
Reputation: 82929
This seems to work, assuming no 'mixed' lists like [1,2,[3]]
:
def get_inner(nested):
if all(type(x) == list for x in nested):
for x in nested:
for y in get_inner(x):
yield y
else:
yield nested
Output of list(get_inner(nested_list))
:
[[1, 3, 4, 5], [1, 3, 8], [1, 7, 8], [6, 7, 8], [9]]
Or even shorter, without generators, using sum
to combine the resulting lists:
def get_inner(nested):
if all(type(x) == list for x in nested):
return sum(map(get_inner, nested), [])
return [nested]
Upvotes: 33
Reputation: 10180
More efficient than recursion:
result = []
while lst:
l = lst.pop(0)
if type(l[0]) == list:
lst += [sublst for sublst in l if sublst] # skip empty lists []
else:
result.insert(0, l)
Upvotes: 5
Reputation: 369194
Using itertools.chain.from_iterable
:
from itertools import chain
def get_inner_lists(xs):
if isinstance(xs[0], list): # OR all(isinstance(x, list) for x in xs)
return chain.from_iterable(map(get_inner_lists, xs))
return xs,
used isinstance(xs[0], list)
instead of all(isinstance(x, list) for x in xs)
, because there's no mixed list / empty inner list.
>>> list(get_inner_lists([[[[[1, 3, 4, 5]], [1, 3, 8]], [[1, 7, 8]]], [[[6, 7, 8]]], [9]]))
[[1, 3, 4, 5], [1, 3, 8], [1, 7, 8], [6, 7, 8], [9]]
Upvotes: 13