Access all elements of nested lists

Question

I am trying to find a short and elegant way in order to access all individual elements in nested lists. For example:

lst1 = ['1', '2', '3']
lst2 = ['4', '5']

outer = [lst1, lst2]

Is there a list comprehension that would return ['1', '2', '3', '4', '5']?

See [Making a flat list out of list of lists in Python](http://stackoverflow.com/questions/952914/making-a-flat-list-out-of-list-of-lists-in-python). I think that's what you're asking — roganjosh, Oct 04 '16 at 17:03
Are you interested in arbitrarily nested lists, or only those nested to a single degree (as your example is)? For flattening arbitrarily-deep structures, see http://rightfootin.blogspot.com/2006/09/more-on-python-flatten.html — Robᵩ, Oct 04 '16 at 17:08
Thank you all, I am looking into arbitrarily nested lists, trying to minimize the number of costly for loops. Flattening is more along the lines of what I'm interested in. — nvergos, Oct 04 '16 at 17:10
In Python 2.7 you can also use `from compiler.ast import flatten` and then just call `flatten()` on a nested list. That also leads to [Python 3 replacement for deprecated compiler.ast flatten function](http://stackoverflow.com/questions/16176742/python-3-replacement-for-deprecated-compiler-ast-flatten-function) which might be of interest. There seem to be lots of ways to do it, not sure why `from compiler.ast import flatten` was deprecated as I never tested efficiency — roganjosh, Oct 04 '16 at 17:13

score 0 · Answer 1 · answered Oct 04 '16 at 17:13

0

import itertools

lst1 = ['1', '2', '3']
lst2 = ['4', '5']
outer = [lst1, lst2]

flattened = list(itertools.chain(*outer))
['1', '2', '3', '4', '5']

answered Oct 04 '16 at 17:13

JoshuaBox

score 0 · Accepted Answer · answered Oct 04 '16 at 17:17

0

There are two short similar ways to do it:

import itertools

# with unpacking
list(itertools.chain(*outer))  

# without unpacking
list(itertools.chain.from_iterable(outer))

answered Oct 04 '16 at 17:17

Artem Selivanov

2 Answers2