how to flatten a 2D list to 1D without using numpy? [duplicate]

Question

I have a list looks like this:

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

and I want to flatten it into [1,2,3,1,2,1,4,5,6,7]

is there a light weight function to do this without using numpy?

Answer 1

Without numpy ( ndarray.flatten ) one way would be using chain.from_iterable which is an alternate constructor for itertools.chain :

>>> list(chain.from_iterable([[1,2,3],[1,2],[1,4,5,6,7]])) [1, 2, 3, 1, 2, 1, 4, 5, 6, 7] 

Or as another yet Pythonic approach you can use a list comprehension :

[j for sub in [[1,2,3],[1,2],[1,4,5,6,7]] for j in sub] 

Another functional approach very suitable for short lists could also be reduce in Python2 and functools.reduce in Python3 (don't use this for long lists):

In [4]: from functools import reduce # Python3  In [5]: reduce(lambda x,y :x+y ,[[1,2,3],[1,2],[1,4,5,6,7]]) Out[5]: [1, 2, 3, 1, 2, 1, 4, 5, 6, 7] 

To make it slightly faster you can use operator.add, which is built-in, instead of lambda:

In [6]: from operator import add  In [7]: reduce(add ,[[1,2,3],[1,2],[1,4,5,6,7]]) Out[7]: [1, 2, 3, 1, 2, 1, 4, 5, 6, 7]  In [8]: %timeit reduce(lambda x,y :x+y ,[[1,2,3],[1,2],[1,4,5,6,7]]) 789 ns ± 7.3 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)  In [9]: %timeit reduce(add ,[[1,2,3],[1,2],[1,4,5,6,7]]) 635 ns ± 4.38 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each) 

benchmark:

:~$ python -m timeit "from itertools import chain;chain.from_iterable([[1,2,3],[1,2],[1,4,5,6,7]])" 1000000 loops, best of 3: 1.58 usec per loop :~$ python -m timeit "reduce(lambda x,y :x+y ,[[1,2,3],[1,2],[1,4,5,6,7]])" 1000000 loops, best of 3: 0.791 usec per loop :~$ python -m timeit "[j for i in [[1,2,3],[1,2],[1,4,5,6,7]] for j in i]" 1000000 loops, best of 3: 0.784 usec per loop 

A benchmark on @Will's answer that used sum (its fast for short list but not for long list) :

:~$ python -m timeit "sum([[1,2,3],[4,5,6],[7,8,9]], [])" 1000000 loops, best of 3: 0.575 usec per loop :~$ python -m timeit "sum([range(100),range(100)], [])" 100000 loops, best of 3: 2.27 usec per loop :~$ python -m timeit "reduce(lambda x,y :x+y ,[range(100),range(100)])" 100000 loops, best of 3: 2.1 usec per loop 

Answer 2

For just a list like this, my favourite neat little trick is just to use sum;

sum has an optional argument: sum(iterable [, start]), so you can do:

list_of_lists = [[1,2,3], [4,5,6], [7,8,9]] print sum(list_of_lists, []) # [1,2,3,4,5,6,7,8,9] 

this works because the + operator happens to be the concatenation operator for lists, and you've told it that the starting value is [] - an empty list.

but the documentaion for sum advises that you use itertools.chain instead, as it's much clearer.