How do I make a flat list out of a list of lists?

A list of lists named xss can be flattened using a nested list comprehension:

flat_list = [
    x
    for xs in xss
    for x in xs
]

The above is equivalent to:

flat_list = []

for xs in xss:
    for x in xs:
        flat_list.append(x)

Here is the corresponding function:

def flatten(xss):
    return [x for xs in xss for x in xs]

This is the fastest method. As evidence, using the timeit module in the standard library, we see:

$ python -mtimeit -s'xss=[[1,2,3],[4,5,6],[7],[8,9]]*99' '[x for xs in xss for x in xs]'
10000 loops, best of 3: 143 usec per loop

$ python -mtimeit -s'xss=[[1,2,3],[4,5,6],[7],[8,9]]*99' 'sum(xss, [])'
1000 loops, best of 3: 969 usec per loop

$ python -mtimeit -s'xss=[[1,2,3],[4,5,6],[7],[8,9]]*99' 'reduce(lambda xs, ys: xs + ys, xss)'
1000 loops, best of 3: 1.1 msec per loop

Explanation: the methods based on + (including the implied use in sum) are, of necessity, O(L**2) when there are L sublists -- as the intermediate result list keeps getting longer, at each step a new intermediate result list object gets allocated, and all the items in the previous intermediate result must be copied over (as well as a few new ones added at the end). So, for simplicity and without actual loss of generality, say you have L sublists of M items each: the first M items are copied back and forth L-1 times, the second M items L-2 times, and so on; total number of copies is M times the sum of x for x from 1 to L excluded, i.e., M * (L**2)/2.

The list comprehension just generates one list, once, and copies each item over (from its original place of residence to the result list) also exactly once.

You can use itertools.chain():

>>> import itertools
>>> list2d = [[1,2,3], [4,5,6], [7], [8,9]]
>>> merged = list(itertools.chain(*list2d))

Or you can use itertools.chain.from_iterable() which doesn't require unpacking the list with the * operator:

>>> import itertools
>>> list2d = [[1,2,3], [4,5,6], [7], [8,9]]
>>> merged = list(itertools.chain.from_iterable(list2d))

This approach is arguably more readable than [item for sublist in l for item in sublist] and appears to be faster too:

$ python3 -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99;import itertools' 'list(itertools.chain.from_iterable(l))'
20000 loops, best of 5: 10.8 usec per loop
$ python3 -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' '[item for sublist in l for item in sublist]'
10000 loops, best of 5: 21.7 usec per loop
$ python3 -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'sum(l, [])'
1000 loops, best of 5: 258 usec per loop
$ python3 -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99;from functools import reduce' 'reduce(lambda x,y: x+y,l)'
1000 loops, best of 5: 292 usec per loop
$ python3 --version
Python 3.7.5rc1

Note from the author: This is very inefficient. But fun, because monoids are awesome.

>>> xss = [[1, 2, 3], [4, 5, 6], [7], [8, 9]]
>>> sum(xss, [])
[1, 2, 3, 4, 5, 6, 7, 8, 9]

sum sums the elements of the iterable xss, and uses the second argument as the initial value [] for the sum. (The default initial value is 0, which is not a list.)

Because you are summing nested lists, you actually get [1,3]+[2,4] as a result of sum([[1,3],[2,4]],[]), which is equal to [1,3,2,4].

Note that only works on lists of lists. For lists of lists of lists, you'll need another solution.

Using functools.reduce, which adds an accumulated list xs to the next list ys:

from functools import reduce
xss = [[1,2,3], [4,5,6], [7], [8,9]]
out = reduce(lambda xs, ys: xs + ys, xss)

Output:

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

A faster way using operator.concat:

from functools import reduce
import operator
xss = [[1,2,3], [4,5,6], [7], [8,9]]
out = reduce(operator.concat, xss)

Output:

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