How to make a flat list out of a list of lists – Dev

The best answers to the question “How to make a flat list out of a list of lists” in the category Dev.

QUESTION:

Is there a shortcut to make a simple list out of a list of lists in Python?

I can do it in a for loop, but is there some cool “one-liner”?

I tried it with functools.reduce():

from functools import reduce
l = [[1, 2, 3], [4, 5, 6], [7], [8, 9]]
reduce(lambda x, y: x.extend(y), l)

But I get this error:

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 1, in <lambda>
AttributeError: 'NoneType' object has no attribute 'extend'

ANSWER:

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:

merged = list(itertools.chain.from_iterable(list2d))

ANSWER:

Given a list of lists t,

flat_list = [item for sublist in t for item in sublist]

which means:

flat_list = []
for sublist in t:
    for item in sublist:
        flat_list.append(item)

is faster than the shortcuts posted so far. (t is the list to flatten.)

Here is the corresponding function:

def flatten(t):
    return [item for sublist in t for item in sublist]

As evidence, you can use the timeit module in the standard library:

$ python -mtimeit -s't=[[1,2,3],[4,5,6], [7], [8,9]]*99' '[item for sublist in t for item in sublist]'
10000 loops, best of 3: 143 usec per loop
$ python -mtimeit -s't=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'sum(t, [])'
1000 loops, best of 3: 969 usec per loop
$ python -mtimeit -s't=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'reduce(lambda x,y: x+y,t)'
1000 loops, best of 3: 1.1 msec per loop

Explanation: the shortcuts based on + (including the implied use in sum) are, of necessity, O(T**2) when there are T 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 T sublists of k items each: the first k items are copied back and forth T-1 times, the second k items T-2 times, and so on; total number of copies is k times the sum of x for x from 1 to T excluded, i.e., k * (T**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.

ANSWER:

I tested most suggested solutions with perfplot (a pet project of mine, essentially a wrapper around timeit), and found

import functools
import operator
functools.reduce(operator.iconcat, a, [])

to be the fastest solution, both when many small lists and few long lists are concatenated. (operator.iadd is equally fast.)

A simpler and also acceptable variant is

out = []
for sublist in a:
    out.extend(sublist)

If the number of sublists is large, this performs a little worse than the above suggestion.

enter image description here

enter image description here


Code to reproduce the plot:

import functools
import itertools
import operator

import numpy as np
import perfplot


def forfor(a):
    return [item for sublist in a for item in sublist]


def sum_brackets(a):
    return sum(a, [])


def functools_reduce(a):
    return functools.reduce(operator.concat, a)


def functools_reduce_iconcat(a):
    return functools.reduce(operator.iconcat, a, [])


def itertools_chain(a):
    return list(itertools.chain.from_iterable(a))


def numpy_flat(a):
    return list(np.array(a).flat)


def numpy_concatenate(a):
    return list(np.concatenate(a))


def extend(a):
    out = []
    for sublist in a:
        out.extend(sublist)
    return out


b = perfplot.bench(
    setup=lambda n: [list(range(10))] * n,
    # setup=lambda n: [list(range(n))] * 10,
    kernels=[
        forfor,
        sum_brackets,
        functools_reduce,
        functools_reduce_iconcat,
        itertools_chain,
        numpy_flat,
        numpy_concatenate,
        extend,
    ],
    n_range=[2 ** k for k in range(16)],
    xlabel="num lists (of length 10)",
    # xlabel="len lists (10 lists total)"
)
b.save("out.png")
b.show()

ANSWER:

Note from the author: This is inefficient. But fun, because monoids are awesome. It’s not appropriate for production Python code.

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

This just sums the elements of iterable passed in the first argument, treating second argument as the initial value of the sum (if not given, 0 is used instead and this case will give you an error).

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.