# python - How to make a flat list out of a list of lists

ID : 79

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### Top 5 Answer for python - How to make a flat list out of a list of lists

94

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.

81

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)) ``

77

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.

62

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.

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() ``

50

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

The `extend()` method in your example modifies `x` instead of returning a useful value (which `functools.reduce()` expects).

A faster way to do the `reduce` version would be

``>>> import operator >>> l = [[1,2,3], [4,5,6], [7], [8,9]] >>> reduce(operator.concat, l) [1, 2, 3, 4, 5, 6, 7, 8, 9] ``