News from this site

 Rental advertising space, please contact the webmaster if you need cooperation


+focus
focused

classification  

no classification

tag  

no tag

date  

no datas

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

posted on 2023-05-03 19:31     read(601)     comment(0)     like(3)     collect(1)


I have a list of lists like

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

How can I flatten it to get [1, 2, 3, 4, 5, 6, 7, 8, 9]?


If your list of lists comes from a nested list comprehension, the problem can be solved more simply/directly by fixing the comprehension; please see How can I get a flat result from a list comprehension instead of a nested list?.

The most popular solutions here generally only flatten one "level" of the nested list. See Flatten an irregular (arbitrarily nested) list of lists for solutions that completely flatten a deeply nested structure (recursively, in general).


solution


Given a list of lists l,

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

which means:

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

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

Here is the corresponding function:

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

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

$ python -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 3: 143 usec per loop
$ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'sum(l, [])'
1000 loops, best of 3: 969 usec per loop
$ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'reduce(lambda x,y: x+y,l)'
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(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 I items each: the first I items are copied back and forth L-1 times, the second I items L-2 times, and so on; total number of copies is I times the sum of x for x from 1 to L excluded, i.e., I * (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.



Category of website: technical article > Q&A

Author:qs

link:http://www.pythonblackhole.com/blog/article/146/09feabd107fe226e28c0/

source:python black hole net

Please indicate the source for any form of reprinting. If any infringement is discovered, it will be held legally responsible.

3 0
collect article
collected

Comment content: (supports up to 255 characters)