Flattening a shallow list in Python -

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• making flat list out of list of lists in python 25 answers

is there simple way flatten list of iterables list comprehension, or failing that, consider best way flatten shallow list this, balancing performance , readability?

i tried flatten such list nested list comprehension, this:

[image image in menuitem menuitem in list_of_menuitems] 

but in trouble of nameerror variety there, because name 'menuitem' not defined. after googling , looking around on stack overflow, got desired results reduce statement:

reduce(list.__add__, map(lambda x: list(x), list_of_menuitems)) 

but method unreadable because need list(x) call there because x django queryset object.

conclusion:

thanks contributed question. here summary of learned. i'm making community wiki in case others want add or correct these observations.

my original reduce statement redundant , better written way:

>>> reduce(list.__add__, (list(mi) mi in list_of_menuitems)) 

this correct syntax nested list comprehension (brilliant summary df!):

>>> [image mi in list_of_menuitems image in mi] 

but neither of these methods efficient using itertools.chain:

>>> itertools import chain >>> list(chain(*list_of_menuitems)) 

and @cdleary notes, it's better style avoid * operator magic using chain.from_iterable so:

>>> chain = itertools.chain.from_iterable([[1,2],[3],[5,89],[],[6]]) >>> print(list(chain)) >>> [1, 2, 3, 5, 89, 6] 

if you're looking iterate on flattened version of data structure , don't need indexable sequence, consider itertools.chain , company.

>>> list_of_menuitems = [['image00', 'image01'], ['image10'], []] >>> import itertools >>> chain = itertools.chain(*list_of_menuitems) >>> print(list(chain)) ['image00', 'image01', 'image10'] 

it work on that's iterable, should include django's iterable querysets, appears you're using in question.

edit: reduce anyway, because reduce have same overhead copying items list that's being extended. chain incur (same) overhead if run list(chain) @ end.

meta-edit: actually, it's less overhead question's proposed solution, because throw away temporary lists create when extend original temporary.

edit: j.f. sebastian says itertools.chain.from_iterable avoids unpacking , should use avoid * magic, the timeit app shows negligible performance difference.