python - Check if two unordered lists are equal -

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• how efficiently compare 2 unordered lists (not sets) in python? 9 answers

i'm looking easy (and quick) way determine if 2 unordered lists contain same elements:

for example:

['one', 'two', 'three'] == ['one', 'two', 'three'] :  true ['one', 'two', 'three'] == ['one', 'three', 'two'] :  true ['one', 'two', 'three'] == ['one', 'two', 'three', 'three'] :  false ['one', 'two', 'three'] == ['one', 'two', 'three', 'four'] :  false ['one', 'two', 'three'] == ['one', 'two', 'four'] :  false ['one', 'two', 'three'] == ['one'] :  false 

i'm hoping without using map.

python has built-in datatype unordered collection of (hashable) things, called set. if convert both lists sets, comparison unordered.

set(x) == set(y) 

documentation on set

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edit: @mdwhatcott points out want check duplicates. set ignores these, need similar data structure keeps track of number of items in each list. called multiset; best approximation in standard library collections.counter:

>>> import collections >>> compare = lambda x, y: collections.counter(x) == collections.counter(y) >>>  >>> compare([1,2,3], [1,2,3,3]) false >>> compare([1,2,3], [1,2,3]) true >>> compare([1,2,3,3], [1,2,2,3]) false >>>