I'm looking for an easy (and quick) way to determine if two unordered lists contain the 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 to do this without using a map.
Using Counter() , we usually are able to get frequency of each element in list, checking for it, for both the list, we can check if two lists are identical or not. But this method also ignores the ordering of the elements in the list and only takes into account the frequency of elements.
We can test if two sets are equal using the normal “==” operator. Notice – since sets are unordered, it doesn't matter that the set other_nums looks like its elements are in a different order. Set equality comparison just takes into account if the two sets share the exact same elements, regardless of order.
Python has a built-in datatype for an unordered collection of (hashable) things, called a set
. If you convert both lists to sets, the comparison will be unordered.
set(x) == set(y)
Documentation on set
EDIT: @mdwhatcott points out that you want to check for duplicates. set
ignores these, so you need a similar data structure that also keeps track of the number of items in each list. This is called a multiset; the best approximation in the standard library is a 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
>>>
If elements are always nearly sorted as in your example then builtin .sort()
(timsort) should be fast:
>>> a = [1,1,2]
>>> b = [1,2,2]
>>> a.sort()
>>> b.sort()
>>> a == b
False
If you don't want to sort inplace you could use sorted()
.
In practice it might always be faster then collections.Counter()
(despite asymptotically O(n)
time being better then O(n*log(n))
for .sort()
). Measure it; If it is important.
sorted(x) == sorted(y)
Copying from here: Check if two unordered lists are equal
I think this is the best answer for this question because
You want to see if they contain the same elements, but don't care about the order.
You can use a set:
>>> set(['one', 'two', 'three']) == set(['two', 'one', 'three'])
True
But the set object itself will only contain one instance of each unique value, and will not preserve order.
>>> set(['one', 'one', 'one']) == set(['one'])
True
So, if tracking duplicates/length is important, you probably want to also check the length:
def are_eq(a, b):
return set(a) == set(b) and len(a) == len(b)
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