How to find if two Numpy arrays have same items?

Question

I have two Numpy arrays A and B, I want to find whether sorted(A) == sorted(B) or not? For ex: If A = [5,3,2,4] and B = [3,2,5,4], then it should evaluate to True.

Can this be achieved in linear time?

set doesn't work if the lists have repeat values that must match up. — user11563547
– user11563547, Commented Aug 2, 2019 at 14:28
Ya right @MikeSperry suppose if A = [1,1,2,3] and B = [3,1,2], then set(A)==set(B) would return TRUE but the answer is actually False. @Lamanus — anon
– anon, Commented Aug 2, 2019 at 14:31
What's wrong with sorted(A) == sorted(B) in your case? Can you tell us more about your constraints? — user11563547
– user11563547, Commented Aug 2, 2019 at 14:31

alexanderhurst · Accepted Answer · 2019-08-02 14:56:30Z

5

How about Counter from collections

>>> import collections
>>> a = collections.Counter([1,1,2,3])
>>> b = collections.Counter([3,1,2])
>>> a == b
False

>>> a = collections.Counter([1,1,2,3])
>>> b = collections.Counter([3,1,2,1])
>>> a == b
True

Construction happens in O(n) time for each list

edit: If anyone is unfamiliar with what counter does. It goes through the list and creates what is essentially a dictionary mapping of value to the number of occurrences that value.
So once we create these dictionaries of the form a = {'1': 2, '2': 1...} Which takes up to O(n) time we can compare the two dictionaries which also takes O(n) time.

edited Aug 2, 2019 at 14:56

answered Aug 2, 2019 at 14:50

alexanderhurst

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Comments

Fokke · Accepted Answer · 2019-08-02 14:34:25Z

1

In this case adding brackets works quite fast:

A = [5,3,2,4]
B = [3,2,5,4]
print((sorted(A) == sorted(B)))

Which yields:

True

edited Aug 2, 2019 at 14:34

answered Aug 2, 2019 at 14:33

Fokke

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1 Comment

Daeho Ro Over a year ago

if there is a dupe for ex, A = [1,1,2,3] and B = [1,2,3] it will not be same.

Daeho Ro · Accepted Answer · 2019-08-02 14:40:14Z

0

Try this. If a and b are

a = np.array([1,2,3,3,2,3,4])
b = np.array([1,2,3,4])

then,

np.array_equal(np.unique(a), np.unique(b))
True

answered Aug 2, 2019 at 14:40

Daeho Ro

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