CODEWITHSUNDEEP
pythonnumpy
dward4
dward4

Reputation: 1942

check two arrays for matching elements given the same index and return number of matches

Given two arrays, is there a numpy non-loop way to check if each ith index matches between the arrays, aka check for every i if a[i]==b[i]?

a = np.array([1,2,3,4,5,6,7,8])
b = np.array([2,3,4,5,6,7,8,9])
Output:  0 matches

I expect this has already been asked but I could not find what I was looking for, apologies if it is.

Upvotes: 2

Views: 29540

Answers (4)

Sunil
Sunil

Reputation: 181

You could try something like:

a = np.array([1,2,3,2,3,4,3,4,5,6])

b = np.array([8,2,10,2,7,4,10,4,9,8])

np.where(a == b)

(array([1, 3, 5, 7]),)

Upvotes: 0

James
James

Reputation: 36746

Try this:

np.arange(len(a))[a==b]

It creates a new array from 0 to length a representing the indices. Then use a==b to slice the array, returning the indices where a and b are the same.

Additionally from @Reblochon-Masque:

You can use numpy.where to extract the indices where two values meet a specified condition:

import numpy

a = numpy.array([0, 1, 2, 3, 4, 5, 6])
b = numpy.array([6, 5, 4, 3, 2, 1, 6])
numpy.where(a==b)

output:

(array([3, 6]),)

Upvotes: 8

AGN Gazer
AGN Gazer

Reputation: 8378

Another variation to other answers:

np.flatnonzero(a == b)

Upvotes: 3

Reblochon Masque
Reblochon Masque

Reputation: 36722

You can use numpy.where to extract the indices where two values meet a specified condition:

import numpy

a = numpy.array([0, 1, 2, 3, 4, 5, 6])
b = numpy.array([6, 5, 4, 3, 2, 1, 6])
numpy.where(a==b)

output:

(array([3, 6]),)

Upvotes: 7

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