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Assume the folowing numpy array:

[1,2,3,1,2,3,1,2,3,1,2,2]

I want to count([1,2]) to count all occurrences of 1 and 2, in a single run, yielding something like

[4, 5]

corresponding to a [1, 2] input.

Is it supported in numpy?

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  • stackoverflow.com/a/28663910/4940954 Could you do something like this? This will give you the counts of each number. Commented Nov 27, 2019 at 20:55
  • yes, that's it. Thanks! Commented Nov 27, 2019 at 21:00
  • Use np.bincount(a)[[1, 2]] Commented Nov 27, 2019 at 21:08

1 Answer 1

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# Setting your input to an array array = np.array([1,2,3,1,2,3,1,2,3,1,2,2]) # Find the unique elements and get their counts unique, counts = np.unique(array, return_counts=True) # Setting the numbers to get counts for as a set search = {1, 2} # Gets the counts for the elements in search search_counts = [counts[i] for i, x in enumerate(unique) if x in search]

This will output [4, 5]

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