Efficiently get unique elements count - NumPy / Python-CodePudding

I'm looking for an extension to this question Efficiently counting number of unique elements - NumPy / Python that can also return the count of each unique element (i.e. how many times it occurs in the array).

import numpy as np
max_a = 20_000
a = np.random.randint(max_a, size=10_000).astype(np.int32)

# Using np.unique
u,counts = np.unique(a,return_counts=True)

# Faster alternative suggested in the post above
q = np.zeros(max_a, dtype=int)
q[a] = 1
v = np.nonzero(q)[0]

I can verify that u and v are the same, and the faster method using q is definitely faster. However, I also want the counts that are returned by the np.unique() call. Is there a way to modify the example here to obtain those?

NB the elements of a will always be of type np.int32 so they can be used for indexing.

CodePudding user response：

In the case where a is of type uint, you can use np.bincount which offers a fast way to get the counts. If you do have negative int's, perhaps you can add an offset to make all positive?

For example:

counts = np.bincount(a) # includes 0-counts
unique = np.flatnonzero(counts)
counts = counts[unique] # remove 0-counts