numpy: find the count of each unique element in an array

Question

I have an array which has many repeating elements and I want to find the count and the value of the non-zero element that repeats the most. I sort of have it working but I am not sure if this is correct way to do it in python.

Some test code:

import numpy as np

x = np.array([0, 0, 0, 10, 10, 10, 10, 10])

l = list()
for i in np.unique(x):
    l.extend((np.count_nonzero(x[x==i]),))

maximum_times = np.max(l)

This tells me the number of times the maximum element has repeated but it does not tell me what that element is and perhaps using the for loop is not a good idea for this but I could not come up with any other pythonic solution.

Numpy version? In 1.9 you can pass in return_counts=True to unique(). — Ashwini Chaudhary
– Ashwini Chaudhary, Commented Jan 9, 2015 at 10:14

Psirus · Accepted Answer · 2015-01-09 11:02:05Z

5

As Ashwini noted, you can use return_counts=True for this.

x = np.array([0, 0, 0, 10, 10, 10, 10, 10])

elements, repeats = np.unique(x, return_counts=True)
index = repeats.argmax()
elem = elements[index]

print "Max elem is " + str(elem) + " with " + str(repeats[index]) + " repetitions."

edited Jan 9, 2015 at 11:02

answered Jan 9, 2015 at 10:48

Psirus

1,4932 gold badges14 silver badges23 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Luca Over a year ago

This is much more elegant than my try. Thanks.

Collectives™ on Stack Overflow

numpy: find the count of each unique element in an array

1 Answer 1

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related