# [SOLVED] count the number of occurance of each one hot code

## Issue

I have a list of numpy arrays (one-hot represantation) like the example bellow, I want to count the number of occurances of each one-hot code.

``````[0 0 1 0 0 0 0 0 0 0]
[0 0 1 0 0 0 0 0 0 0]
[0 1 0 0 0 0 0 0 0 0]
[0 0 0 0 0 1 0 0 0 0]
[0 1 0 0 0 0 0 0 0 0]
[0 0 0 0 1 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 1]
[0 0 0 0 1 0 0 0 0 0]
[1 0 0 0 0 0 0 0 0 0]
[0 0 0 1 0 0 0 0 0 0]
[0 1 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 1]
``````

Edit :
Expected output :

``````[1 0 0 0 0 0 0 0 0 0] ==> 1 occurrence
[0 0 1 0 0 0 0 0 0 0] ==> 2 occurrences
[0 1 0 0 0 0 0 0 0 0] ==> 3 occurrences
[0 0 0 0 0 1 0 0 0 0] ==> 1 occurrence
[0 0 0 0 1 0 0 0 0 0] ==> 2 occurrences
[0 0 0 0 0 0 0 0 0 1] ==> 2 occurrences
``````

## Solution

I think you can get the result you seek:

``````[1 3 2 1 2 1 0 0 0 2]
``````

indicating the count of occurrences of one hot in that position via a simple column-wise sum using `ndarray.sum()`:

``````import numpy
data = numpy.array([
[0, 0, 1, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 1, 0, 0, 0, 0, 0, 0, 0],
[0, 1, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 1, 0, 0, 0, 0],
[0, 1, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[1, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 1, 0, 0, 0, 0, 0, 0],
[0, 1, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
])
print(numpy.ndarray.sum(data, axis=0))
``````

or more compactly as just:

``````print(data.sum(axis=0))
``````

both should give you:

``````[1 3 2 1 2 1 0 0 0 2]
``````

Answered By – JonSG

Answer Checked By – Jay B. (BugsFixing Admin)