Count number array python

Question

Here is an array in python:

T = np.array([[1,1,2],[2,1,1],[3,3,3]])

print np.where(T==1)

I want to find the number of appearance of each element. I tried to use np.where and then len(np.where). However the output of np.where doesn't allow to use the len() function.

was any of these answers useful for your problem? – Dalek Feb 08 '16 at 13:36 — Dalek, Feb 08 '16 at 13:36

Zachi Shtain · Answer 1 · 2016-02-07T17:21:52.747

0

use numpy.unique, it returns both the unique values and the number of appearances in the array. The method has a flag called return_counts which is default value is false like this:

uniqueVals, count = numpy.unique(T, return_counts = true)

edited Feb 07 '16 at 17:21

answered Feb 07 '16 at 16:20

Zachi Shtain

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`numpy.unique` will work without `reshape`, but what parameter do you pass to get the number of appearances? – Akavall Feb 07 '16 at 16:46
@Akavall `return_counts = true` – Zachi Shtain Feb 07 '16 at 17:06
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Ahh. Nice! Maybe you should include that in your answer. – Akavall Feb 07 '16 at 17:18

score 0 · Answer 2 · answered Feb 07 '16 at 16:28

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You can do as following to show the number of times that the element 1 is repeated in the array T:

>>> (T == 1).sum()
4
>>> (T == 2).sum()
2

answered Feb 07 '16 at 16:28

Dalek

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score 0 · Answer 3 · answered Feb 07 '16 at 16:53

If range of your integers in the array is small you can use np.bincount

In [25]: T = np.array([[1,1,2],[2,1,1],[3,3,3]])

In [26]: np.bincount(T.reshape(-1))
Out[26]: array([0, 4, 2, 3])  # 0 showed up 0 times
                              # 1 showed up 4 times
                              # 2 showed up 2 times ...

Count number array python

3 Answers3