How to count the number of true elements in a NumPy bool array

I have a NumPy array 'boolarr' of boolean type. I want to count the number of elements whose values are True. Is there a NumPy or Python routine dedicated for this task? Or, do I need to iterate over the elements in my script?

You have multiple options. Two options are the following.

boolarr.sum()
numpy.count_nonzero(boolarr)

Here's an example:

>>> import numpy as np
>>> boolarr = np.array([[0, 0, 1], [1, 0, 1], [1, 0, 1]], dtype=np.bool)
>>> boolarr
array([[False, False,  True],
       [ True, False,  True],
       [ True, False,  True]], dtype=bool)

>>> boolarr.sum()
5

Of course, that is a bool-specific answer. More generally, you can use numpy.count_nonzero.

>>> np.count_nonzero(boolarr)
5

That question solved a quite similar question for me and I thought I should share :

In raw python you can use sum() to count True values in a list :

>>> sum([True,True,True,False,False])
3

But this won't work :

>>> sum([[False, False, True], [True, False, True]])
TypeError...

In terms of comparing two numpy arrays and counting the number of matches (e.g. correct class prediction in machine learning), I found the below example for two dimensions useful:

import numpy as np
result = np.random.randint(3,size=(5,2)) # 5x2 random integer array
target = np.random.randint(3,size=(5,2)) # 5x2 random integer array

res = np.equal(result,target)
print result
print target
print np.sum(res[:,0])
print np.sum(res[:,1])

which can be extended to D dimensions.

The results are:

Prediction:

[[1 2]
 [2 0]
 [2 0]
 [1 2]
 [1 2]]

Target:

[[0 1]
 [1 0]
 [2 0]
 [0 0]
 [2 1]]

Count of correct prediction for D=1: 1

Count of correct prediction for D=2: 2

boolarr.sum(axis=1 or axis=0)

axis = 1 will output number of trues in a row and axis = 0 will count number of trues in columns so

boolarr[[true,true,true],[false,false,true]]
print(boolarr.sum(axis=1))

will be (3,1)

b[b].size

where b is the Boolean ndarray in question. It filters b for True, and then count the length of the filtered array.

This probably isn't as efficient np.count_nonzero() mentioned previously, but is useful if you forget the other syntax. Plus, this shorter syntax saves programmer time.

Demo:

In [1]: a = np.array([0,1,3])

In [2]: a
Out[2]: array([0, 1, 3])

In [3]: a[a>=1].size
Out[3]: 2

In [5]: b=a>=1

In [6]: b
Out[6]: array([False,  True,  True])

In [7]: b[b].size
Out[7]: 2

For 1D array, this is what worked for me:

import numpy as np
numbers= np.array([3, 1, 5, 2, 5, 1, 1, 5, 1, 4, 2, 1, 4, 5, 3, 4, 
                  5, 2, 4, 2, 6, 6, 3, 6, 2, 3, 5, 6, 5])

numbersGreaterThan2= np.count_nonzero(numbers> 2)