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

## Question:

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?

## Answers:

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)
```