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?
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4For pandas: stackoverflow.com/questions/26053849/…Private– Private2017-04-03 08:16:47 +00:00Commented Apr 3, 2017 at 8:16
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6 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
9 Comments
norio
Thanks, David. They look neat. About the method with sum(..), is True always equal to 1 in python (or at least in numpy)? If it is not guaranteed, I will add a check, 'if True==1:' beforehand. About count_nonzero(..), unfortunately, it seems not implemented in my numpy module at version 1.5.1, but I may have a chance to use it in the future.
David Alber
@norio Regarding
bool: boolean values are treated as 1 and 0 in arithmetic operations. See "Boolean Values" in the Python Standard Library documentation. Note that NumPy's bool and Python bool are not the same, but they are compatible (see here for more information).David Alber
@norio Regarding
numpy.count_nonzero not being in NumPy v1.5.1: you are right. According to this release announcement, it was added in NumPy v1.6.0.
chbrown
FWIW,
numpy.count_nonzero is about a thousand times faster, in my Python interpreter, at least. python -m timeit -s "import numpy as np; bools = np.random.uniform(size=1000) >= 0.5" "np.count_nonzero(bools)" vs. python -m timeit -s "import numpy as np; bools = np.random.uniform(size=1000) >= 0.5" "sum(bools)"
mab
@chbrown you are right. But you should compare to
np.sum(bools) instead! However, np.count_nonzero(bools) is still ~12x faster. |
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...
3 Comments
tommy chheng
You should "flatten" the array of arrays first. unfortunately, there's no builtin method, see stackoverflow.com/questions/2158395/…
JJFord3
Thanks Guillaume! Works with Pandas dataframes as well.
Elias Hasle
The raw built-in
sum is much slower for Pandas DataFrames and numpy arrays than their respective sum methods.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
Comments
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
Comments
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)
Comments
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)
