I'm trying to convert piece of matlab code to python.
a=[1 2 3;4 5 6]
b= sum(a<5)
//output :
ans :
2 1 1
Actually return the number of elements in every column which has the condition. Is there any equivalent function in numpy (python) to do this ?
Its the same.
a=np.array([[1, 2, 3],[4, 5, 6]])
b=np.sum(a<5,axis=0) # the only difference is that you need to explicitly set the dimension
Although not made for this purpose, an alternate solution would be
a=np.array([[1, 2, 3],[4, 5, 6]])
np.count_nonzero(a<5, axis=0)
# array([2, 1, 1])
Performance
For small arrays, np.sum seems to be slightly faster
x = np.repeat([1, 2, 3], 100)
y = np.repeat([4, 5, 6], 100)
a=np.array([x,y])
%timeit np.sum(a<5, axis=0)
# 7.18 µs ± 669 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
%timeit np.count_nonzero(a<5, axis=0)
# 11.8 µs ± 386 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
For very large arrays, np.count_nonzero seems to be slightly faster
x = np.repeat([1, 2, 3], 5000000)
y = np.repeat([4, 5, 6], 5000000)
a=np.array([x,y])
%timeit np.sum(a<5, axis=0)
# 126 ms ± 6.92 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit np.count_nonzero(a<5, axis=0)
# 100 ms ± 6.72 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
(a<5).sum(axis=0)