Set numpy array elements to zero if they are above a specific threshold

Question

Say, I have a numpy array consists of 10 elements, for example:

a = np.array([2, 23, 15, 7, 9, 11, 17, 19, 5, 3])

Now I want to eff

Answer 1

In [7]: a = np.array([2, 23, 15, 7, 9, 11, 17, 19, 5, 3])

In [8]: a[a > 10] = 0

In [9]: a
Out[9]: array([2, 0, 0, 7, 9, 0, 0, 0, 5, 3])

Answer 2

If you don't want to change your original array

In [1]: import numpy as np


In [2]: a = np.array([2, 23, 15, 7, 9, 11, 17, 19, 5, 3])


In [3]: b = np.where(a > 10, 0, a)


In [4]: a

Out[4]: array([ 2, 23, 15,  7,  9, 11, 17, 19,  5,  3])


In [5]: b

Out[5]: array([2, 0, 0, 7, 9, 0, 0, 0, 5, 3])

Answer 3

Generally, list comprehensions are faster than for loops in python (because python knows that it doesn't need to care for a lot of things that might happen in a regular for loop):

a = [0 if a_ > thresh else a_ for a_ in a]

but, as @unutbu correctly pointed out, numpy allows list indexing, and element-wise comparison giving you index lists, so:

super_threshold_indices = a > thresh
a[super_threshold_indices] = 0

would be even faster.

Generally, when applying methods on vectors of data, have a look at numpy.ufuncs, which often perform much better than python functions that you map using any native mechanism.