How to yield nothing if condition is not fulfilled instead of an alternative using np.where?

user_17530552 0
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I have a question concerning numpy.where. Beside a condition, the parameters of np.where require both x and y.

  • x: x is yielded where the condition is true.
  • y: y is yielded/used where the condition is not fulfilled.

The part of the related code could look like the following:

BypassZero = np.where(f > 0.00000000000001, f, -8.5)

My question is: is it possible to tell np.where that "nothing" for y instead of -8.5 should be used? An example: if the condition is to use values only higher than 0.00000000000001, as shown above in the code, and this condition is not fulfilled in one numeric value out of a row (of many values), np.where should ignore this step and directly jump to the next numeric value in the row, instead of using -8.5.

numpy python
1 Answers

Edit: After some suggestions from the comments on my original answer and clarification from the OP, the task is to select those values which satisfy a condition and not include other values.

e.g. If I wanted to select all even numbers and exclude all odd numbers, I'd do it this way:

import numpy as np

x = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9, 10])
evens = x[x % 2 == 0]  # select even numbers only
print(evens)

The output:

[ 2  4  6  8 10]

In your case:

result = f[f > 0.00000000000001]

As Timus pointed out, np.where is not necessary at all.

user_98600330
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