How can I logically test the output of a np.where result?

Question

I was trying to scan an array for values and take action depending on the result. However, when I had a closer look at what the code was doing I noticed that my logical condition was ill posed.

I will illustrate what I mean with the following example:

#importing numpy
import numpy as np

#creating a test array
a = np.zeros((3,3))

#searching items bigger than 1 in 'a'
index = np.where(a > 1)

I was expecting my index to return an empty list. In fact it returns a tuple object, like:

index
Out[5]: (array([], dtype=int64), array([], dtype=int64))

So, the test I was imposing:

#testing if there are values
#in 'a' that fulfil the where condition 
if index[0] != []:
    print('Values found.')

#testing if there are no values
#in 'a' that fulfil the where condition
if index[0] == []:
    print('No values found.')

Will not achieve its purpose because I was comparing different objects (is that correct to say?).

So what is the correct way to create this test?

Thanks for your time!

Answer 1

For your 2D array, np.where returns a tuple of arrays of indices (one for each axis), so that a[index] gives you an array of the elements fulfilling the condition.

Indeed, you compared an empty list to an empty array. Instead, I would compare the size property (or e.g. len()) of the first element of this tuple:

if index[0].size == 0:
    print('No values found.')