Numpy element-wise comparison using a different array for values

Question

This might be a simple problem but I am stuck with this one and the solution I found is not efficient (I think).

Let's say I have two numpy arrays, one containing indices for each position and the second one the valid indices:

import numpy as np

x = np.array([0, 1, 2, 1, 3, 2])

indices = np.array([True, True, False, False])

I would like to get an boolean array indicating where in the 1st array the value is the same as the ones contained in the second array. My solution which as I mentioned is not efficient is this:

indices2 = np.where(indices)[0]

y1 = (x == indices2[0]) | (x == indices2[1])

y2 = np.zeros_like(x, dtype=bool)
for i in indices2:
    y2 = (x == i) | y2

np.all(y2 == y1)

True

y1

array([ True, True, False, True, False, False])

So, is there a more efficient, more numpy-style way to achieve this (without the for loop for example)?

Edit:
corrected some bugs and replace the example with a smaller one as mentioned in comments.

(1) please make x much smaller for demonstrating, like 5 values, so it's easy to understand. (2) why is this not just x1 == x2? (3) if the first array in your question is x, what is the second array? Is it indices? Because none of the values in x are in indices since they have different types. — Dan
– Dan, Commented Jul 31, 2020 at 11:58
@Roberto: Well, that seems to work. I expected a compact answer but this excels all! — Eypros
– Eypros, Commented Jul 31, 2020 at 12:05

Roberto Trani · Accepted Answer · 2020-07-31 12:05:53Z

2

It seems that you want to index indices with the indices in x. Try with:

indices[x]

answered Jul 31, 2020 at 12:05

Roberto Trani

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Comments

Sabito · Accepted Answer · 2020-07-31 12:19:49Z

Here is my approach to the problem:

import numpy as np

x = np.array([0, 0, 2, 0, 3, 0, 1, 0, 0, 0, 0, 0, 3, 0, 3, 1, 0, 0, 0, 0, 0, 0,
       0, 0, 1, 0, 0, 0, 3, 0, 3, 0, 0, 1, 0, 0, 3, 0, 0, 0, 0, 0, 1, 0,
       0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0,
       0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 2, 0, 1, 0, 0, 0, 2, 0, 2, 2, 0, 2,
       0, 0, 2, 2, 0, 2, 2, 0, 2, 0, 0, 2])

indices = [True, True, False, False]

x2 = np.copy(x)
x2 = np.array(x2,dtype='object')

for i,val in enumerate(x):
    x2[i] = indices[val]
    
print(x2)

In case you don't need x later on then you can just do x = np.array(x,dtype='object') instead of using a seperate variable x2.

Output:

[True True False True False True True True True True True True False True
 False True True True True True True True True True True True True True
 False True False True True True True True False True True True True True
 True True True True True True True True True True True True True True
 True True True True True True True True True True True True True True
 True True True False True True False True True True True True False True
 False False True False True True False False True False False True False
 True True False]

Edit:

The above code, on average, took 0.12967189999994844 seconds and Roberto's code took 0.05755720000001929 seconds when tested 5000 times. So you should prefer his code instead.

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