Comparing two same-shape Numpy arrays, one against a set of conditions to then change the other one

Question

array_one and array_two can be assumed to be the same shape.

I want to test array_two against two conditions (that it's not "invalid_data" and the value is one of the numbers in "list_of_desired_values").

For the index where those conditions were deemed true for array_two, I want to then change array_one at that same index to -9999 in this case.

I can't get this to work, and ideally I'd like to be able to do this within Numpy rather than a for loop for speed. Does anyone know an effective way to do this?

    invalid_data = 5 #example
    list_of_desired_values = [11, 2]
    array_one = numpy.array([[2, 4, 6], [6, 8, 10]], numpy.int32)
    array_two = numpy.array([[5, 2, 1], [7, 11, 55]], numpy.int32)
    for x in array_two.flatten():
        if array_two.flatten()[x] != invalid_data:
            if array_two.flatten()[x] in list_of_desired_values:
                array_one.flatten()[x] = -9999

Answer 1

Your code has 2 flaws:

1. In for x in array_two.flatten(): x is set to the value of each element, so there is no sense to use it as an index (index out of bounds exception is likely to occur).

2. array_one.flatten()[x] = -9999 has also no sense, since the result of array_one.flatten() is another (temporary) array, which then immediately disappears ( flatten() returns a copy of the source array, collapsed into one dimension).

Your code could be reworked to:

for idx, x in np.ndenumerate(array_two):
    if x != invalid_data:
        if x in list_of_desired_values:
            array_one[idx] = -9999

But the proper, numpythonic solution is to refer just to array_one with boolean indexing (indices meeting your criteria) and set there your target value:

array_one[np.logical_and(np.not_equal(array_two, invalid_data),
    np.isin(array_two, list_of_desired_values))] = -9999

The result ( array_one after this operation) is:

array([[   2, -9999,    6],
       [   6, -9999,   10]])