replace element in numpy array pending a different element's value

Question

I have the following numpy array (as an example):

my_array = [[3, 7, 0]
            [20, 4, 0]
            [7, 54, 0]]

I want to replace the 0's in the 3rd column of each row with a value of 5 only if the first index is odd. So the expected outcome would be:

my_array = [[3, 7, 5]
            [20, 4, 0]
            [7, 54, 5]]

I tried numpy.where and numpy.place, but couldn't get the expected results.

Is there an elegant way to do this with numpy functions?

np.place(my_array[:,2], my_array[:,0] % 2, 5). Please provide valid python examples with the expected type. — Michael Szczesny
– Michael Szczesny, Commented Jun 4, 2022 at 22:16

Ali_Sh · Accepted Answer · 2022-06-04 21:06:14Z

2

you can do this by indexing as:

my_array[my_array[:, 0] % 2 != 0, 2] = 5

# my_array[:, 0] % 2 != 0   --- Boolean shows modifying rows --> [ True False  True]

answered Jun 4, 2022 at 21:06

Ali_Sh

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