perform operation on select indices of a numpy array based on values in another array

Question

I have an array A. I have an another array which I call switch. Array switch, is an array of boolean values (and np.nan which correspond to location of np.nan 3 x 3 arrays in A) that determines which 3 x 3 array in A gets selected for an operation.

Example:

switch = np.array([ True, False, False, False, np.nan])

a = np.array([[[ 0.,  0.,  0.],
        [ 9.,  0.,  0.],
        [ 3., 15.,  0.]],

       [[ 0.,  0.,  0.],
        [ 9.,  0.,  0.],
        [ 3., 15.,  0.]],

       [[ 0., 27.,  0.],
        [ 0.,  0.,  0.],
        [12., 18.,  0.]],

       [[ 0.,  0.,  6.],
        [12.,  0., 27.],
        [ 0.,  0.,  0.]],

       [[ np.nan, np.nan, np.nan],
        [np.nan, np.nan, np.nan],
        [ np.nan, np.nan, np.nan]]])

The final selected array based on the example above should be:

np.array([[[ 0.,  0.,  0.],
            [ 9.,  0.,  0.],
            [ 3., 15.,  0.]],

       [[ np.nan, np.nan, np.nan],
        [np.nan, np.nan, np.nan],
        [ np.nan, np.nan, np.nan]],

       [[ np.nan, np.nan, np.nan],
        [np.nan, np.nan, np.nan],
        [ np.nan, np.nan, np.nan]],

       [[ np.nan, np.nan, np.nan],
        [np.nan, np.nan, np.nan],
        [ np.nan, np.nan, np.nan]],

       [[ np.nan, np.nan, np.nan],
        [np.nan, np.nan, np.nan],
        [ np.nan, np.nan, np.nan]]])

I can do this by using a for loop and iterating through each 3 x 3 array in A with the use of if/else control statements but I am wondering if there is a way to do this using the built in functionality within numpy. For loops and if/else statements is likely much slower than array manipulation. I've been tinkering but so far have not been able to find an optimal numpy solution.

I can restructure array switch but array A I cannot change the structure/shape.

What have you actually tried? Numpy supports direct boolean indexing — Mad Physicist
– Mad Physicist, Commented Nov 19, 2020 at 0:11

Quang Hoang · Accepted Answer · 2020-11-18 23:46:43Z

1

You can use np.where:

out = np.where(switch[:,None,None], a, np.nan)

Or use masking:

out = a.copy()
out[switch!=True] = np.nan

Output:

array([[[ 0.,  0.,  0.],
        [ 9.,  0.,  0.],
        [ 3., 15.,  0.]],

       [[nan, nan, nan],
        [nan, nan, nan],
        [nan, nan, nan]],

       [[nan, nan, nan],
        [nan, nan, nan],
        [nan, nan, nan]],

       [[nan, nan, nan],
        [nan, nan, nan],
        [nan, nan, nan]],

       [[nan, nan, nan],
        [nan, nan, nan],
        [nan, nan, nan]]])

answered Nov 18, 2020 at 23:46

Quang Hoang

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2 Comments

codingknob Over a year ago

The first method doesn't work but the second method does work. Any thoughts why the first doesn't work? I know the nan is not a proper bool mask but thats how my data arrives.

Quang Hoang Over a year ago

It works on my system with numpy 1.19.2. Maybe explicitly use switch==True instead of switchhelps.

Mad Physicist · Accepted Answer · 2020-11-19 00:07:08Z

1

This would be much easier if you used a proper boolean mask. Numpy allows for boolean indexing, but placing an nan in the array turns it into a float.

For this particular example, it doesn't matter which value NaNs take, so you can do

switch = switch.astype(bool)
result = np.full_like(a, np.nan)
result[switch] = a[switch]

answered Nov 19, 2020 at 0:07

Mad Physicist

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