Collapse mask array along axis - Numpy in Python

Question

I have a 2D array of masks that I want to collapse along axis 0 using logical OR operation for values that are True. I was wondering whether there was a numpy function to do this process. My code looks something like:

>>> all_masks
array([[False, False, False, ..., False, False, False],
       [False, False, False, ..., False, False, False],
       [False, False, False, ..., False, False, False],
       [False,  True, False, ..., False,  True, False],
       [False, False, False, ..., False, False, False],
       [False,  True, False, ..., False,  True, False]])

>>> all_masks.shape
(6, 870)

>>> output_mask
array([False, True, False, ..., False, True, False])

>>> output_mask.shape
(870,)

I have achieved output_mask this process through using a for loop. However I know using a for loop makes my code slower (and kinda messy) so I was wondering whether this process could be completed through a function of numpy or likewise?

Code for collapsing masks using for loop:

mask_out = np.zeros(all_masks.shape[1], dtype=bool)
for mask in all_masks:
    mask_out = mask_out | mask

return mask_out

Answer 1

You can use ndarray.any:

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

all_masks.any(axis=0)

Output:

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