Numpy: Overlay Boolean Array on "True"s of other boolean array

Question

I have a bool 2D-array A with the numbers of True being the dimension of bool 2D-array B.

A = np.array([[False, True, True, False, True],[False, False, False, False, False],[False, True, True, False, True]])
B = np.array([[True, False, True],[True, True, True]])

A =[[False, True,  True,  False, True],
    [False, False, False, False, False],
    [False, True,  True,  False, True]]
B =[[True, False, True],
    [True, False, True]]

I want to "overlay" B on the "True"-Array of A, so that I'd get

C =  
[[False, **True**,  **False**,  False, **True**],  
[False, False, False, False, False],  
[False, **True**,  **False**,  False, **True**]]

My ultimate goal is to manipulate an array

arr = [[1, 2, 3, 4, 5], [6,7,8,9,10], [11, 12, 13, 14, 15]]

with something like

arr[A] = arr[A] + B*2

to get

arr = [[1, 4, 3, 4, 7], [6,7,8,9,10], [11, 14, 13, 14, 17]]

Thanks in advance.

dwjbosman · Accepted Answer · 2020-02-23 12:28:49Z

1

# get the indexes that are True
Xi = np.nonzero(A)

# convert to an array of 1D
B1 = np.ndarray.flatten(B)

# use Xi for dynamic indexing
A[Xi]=B1

answered Feb 23, 2020 at 12:28

dwjbosman

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Comments

Jonas Jo · Accepted Answer · 2020-02-23 12:45:15Z

0

The solution I came up with was (works only if B is quadratic):

arr[A] = (arr[A].reshape(len(B), len(B)) + 2 * B).ravel()

answered Feb 23, 2020 at 12:45

Jonas Jo

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