Alternately fill numpy array between non-zero values-CodePudding

I have a binary numpy array, mostly zero-valued, and I want to fill the gaps bewteen non-zero values with a given value, but in an alternate way. For example:

[0,0,1,0,0,0,0,1,0,0,1,1,0,0,0,0,0,1,0,1,0,0,0,0,1,0,0,1,0,0]

should result in either

[0,0,1,1,1,1,1,1,0,0,1,1,0,0,0,0,0,1,1,1,0,0,0,0,1,1,1,1,0,0]

[1,1,1,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,0,0,1,1,1]

The idea is: while scanning the array left to right, fill 0 values with 1 up the next 1, if you didn't do it up to the previous 1. I can do this iteratively and in this way

A = np.array([0,0,1,0,0,0,0,1,0,0,1,1,0,0,0,0,0,1,0,1,0,0,0,0,1,0,0,1,0,0])
ones_index = np.where(A == 1)[0]
begins = ones_index[::2] # beginnings of filling section
ends = ones_index[1::2]  # ends of filling sections

from itertools import zip_longest

# fill those sections
for begin, end in zip_longest(begins, ends, fillvalue=len(A)):
    A[begin:end] = 1

but I'm looking for a more efficent solution, maybe with numpy broadcasting. Any ideas?

CodePudding user response：

One nice answer to this question is that we can produce the first result via np.logical_xor.accumulate(arr) | arr and the second via ~np.logical_xor.accumulate(arr) | arr. A quick demonstration:

A = np.array([0,0,1,0,0,0,0,1,0,0,1,1,0,0,0,0,0,1,0,1,0,0,0,0,1,0,0,1,0,0])
print(np.logical_xor.accumulate(A) | A)
print(~np.logical_xor.accumulate(A) | A)

The resulting output:

[0 0 1 1 1 1 1 1 0 0 1 1 0 0 0 0 0 1 1 1 0 0 0 0 1 1 1 1 0 0]
[1 1 1 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 0 0 1 1 1]

CodePudding user response：

np.where(arr.cumsum() % 2 == 1, 1, arr)
# array([0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0,
#        0, 0, 1, 1, 1, 1, 0, 0])