NumPy: find indices of strictly increasing subsequence within an array

Question

I am NOT looking to find the largest increasing subsequence.

I have a NumPy array similar to the following:

[0, 1, 5, 2, 4, 8, 8, 6, 10]

I need to find the indices of the elements that form a strictly increasing subsequence, with priority being given to earlier indices. So, for the above input, I would expect the output to be the following NumPy array:

[0, 1, 2, 5, 8]

I imagine the numbers representing the heights of buildings in a row, and I want to find the indices of the buildings I can see from one end.

I can easily do this with a for loop over the array elements, keeping track of the maximum value seen so far. But I was hoping that there exists a NumPy-based one line solution that would directly generate a NumPy int array.

Answer 1

I don't know if you consider this "NumPy-based", but how about this:

import numpy as np
arr = np.array([0,1,5,2,4,8,8,6,10])
np.array([i for i in range(len(arr)) if all(arr[:i] < arr[i])])

Answer 2

Maybe this could help

x = [0, 1, 5, 2, 4, 8, 8, 6, 10]
u, idx = np.unique(np.maximum.accumulate(x), return_index=True)

or might be more efficient

k = np.nonzero(np.diff(np.maximum.accumulate(x)))
idx = np.append(k[0][0],np.add(k,1))

where idx gives the indices

[0 1 2 5 8]