issue_comments: 991714854

@mathause Thanks for the many excellent suggestions! After having removed the for loop the way you suggested, I tried to address this:

The algorithm is quite clever but it multiplies all elements (except 2) with 0 - this could maybe be sped up by only using the relevant elements.

At first I thought that something like this could work:

python w = np.diff(v) nz = np.nonzero(w) d2 = np.tile(data, (n, 1)) r = w[nz] * d2[nz] r = r[::2] + r[1::2]

The problem however is that it turns out that w's rows sometimes have one element only, instead of two (which is when an h coincides exactly with a weight value, instead of lying between two). Given that difficulty, my impression is that it's not really solvable, or at least not in a way that would result in a more efficient version.