I have a 1D NumPy array, and I want to find the index of the last non-zero element. The vector consists only of 1s and 0s, not sure if this is relevant information. For example: import numpy as np vec = np.array([0, 1, 1, 0, 0, 1, 0, 0]) In this case, the result should be 5 because vec[5] == 1 is the last non-zero value. What is the most efficient way to do this using NumPy? I am forced to use Python, but this will run a few million times with arrays that are quite a bit larger than the example and I need this to be as performant as possible. Any little edge helps. Here are two approaches I currently have. One using standard python loops, the other one trying to leverage what NumPy gives me. I feel like there might be some elegant way of making this even more performant, but I am constrained by only having an Economics degree. def last_nonzero_index(vec): for i in range(len(vec) - 1, -1, -1): if vec != 0: return i return None def last_nonzero_index(vec): rev_idx = np.flatnonzero(vec[::-1]) return len(vec) - 1 - rev_idx[0] if rev_idx.size else None Right now the native python loop does beat everything else. Continue reading...
