Hi Alan, You can abuse np.argmax to calculate the first nonzero element in a vectorized manner:
import numpy as np A = (2 * np.random.rand(100, 50, 50)).astype(int) Compare: np.argmax(A != 0, axis=0) np.array([[np.flatnonzero(A[:,i,j])[0] for j in range(50)] for i in range(50)]) You'll also want to check for all zero arrays with np.all: np.all(A == 0, axis=0) Cheers, Stephan On Thu, Apr 17, 2014 at 9:32 AM, Alan G Isaac <alan.is...@gmail.com> wrote: > Given an array A of shape m x n x n > (i.e., a stack of square matrices), > I want an n x n array that gives the > minimum "depth" to a nonzero element. > E.g., the 0,0 element of the result is > np.flatnonzero(A[:,0,0])[0] > Can this be vectorized? > (Assuming a nonzero element exists is ok, > but dealing nicely with its absence is even better.) > > Thanks, > Alan Isaac > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@scipy.org > http://mail.scipy.org/mailman/listinfo/numpy-discussion >
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