NumPy array indexing a 4D array
17,250
Yes you can do this in a vectorized form:
p,m,n,r = a.shape
a.reshape(-1,r)[np.arange(p*m*n),b.ravel()] = 1
This should generalize more easily to higher order ndarrays.
Author by
Babak
Updated on June 07, 2022Comments
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Babak almost 2 years
I have a 4D array 'a' of size (2,3,4,4) filled with zeros.
import numpy as np a = np.zeros((2,3,4,4))
I also have a 3D array 'b' of size(2,3,4) that carries some index values (all between 0 and 3).
What I want to do is replace the element of every last array in 'a' (the 4th dimension of 'a') that corresponds to the index in 'b', with 1.
I can do this with 3 for loops, as shown below:
for i in a.shape[0]: for j in a.shape[1]: for z in a.shape[2]: a[i,j,z][b[i,j,z]] = 1
But I was wondering if there is anyway I can avoid looping at all. Something similar to:
a[b] = 1