CODEWITHSUNDEEP
pythonnumpymatrix-indexing
Coffee_Table
Coffee_Table

Reputation: 284

Efficiently getting i-th column of i-th 2D slice of 3D NumPy array, for all i

Suppose I have a NumPy array A of shape (N,N,N). From this, I form a 2D array B of shape (N,N) as follows:

B = np.column_stack( tuple(A[i,:,i] for i in range(N)) )

In other words, for the i-th 2D slice of A, I take it's i-th column; I then stack these columns to form B.

My question is:

Is there a more efficient way (NumPy indexing/slicing) to construct B from A; mainly, is it possible to eliminate the inner for loop over the 2D slices of A?

Upvotes: 1

Views: 64

Answers (1)

akuiper
akuiper

Reputation: 214927

You can use advanced indexing:

idx = np.arange(N)  # or idx = range(N)
A[idx,:,idx].T

Example:

import numpy as np
A = np.arange(27).reshape(3,3,3)

idx = np.arange(3)
A[idx,:,idx].T
#array([[ 0, 10, 20],
#       [ 3, 13, 23],
#       [ 6, 16, 26]])

np.column_stack( tuple(A[i,:,i] for i in range(3)) )
#array([[ 0, 10, 20],
#       [ 3, 13, 23],
#       [ 6, 16, 26]])

Timing: it is faster for a large array

def adv_index(N):
    idx = range(N)
    return A[idx,:,idx].T

N = 100
import numpy as np
A = np.arange(N*N*N).reshape(N,N,N)
​    
%timeit np.column_stack(tuple(A[i,:,i] for i in range(N)))
# The slowest run took 4.01 times longer than the fastest. This could mean that an intermediate result is being cached.
# 1000 loops, best of 3: 210 µs per loop

%timeit adv_index(N)
# The slowest run took 5.87 times longer than the fastest. This could mean that an intermediate result is being cached.
# 10000 loops, best of 3: 51.1 µs per loop

(np.column_stack(tuple(A[i,:,i] for i in range(N))) == adv_index(N)).all()
# True

Upvotes: 1

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