Numpy: Find the Euclidean distance between two 3-D arrays

Question

Given that two three-dimensional arrays of sizes (2,2,2):

A = [[[ 0,  0],
    [92, 92]],

   [[ 0, 92],
    [ 0, 92]]]

B = [[[ 0,  0],
    [92,  0]],

   [[ 0, 92],
    [92, 92]]]

How do you find the Euclidean distance for each vector A and B efficiently?

I tried for-loops, but they are slow and I work with 3-D arrays in order (→ 2, → 2, 2).

Ultimately, I want to get the form matrix:

C = [[d1, d2],
     [d3, d4]]

Edit:

I tried the next cycle, but the biggest problem with it is the losses that I want to save. But the distances are correct.

[numpy.sqrt((A[row, col][0] - B[row, col][0])**2 + (B[row, col][1] -A[row, col][1])**2) for row in range(2) for col in range(2)]

Answer 1

NumPy, , , , . , -

np.sqrt(((A - B)**2).sum(-1))

np.einsum , , , ,

subs = A - B
out = np.sqrt(np.einsum('ijk,ijk->ij',subs,subs))

numexpr module -

import numexpr as ne
np.sqrt(ne.evaluate('sum((A-B)**2,2)'))

2 , evaluate. , . , -

a0 = A[...,0]
a1 = A[...,1]
b0 = B[...,0]
b1 = B[...,1]
out = ne.evaluate('sqrt((a0-b0)**2 + (a1-b1)**2)')

-

def sqrt_sum_sq_based(A,B):
    return np.sqrt(((A - B)**2).sum(-1))

def einsum_based(A,B):
    subs = A - B
    return np.sqrt(np.einsum('ijk,ijk->ij',subs,subs))

def numexpr_based(A,B):
    return np.sqrt(ne.evaluate('sum((A-B)**2,2)'))

def numexpr_based_with_slicing(A,B):
    a0 = A[...,0]
    a1 = A[...,1]
    b0 = B[...,0]
    b1 = B[...,1]
    return ne.evaluate('sqrt((a0-b0)**2 + (a1-b1)**2)')

-

In [288]: # Setup input arrays
     ...: dim = 2
     ...: N = 1000
     ...: A = np.random.rand(N,N,dim)
     ...: B = np.random.rand(N,N,dim)
     ...: 

In [289]: %timeit sqrt_sum_sq_based(A,B)
10 loops, best of 3: 40.9 ms per loop

In [290]: %timeit einsum_based(A,B)
10 loops, best of 3: 22.9 ms per loop

In [291]: %timeit numexpr_based(A,B)
10 loops, best of 3: 18.7 ms per loop

In [292]: %timeit numexpr_based_with_slicing(A,B)
100 loops, best of 3: 8.23 ms per loop

In [293]: %timeit np.linalg.norm(A-B, axis=-1) #@dnalow soln
10 loops, best of 3: 45 ms per loop

Answer 2

:

np.linalg.norm(A-B, axis=-1)