numpy:找到两个3-D阵列之间的欧式距离

问题描述

给出两个尺寸为(2,2,2)的3-D数组:

Given, two 3-D arrays of dimensions (2,2,2):

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

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

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

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

您如何有效地找到A和B中每个向量的欧几里得距离?

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

我尝试了for循环，但是速度很慢，并且我正在按(>> 2，>> 2，2)的顺序处理3-D数组.

I have tried for-loops but these are slow, and I'm working with 3-D arrays in the order of (>>2, >>2, 2).

最终我想要一个如下形式的矩阵:

Ultimately I want a matrix of the form:

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

我尝试了以下循环，但是最大的问题是失去了我想要保留的尺寸.但是距离是正确的.

I've tried the following loop, but the biggest issue with it is that loses the dimensions I want to keep. 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)]

推荐答案

以NumPy向量化的方式进行思考，该方法将对元素进行微分，并沿最后一个轴进行平方和求和，最后得到平方根.因此，简单的实现方式是-

Thinking in a NumPy vectorized way that would be performing element-wise differentiation, squaring and summing along the last axis and finally getting square root. So, the straight-forward implementation would be -

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

我们可以使用 np.einsum ，从而使其更加高效-

We could perform the squaring and summing along the last axis in one go with np.einsum and thus make it more efficient, like so -

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

使用 numexpr模块-

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

由于我们沿最后一条轴的长度为2，因此我们可以将其切成薄片并将其输入evaluate方法.请注意，在评估字符串中无法切片.因此，修改后的实现将是-

Since, we are working with a length of 2 along the last axis, we could just slice those and feed it to evaluate method. Please note that slicing isn't possible inside the evaluate string. So, the modified implementation would be -

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

运行时测试

函数定义-

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)')

时间-

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's soln
10 loops, best of 3: 45 ms per loop

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