python - 具有 3D 向量的两个 numpy 数组的点积

Question

我的目标是找到距离单个点最近的线段(在线段数组中)。 获取 2D 坐标数组之间的点积有效，但使用 3D 坐标会出现以下错误:

*ValueError: matmul: Input operand 1 has a mismatch in its core dimension 0, with gufunc signature (n?,k),(k,m?)->(n?,m?) (size 2 is different from 3)*


A = np.array([[1,1,1],[2,2,2]])
B = np.array([[3,3,3], [4,4,4]])

dp = np.dot(A,B)

dp 应该返回 2 个值， [1,1,1]@[3,3,3] 和 [2,2,2]@[4,4,4] 的点积>/p>

//谢谢大家

这是找到离单个点最近的线段的最终解决方案。
欢迎任何优化。

import numpy as np
import time

#find closest segment to single point

then = time.time()

#random line segment
l1 = np.random.rand(1000000, 3)*10   
l2 = np.random.rand(1000000, 3)*10

#single point
p = np.array([5,5,5]) #only single point

#set to origin
line = l2-l1
pv = p-l1  

#length of line squared
len_sq = np.sum(line**2, axis = 1) #len_sq = numpy.einsum("ij,ij->i", line, line)

#dot product of 3D vectors with einsum
dot = np.einsum('ij,ij->i',line,pv) #np.sum(line*pv,axis=1)


#percentage of line the pv vector travels in
param = np.array([dot/len_sq])

#param<0 projected point=l1, param>1 pp=l2
clamped_param = np.clip(param,0,1)

#add line fraction to l1 to get projected point
pp = l1+(clamped_param.T*line)

##distance vector between single point and projected point
pp_p = pp-p

#sort by smallest distance between projected point and l1
index_of_mininum_dist = np.sum(pp_p**2, axis = 1).argmin()

print(index_of_mininum_dist)
print("FINISHED IN: ", time.time()-then)

Answer 1

np.dot 仅适用于向量，不适用于矩阵。当传递矩阵时，它期望进行矩阵乘法，但由于传递的维数会失败。

在矢量上它会像你预期的那样工作:

np.dot(A[0,:],B[0,:])
np.dot(A[1,:],B[1,:])

一次完成:

np.sum(A*B,axis=1)