两个 numpy arrays 与 3D 向量的点积

Question

My goal is finding the closest Segment (in an array of segments) to a single point. Getting the dot product between arrays of 2D coordinates work, but using 3D coordinates gives the following error:

*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 should return 2 values, The dot product of [1,1,1]@[3,3,3] and [2,2,2]@[4,4,4]

// Thanks everyone.

Here is the final solution to find the closest line segment to a single point.
Any optimization is welcome.

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 works only on vectors, not matrices. When passing matrices it expects to do a matrix multiplication, which will fail because of the dimensions passed.

On a vector it will work like you expected:

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

To do it in one go:

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

Answer 2

Do you mean this:

np.einsum('ij,ij->i',A,B)

output:

[ 9 24]

However, if you want the dot product of every row in A with every row in B, you should do:

A@B.T

output:

[[ 9 12]
 [18 24]]

Answer 3

The dot product is numpy is not designed to be used with arrays apparently. It's pretty easy to write some wrapper around it. Like this for example:

def array_dot(A, B):
    return [A[i]@B[i] for i in range(A.shape[0])]

Answer 4

In [265]: A = np.array([[1,1,1],[2,2,2]]) 
     ...: B = np.array([[3,3,3], [4,4,4]]) 

Element wise multiplication followed by sum works fine:

In [266]: np.sum(A*B, axis=1)                                                                        
Out[266]: array([ 9, 24])

einsum also makes expressing this easy:

In [267]: np.einsum('ij,ij->i',A,B)                                                                  
Out[267]: array([ 9, 24])

dot with 2d arrays (here (2,3) shaped), performs matrix multiplication, the classic across rows, down columns. In einsum notation this is 'ij,jk->ik'.

In [268]: np.dot(A,B)                                                                                
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-268-189f80e2c351> in <module>
----> 1 np.dot(A,B)

<__array_function__ internals> in dot(*args, **kwargs)

ValueError: shapes (2,3) and (2,3) not aligned: 3 (dim 1) != 2 (dim 0)

With a transpose, dimensions match (2,3) with (3,2),but the result is (2,2):

In [269]: np.dot(A,B.T)                                                                              
Out[269]: 
array([[ 9, 12],
       [18, 24]])

The desired values are on the diagonal.

One way to think of the problem is that we want to do a batch of 1d products. matmul/@ was added to perform batch matrix multiplication (which dot can't do). But the arrays have to be expanded to 3d, so the batch dimension is the leading one (and the 3 is on the respective last and 2nd to the last dimensions):

In [270]: A[:,None,:]@B[:,:,None]       # (2,1,3) with (2,3,1)                                                              
Out[270]: 
array([[[ 9]],

       [[24]]])

But the result is (2,1,1) shaped. The right numbers are there, but we have to squeeze out the extra dimensions.

Overall then the first 2 solutions are simplest - sum or product or einsum equivalent.

Answer 5

Function numpy.dot works with numpy.ndarray objects. The example below works.

import numpy as np

A = np.zeros((3, 3))
B = np.zeros((3, 3))
print(type(A))
# <class 'numpy.ndarray'>
C = np.dot(A, B)