How do I do the following dot product in 3 dimensions with numpy?

I tried:

x = np.array([[[-1, 2, -4]], [[-1, 2, -4]]])
W = np.array([[[2, -4, 3], [-3, -4, 3]], 
              [[2, -4, 3], [-3, -4, 3]]])
y = np.dot(W, x.transpose())

but received this error message:

    y = np.dot(W, x)
ValueError: shapes (2,2,3) and (2,1,3) not aligned: 3 (dim 2) != 1 (dim 1)

It's 2 dimensions equivalent is:

x = np.array([-1, 2, -4])
W = np.array([[2, -4, 3],
              [-3, -4, 3]])
y = np.dot(W,x)
print(f'{y=}')

which will return:

y=array([-22, -17])

Also, y = np.dot(W,x.transpose()) will return the same answer.

How do I do the following dot product in 3 dimensions with numpy?

I tried:

x = np.array([[[-1, 2, -4]], [[-1, 2, -4]]])
W = np.array([[[2, -4, 3], [-3, -4, 3]], 
              [[2, -4, 3], [-3, -4, 3]]])
y = np.dot(W, x.transpose())

but received this error message:

    y = np.dot(W, x)
ValueError: shapes (2,2,3) and (2,1,3) not aligned: 3 (dim 2) != 1 (dim 1)

It's 2 dimensions equivalent is:

x = np.array([-1, 2, -4])
W = np.array([[2, -4, 3],
              [-3, -4, 3]])
y = np.dot(W,x)
print(f'{y=}')

which will return:

y=array([-22, -17])

Also, y = np.dot(W,x.transpose()) will return the same answer.

• python
• numpy

Share Improve this question Follow edited Feb 10 at 13:50 ThomasIsCoding 104k99 gold badges3737 silver badges103103 bronze badges asked Feb 10 at 10:54 Sun BearSun Bear 8,3531313 gold badges6868 silver badges114114 bronze badges 1

• When you tried x.T did you get exactly the same error? I think it couplained about a (3,1,2) aray. The docs for transpose, dot, matmul/@ (and einsum) are all relevant. They are all clear about how axes are matched. – hpaulj Commented Feb 10 at 19:00

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5 Answers 5

Sorted by: Reset to default 6

The issue comes from the 3D transposition which does not transpose the axes you want by default. You need to specify the right axes during this call:

W @ x.transpose(0, 2, 1)

# Output:
# array([[[-22],
#         [-17]],
# 
#        [[-22],
#         [-17]]])

You might also want to use einsum, this makes the matching dimensions more explicit:

np.einsum('ijk,ilk->ijl', W, x)

Output:

array([[[-22],
        [-17]],

       [[-22],
        [-17]]])

First of all, I think @Jérôme Richard and @mozway have provided the most concise and efficient solutions. I cannot imagine better answers than those.

My approach is just for fun, if you would like to play with zip + map

np.array(list(map(lambda u: np.inner(*u), zip(W,x))))

gives

array([[[-22],
        [-17]],

       [[-22],
        [-17]]])

Another possible solution:

np.sum(W * x, axis=2, keepdims=True)

By multiplying W (shape (2, 2, 3)) with x (shape (2, 1, 3)), numpy automatically broadcasts x to match W's shape, resulting in an element-wise product of shape (2, 2, 3). Summing this product along the last axis (axis=2) with keepdims=True yields the desired output shape (2, 2, 1).

Output:

array([[[-22],
        [-17]],

       [[-22],
        [-17]]])

from einops import einsum
import numpy as np

x = np.array([[[-1, 2, -4]], 
              [[-1, 2, -4]]])

W = np.array([[[ 2,  -4, 3], 
               [-3, -4,  3]], 
              
              [[2, -4, 3], 
               [-3, -4, 3]]])


res = einsum(W, x, 'b i j, b k j -> b i k') 
res = np.squeeze(res)

'''
[[-22 -17]
 [-22 -17]]
'''