I'm trying to do the multiplying of 3d matrix and 3d matrix, my matrix is as follows:
Z = np.array([
[[0,0,0.25],[0.25,0.5,0.75],[0,0,0.25],[0.75,1.0,1.0],[0.75,1.0,1.0]],
[[0,0,0.25],[0,0,0.25],[0.5,0.75,1.0],[0,0,0.25],[0,0,0.25]],
[[0,0,0.25],[0,0,0.25],[0,0,0.25],[0,0.25,0.5],[0,0,0.25]],
[[0,0,0.25],[0.25,0.5,0.75],[0,0,0.25],(0,0,0.25),[0,0,0.25]],
[[0,0,0.25],[0,0,0.25],[0,0,0.25],[0,0,0.25],[0,0,0.25]]
])
print(Z)
print(type(Z))
print("np.shape = ",np.shape(Z))
The shape is (5,5,3), I want to do the multiplying like np.dot(Z,Z)
,but it can't work in 3d matrix.
I've seen about using np.tensordot(Z,Z,axes=?)
, but I don't know how to set axes.
CodePudding user response:
I suggest you take a look at the documentation of the tensordot()
function to actually understand what it is doing with the matrices:
import numpy as np
Z = np.array([
[[0,0,0.25],[0.25,0.5,0.75],[0,0,0.25],[0.75,1.0,1.0],[0.75,1.0,1.0]],
[[0,0,0.25],[0,0,0.25],[0.5,0.75,1.0],[0,0,0.25],[0,0,0.25]],
[[0,0,0.25],[0,0,0.25],[0,0,0.25],[0,0.25,0.5],[0,0,0.25]],
[[0,0,0.25],[0.25,0.5,0.75],[0,0,0.25],(0,0,0.25),[0,0,0.25]],
[[0,0,0.25],[0,0,0.25],[0,0,0.25],[0,0,0.25],[0,0,0.25]]
])
B = np.tensordot(Z, Z, axes=[1, 0])
print(B)
Output:
[[[[0. 0. 0.4375]
[0.1875 0.375 0.8125]
[0.125 0.1875 0.625 ]
[0. 0. 0.4375]
[0. 0. 0.4375]]
[[0. 0. 0.625 ]
[0.25 0.5 1.125 ]
[0.25 0.375 1. ]
[0. 0. 0.625 ]
[0. 0. 0.625 ]]
[[0. 0. 0.8125]
[0.3125 0.625 1.4375]
[0.375 0.5625 1.375 ]
[0.1875 0.3125 1.0625]
[0.1875 0.25 1. ]]]
[[[0. 0. 0.125 ]
[0. 0. 0.125 ]
[0. 0. 0.125 ]
[0. 0.125 0.25 ]
[0. 0. 0.125 ]]
[[0. 0. 0.1875]
[0. 0. 0.1875]
[0. 0. 0.1875]
[0. 0.1875 0.375 ]
[0. 0. 0.1875]]
[[0. 0. 0.5 ]
[0.125 0.25 0.75 ]
[0.125 0.1875 0.6875]
[0.1875 0.5 0.9375]
[0.1875 0.25 0.6875]]]
[[[0. 0. 0. ]
[0. 0. 0. ]
[0. 0. 0. ]
[0. 0. 0. ]
[0. 0. 0. ]]
[[0. 0. 0.0625]
[0.0625 0.125 0.1875]
[0. 0. 0.0625]
[0. 0. 0.0625]
[0. 0. 0.0625]]
[[0. 0. 0.375 ]
[0.1875 0.375 0.75 ]
[0.125 0.1875 0.5625]
[0.1875 0.3125 0.625 ]
[0.1875 0.25 0.5625]]]
[[[0. 0. 0.0625]
[0. 0. 0.0625]
[0.125 0.1875 0.25 ]
[0. 0. 0.0625]
[0. 0. 0.0625]]
[[0. 0. 0.125 ]
[0. 0. 0.125 ]
[0.25 0.375 0.5 ]
[0. 0. 0.125 ]
[0. 0. 0.125 ]]
[[0. 0. 0.4375]
[0.125 0.25 0.6875]
[0.375 0.5625 1. ]
[0.1875 0.3125 0.6875]
[0.1875 0.25 0.625 ]]]
[[[0. 0. 0. ]
[0. 0. 0. ]
[0. 0. 0. ]
[0. 0. 0. ]
[0. 0. 0. ]]
[[0. 0. 0. ]
[0. 0. 0. ]
[0. 0. 0. ]
[0. 0. 0. ]
[0. 0. 0. ]]
[[0. 0. 0.3125]
[0.125 0.25 0.5625]
[0.125 0.1875 0.5 ]
[0.1875 0.3125 0.5625]
[0.1875 0.25 0.5 ]]]]
CodePudding user response:
With a 3d array, there are various ways of doing a matrix product:
In [365]: Z.shape
Out[365]: (5, 5, 3)
You say np.dot
isn't valid, but don't show why:
In [366]: np.dot(Z,Z).shape
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
Input In [366], in <cell line: 1>()
----> 1 np.dot(Z,Z).shape
File <__array_function__ internals>:5, in dot(*args, **kwargs)
ValueError: shapes (5,5,3) and (5,5,3) not aligned: 3 (dim 2) != 5 (dim 1)
If you take time to read the np.dot
docs, you see that it trys to do the sum-of-products on the last axis of A and 2nd to the last of B, hence the mismatch between 3 and 5.
You'd get the same error is Z
was (5,3) shape.
One way around this is to change the 2nd Z
to (5,3,5) shape:
In [367]: np.dot(Z,Z.transpose(0,2,1)).shape
Out[367]: (5, 5, 5, 5)
But dot
does a kind of outer-product on the leading dimensions.
I think the tensordot
in the other answer does this as well. tensordot
just reduces the calculation down to a dot
call, with some reshape and transposes.
matmul/@
treats the leading dimensions as 'batch' and applies normal broadcasting
rules:
In [368]: np.matmul(Z,Z.transpose(0,2,1)).shape
Out[368]: (5, 5, 5)
With einsum
we can specify other combinations of axes.
In [369]: np.einsum('ijk,ijk->ij',Z,Z).shape
Out[369]: (5, 5)
In [370]: np.einsum('ijk,ijk->ik',Z,Z).shape
Out[370]: (5, 3)
In [371]: np.einsum('ijk,ilk->ijl',Z,Z).shape
Out[371]: (5, 5, 5)
In [373]: np.einsum('ijk,ijl->ijl',Z,Z).shape
Out[373]: (5, 5, 3)
In [374]: np.einsum('ijk,jlk->ilk',Z,Z).shape
Out[374]: (5, 5, 3)