將 3d 矩陣和 3d 矩陣相乘

Question

我正在嘗試將 3d 矩陣和 3d 矩陣相乘，我的矩陣如下：

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))

形狀是 (5,5,3)，我想像np.dot(Z,Z)那樣做乘法，但它不能在 3d 矩陣中工作。

我見過使用np.tensordot(Z,Z,axes=?) ，但我不知道如何設置軸。

Answer 1

我建議您查看tensordot()函數的文檔，以真正了解它對矩陣的作用：

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)

輸出：

[[[[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   ]]]]

Answer 2

對於 3d 數組，有多種方法可以進行矩陣乘積：

In [365]: Z.shape
Out[365]: (5, 5, 3)

您說np.dot無效，但不說明原因：

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)

如果您花時間閱讀np.dot文檔，您會發現它試圖在 A 的最后一個軸和 B 的最后一個軸上進行乘積之和，因此 3 和 5 之間的不匹配。

你會得到同樣的錯誤是Z是 (5,3) 形狀。

解決此問題的一種方法是將第二個Z更改為 (5,3,5) 形狀：

In [367]: np.dot(Z,Z.transpose(0,2,1)).shape
Out[367]: (5, 5, 5, 5)

但是dot在領先的維度上做了一種外積。

我認為另一個tensordot中的張量點也可以做到這一點。 tensordot只是將計算簡化為一個dot調用，並進行了一些整形和轉置。

matmul/@將前導維度視為“批處理”並應用正常的broadcasting規則：

In [368]: np.matmul(Z,Z.transpose(0,2,1)).shape
Out[368]: (5, 5, 5)

使用einsum ，我們可以指定其他軸組合。

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)