广播np.dot vs tf.matmul以进行张量矩阵乘法(形状必须为2级，但为3级错误) [英] Broadcasting np.dot vs tf.matmul for tensor-matrix multiplication (Shape must be rank 2 but is rank 3 error)

问题描述

假设我有以下张量:

X = np.zeros((3,201, 340))
Y = np.zeros((340, 28))

使用numpy成功制作X，Y的点积，并产生形状为(3，201，28)的张量. 但是使用tensorflow时出现以下错误:Shape must be rank 2 but is rank 3 error ...

Making a dot product of X, Y is successful with numpy, and yields a tensor of shape (3, 201, 28). However with tensorflow I get the following error: Shape must be rank 2 but is rank 3 error ...

最小代码示例:

X = np.zeros((3,201, 340))
Y = np.zeros((340, 28))
print(np.dot(X,Y).shape) # successful (3, 201, 28)
tf.matmul(X, Y) # errornous

有人知道如何通过张量流实现相同的结果吗?

Any idea how to achieve the same result with tensorflow?

推荐答案

由于您正在使用tensors，因此(出于性能考虑)在其中使用tensordot比使用np.dot更好. NumPy允许它(numpy.dot)通过降低性能在tensors上工作，而tensorflow似乎根本不允许它.

Since, you are working with tensors, it would be better (for performance) to use tensordot there than np.dot. NumPy allows it (numpy.dot) to work on tensors through lowered performance and it seems tensorflow simply doesn't allow it.

因此，对于NumPy，我们将使用 np.tensordot -

So, for NumPy, we would use np.tensordot -

np.tensordot(X, Y, axes=((2,),(0,)))

对于tensorflow，将使用 tf.tensordot -

For tensorflow, it would be with tf.tensordot -

tf.tensordot(X, Y, axes=((2,),(0,)))

相关帖子，以了解tensordot .

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