（m，）向量与（m，n）矩阵相乘时的点行为

Question

我已经用Python和numpy工作了几个星期。直到今天我才意识到这一点

a = np.array([1,2,3])
b = np.array([[1,2], [3,4], [5,6]])

这两个计算给出了相同的结果

a @ b
b.T @ a

尽管第一个在代数中没有意义（关于维度）

所以我的问题是，.dot的算法在第一次计算中是如何工作的？或者NoMPy如何考虑1-D和N-D数组？

Answer 1

a = np.array([1,2,3])
b = np.array([[1,2], [3,4], [5,6]])

对于1d和2d数组，dot和matmul做同样的事情，尽管文档的措辞有点不同

来自dot的两个病例：

- If `a` is an N-D array and `b` is a 1-D array, it is a sum product over
  the last axis of `a` and `b`.

- If `a` is an N-D array and `b` is an M-D array (where ``M>=2``), it is a
  sum product over the last axis of `a` and the second-to-last axis of `b`::

你的a是（3，），而b是（3,2）：

In [263]: np.dot(b.T,a)
Out[263]: array([22, 28])

这首先适用于（2,3）和（3，）->；共享大小3维上的和积

In [264]: np.dot(a,b)
Out[264]: array([22, 28])

第二种情况适用，a（3，）和a（3,2）——>；最后一个（3，）和第二个（3,2）和最后一个（3,2）的和积，同样是共享的3

“A的最后一个，B的第二个到最后一个”是基本的矩阵乘法规则。In只需要在B为1d时进行调整，并且没有倒数第二个

matmul规则的表述方式是添加维度，然后删除维度

- If the first argument is 1-D, it is promoted to a matrix by
  prepending a 1 to its dimensions. After matrix multiplication
  the prepended 1 is removed.
- If the second argument is 1-D, it is promoted to a matrix by
  appending a 1 to its dimensions. After matrix multiplication
  the appended 1 is removed.

（3，）与（3,2）=>；（1,3）与（3,2）=>；（1,2）=>；（2，）

（2,3）与（3，）=>；（2,3）与（3,1）=>；（2,1）=>；（2，）

Answer 2

1. 你可能不是在问np.dot，它有不同的广播规则

2. 因为您的两个示例都涉及@操作符，即np.matmul的语法糖，所以我将用np.matmul来回答您的问题

答案很简单，只要引用the documentation of ^{}

The behavior depends on the arguments in the following way.

• If both arguments are 2-D they are multiplied like conventional matrices.
• If either argument is N-D, N > 2, it is treated as a stack of matrices residing in the last two indexes and broadcast accordingly.
• If the first argument is 1-D, it is promoted to a matrix by prepending a 1 to its dimensions. After matrix multiplication the prepended 1 is removed.
• If the second argument is 1-D, it is promoted to a matrix by appending a 1 to its dimensions. After matrix multiplication the appended 1 is removed.

（重点是我的）