Python 中真的有@运算符来计算点积吗？

Question

这个答案是否正确： ?

引用那个答案。

In Python 3.5, there is a new operator for the dot product, so you can write a= A @ B instead of a= numpy.dot(A,B)

它似乎不适合我。

$ python3
Python 3.6.1 (default, Apr  4 2017, 09:40:21) 
[GCC 4.2.1 Compatible Apple LLVM 8.1.0 (clang-802.0.38)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> a = [1, 2, 3]
>>> b = [4, 5, 6]
>>> a @ b
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for @: 'list' and 'list'
>>>

但是链接的答案已经获得了 6 个赞成票，所以我一定遗漏了什么。您能否提供一个完整的示例来说明如何使用 @ 运算符计算点积？

Answer 1

参见what's new in Python 3.5, section matrix mult (PEP 465)：

PEP 465 adds the @ infix operator for matrix multiplication. Currently, no builtin Python types implement the new operator, however, it can be implemented by defining __matmul__(), __rmatmul__(), and __imatmul__() for regular, reflected, and in-place matrix multiplication. The semantics of these methods is similar to that of methods defining other infix arithmetic operators.

因此，您必须自己实施这些方法。

或者，使用已经支持新运算符的numpy>=1.10：

>>> import numpy
>>> x = numpy.ones(3)
>>> m = numpy.eye(3)
>>> x @ m
array([ 1., 1., 1.])

Python 中真的有@运算符来计算点积吗？

Is there really an @ operator in Python to calculate dot product?

python

operators