Python中的’@ =’符号是什么?

问题:Python中的’@ =’符号是什么?

我知道@是给装饰器用的,但是@=Python有什么用呢?只是保留一些未来的想法吗?

这只是我阅读时遇到的许多问题之一tokenizer.py

I know @ is for decorators, but what is @= for in Python? Is it just reservation for some future idea?

This is just one of my many questions while reading tokenizer.py.


回答 0

文档

@(在)操作者意图被用于矩阵乘法。没有内置的Python类型实现此运算符。

@运算符是在Python 3.5中引入的。@=正如您所期望的那样,是矩阵乘法,后跟赋值。它们映射到__matmul____rmatmul____imatmul__类似于如何++=映射__add____radd____iadd__

PEP 465中详细讨论了操作员及其背后的原理。

From the documentation:

The @ (at) operator is intended to be used for matrix multiplication. No builtin Python types implement this operator.

The @ operator was introduced in Python 3.5. @= is matrix multiplication followed by assignment, as you would expect. They map to __matmul__, __rmatmul__ or __imatmul__ similar to how + and += map to __add__, __radd__ or __iadd__.

The operator and the rationale behind it are discussed in detail in PEP 465.


回答 1

@=@是Python 3.5中引入的用于执行矩阵乘法的新运算符。它们的目的是澄清迄今为止​​与运算符之间存在的混淆,该运算符*根据该特定库/代码中采用的约定用于元素方式乘法或矩阵乘法。结果,将来,运营商*只能用于按元素乘法。

PEP0465中所述,引入了两个运算符:

  • 一个新的二进制运算符A @ B,与A * B
  • 就地版本A @= B,与A *= B

矩阵乘法与按元素乘法

为了快速突出区别,对于两个矩阵:

A = [[1, 2],    B = [[11, 12],
     [3, 4]]         [13, 14]]
  • 逐元素乘法将生成:

    A * B = [[1 * 11,   2 * 12], 
             [3 * 13,   4 * 14]]
  • 矩阵乘法将生成:

    A @ B  =  [[1 * 11 + 2 * 13,   1 * 12 + 2 * 14],
               [3 * 11 + 4 * 13,   3 * 12 + 4 * 14]]

在Numpy中使用

到目前为止,Numpy使用以下约定:

@运算符的引入使涉及矩阵乘法的代码更易于阅读。PEP0465举了一个例子:

# Current implementation of matrix multiplications using dot function
S = np.dot((np.dot(H, beta) - r).T,
            np.dot(inv(np.dot(np.dot(H, V), H.T)), np.dot(H, beta) - r))

# Current implementation of matrix multiplications using dot method
S = (H.dot(beta) - r).T.dot(inv(H.dot(V).dot(H.T))).dot(H.dot(beta) - r)

# Using the @ operator instead
S = (H @ beta - r).T @ inv(H @ V @ H.T) @ (H @ beta - r)

显然,最后一种实现更易于阅读和解释为等式。

@= and @ are new operators introduced in Python 3.5 performing matrix multiplication. They are meant to clarify the confusion which existed so far with the operator * which was used either for element-wise multiplication or matrix multiplication depending on the convention employed in that particular library/code. As a result, in the future, the operator * is meant to be used for element-wise multiplication only.

As explained in PEP0465, two operators were introduced:

  • A new binary operator A @ B, used similarly as A * B
  • An in-place version A @= B, used similarly as A *= B

Matrix Multiplication vs Element-wise Multiplication

To quickly highlight the difference, for two matrices:

A = [[1, 2],    B = [[11, 12],
     [3, 4]]         [13, 14]]
  • Element-wise multiplication will yield:

    A * B = [[1 * 11,   2 * 12], 
             [3 * 13,   4 * 14]]
    
  • Matrix multiplication will yield:

    A @ B  =  [[1 * 11 + 2 * 13,   1 * 12 + 2 * 14],
               [3 * 11 + 4 * 13,   3 * 12 + 4 * 14]]
    

Usage in Numpy

So far, Numpy used the following convention:

Introduction of the @ operator makes the code involving matrix multiplications much easier to read. PEP0465 gives us an example:

# Current implementation of matrix multiplications using dot function
S = np.dot((np.dot(H, beta) - r).T,
            np.dot(inv(np.dot(np.dot(H, V), H.T)), np.dot(H, beta) - r))

# Current implementation of matrix multiplications using dot method
S = (H.dot(beta) - r).T.dot(inv(H.dot(V).dot(H.T))).dot(H.dot(beta) - r)

# Using the @ operator instead
S = (H @ beta - r).T @ inv(H @ V @ H.T) @ (H @ beta - r)

Clearly, the last implementation is much easier to read and interpret as an equation.


回答 2

@是Python3.5中新增的矩阵乘法运算符

参考:https : //docs.python.org/3/whatsnew/3.5.html#whatsnew-pep-465

C = A @ B

@ is the new operator for Matrix Multiplication added in Python3.5

Reference: https://docs.python.org/3/whatsnew/3.5.html#whatsnew-pep-465

Example

C = A @ B