在numpy中重新定义* =运算符 [英] Redefine *= operator in numpy

问题描述

如此处提到和

As mentioned here and here, this doesn't work anymore in numpy 1.7+ :

import numpy
A = numpy.array([1, 2, 3, 4], dtype=numpy.int16)
B = numpy.array([0.5, 2.1, 3, 4], dtype=numpy.float64)
A *= B

一种解决方法是:

def mult(a,b):
    numpy.multiply(a, b, out=a, casting="unsafe")

def add(a,b):
    numpy.add(a, b, out=a, casting="unsafe")

mult(A,B)

但这对于每个矩阵运算来说太长了！

but that's way too long to write for each matrix operation!

默认情况下如何覆盖numpy *=运算符以执行此操作?

How can override the numpy *= operator to do this by default?

我应该继承某些东西吗?

Should I subclass something?

推荐答案

您可以使用np.set_numeric_ops覆盖数组算术方法:

You can use np.set_numeric_ops to override array arithmetic methods:

import numpy as np

def unsafe_multiply(a, b, out=None):
    return np.multiply(a, b, out=out, casting="unsafe")

np.set_numeric_ops(multiply=unsafe_multiply)

A = np.array([1, 2, 3, 4], dtype=np.int16)
B = np.array([0.5, 2.1, 3, 4], dtype=np.float64)
A *= B

print(repr(A))
# array([ 0,  4,  9, 16], dtype=int16)

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