脾气暴躁的地方和除以零 [英] Numpy where and division by zero

问题描述

我需要通过以下方式(旧版代码)计算x:

I need to compute x in the following way (legacy code):

x = numpy.where(b == 0, a, 1/b) 

我想它可以在 python-2.x (因为它在 python-2.7 代码)，但在 python-3.x (if b = 0它返回错误).

I suppose it worked in python-2.x (as it was in a python-2.7 code), but it does not work in python-3.x (if b = 0 it returns an error).

我如何使其在python-3.x中工作?

How do I make it work in python-3.x?

错误消息(Python 3.6.3):

error message (Python 3.6.3):

ZeroDivisionError: division by zero

推荐答案

The numpy.where documentation states:

如果给出了x和y并且输入数组为一维，则where为 等同于::

If x and y are given and input arrays are 1-D, where is equivalent to::

    [xv if c else yv for (c,xv,yv) in zip(condition,x,y)]

那么为什么您会看到错误?举个简单的例子:

So why do you see the error? Take this trivial example:

c = 0
result = (1 if c==0 else 1/c)
# 1

到目前为止，一切都很好.首先检查if c==0，结果为1.该代码不会尝试评估1/c.这是因为Python解释器处理 lazy 三元运算符，因此仅求值适当的表达式.

So far so good. if c==0 is checked first and the result is 1. The code does not attempt to evaluate 1/c. This is because the Python interpreter processes a lazy ternary operator and so only evaluates the appropriate expression.

现在让我们将其翻译为numpy.where方法:

Now let's translate this into numpy.where approach:

c = 0
result = (xv if c else yv for (c, xv, yv) in zip([c==0], [1], [1/c]))
# ZeroDivisionError

在应用逻辑之前，在评估zip([c==0], [1], [1/c])时发​​生错误.生成器表达式本身无法求值.作为功​​能，numpy.where不能也确实不能复制Python三元表达式的惰性计算.

The error occurs in evaluating zip([c==0], [1], [1/c]) before even the logic is applied. The generator expression itself can't be evaluated. As a function, numpy.where does not, and indeed cannot, replicate the lazy computation of Python's ternary expression.

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