脾气暴躁的“哪里"函数不能避免评估Sqrt(负) [英] Numpy "Where" function can not avoid evaluate Sqrt(negative)

问题描述

np.where函数似乎先评估所有可能的结果，然后再评估条件.这意味着，就我而言，即使以后不再使用，它​​也会求出-5，-4，-3，-2，-1的平方根.

It seems that the np.where function evaluates all the possible outcomes first, then it evaluates the condition later. This means that, in my case, it will evaluate square root of -5, -4, -3, -2, -1 even though it will not be used later on.

我的代码运行并运行.但是我的问题是警告.我避免使用循环来评估每个元素，因为它的运行速度比np.where慢得多.

My code runs and works. But my problem is the warning. I avoided using a loop to evaluate each element, because it will run much slower than np.where.

所以，我在这里问

1. 有什么办法让np.where首先评估条件?
2. 我可以仅关闭此特定警告吗?怎么样?
3. 如果有更好的建议，另一种更好的方法.

1. Is there any way to make np.where evaluate the condition first?
2. Can I turn off just this specific warning? How?
3. Another better way to do it if you have a better suggestion.

这里有一个简短的示例代码，与我的真实代码相对应，这是巨大的.但是本质上有同样的问题.

Here just a short example code corresponding my real code which is gigantic. But essentially has the same problem.

输入:

import numpy as np

c=np.arange(10)-5
d=np.where(c>=0, np.sqrt(c) ,c )

输出:

RuntimeWarning: invalid value encountered in sqrt
d=np.where(c>=0,np.sqrt(c),c)

推荐答案

有一种很多更好的方法.让我们看看您的代码在做什么，看看为什么.

There is a much better way of doing this. Let's take a look at what your code is doing to see why.

np.where接受三个数组作为输入.数组不支持惰性求值.

np.where accepts three arrays as inputs. Arrays do not support lazy evaluation.

d = np.where(c >= 0, np.sqrt(c), c)

因此，此行等效于

a = (c >= 0)
b = np.sqrt(c)
d = np.where(a, b, c)

请注意，在调用where之前，将立即计算输入.

Notice that the inputs are computed immediately, before where ever gets called.

幸运的是，您根本不需要使用where.相反，只需使用布尔掩码:

Luckily, you don't need to use where at all. Instead, just use a boolean mask:

mask = (c >= 0)
d = np.empty_like(c)
d[mask] = np.sqrt(c[mask])
d[~mask] = c[~mask]

如果您期望有很多负面因素，则可以复制所有元素，而不仅仅是负面元素:

If you expect a lot of negatives, you can copy all the elements instead of just the negative ones:

d = c.copy()
d[mask] = np.sqrt(c[mask])

一个更好的解决方案可能是使用掩码数组:

An even better solution might be to use masked arrays:

d = np.ma(c, c < 0)
d = np.ma.sqrt(d)

要访问整个数据数组，且未更改被遮罩的部分，请使用d.data.

To access the whole data array, with the masked portion unaltered, use d.data.

这篇关于脾气暴躁的“哪里"函数不能避免评估Sqrt(负)的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！