Python:最大/最小内置函数取决于参数顺序 [英] Python: max/min builtin functions depend on parameter order

问题描述

max(float('nan'), 1) 计算结果为 nan

max(1, float('nan')) 计算结果为 1

这是预期的行为吗?

感谢您的回答.

max 在可迭代对象为空时引发异常.当 nan 存在时，为什么 Python 的 max 不会引发异常?或者至少做一些有用的事情，比如返回 nan 或忽略 nan.目前的行为非常不安全，看起来完全不合理.

max raises an exception when the iterable is empty. Why wouldn't Python's max raise an exception when nan is present? Or at least do something useful, like return nan or ignore nan. The current behavior is very unsafe and seems completely unreasonable.

我发现了这种行为的更令人惊讶的后果，所以我刚刚发布了一个 相关问题.

I found an even more surprising consequence of this behavior, so I just posted a related question.

推荐答案

In [19]: 1>float('nan')
Out[19]: False

In [20]: float('nan')>1
Out[20]: False

浮点数nan既不大于也不小于整数1.max 从选择第一个元素开始，只有当它找到一个严格更大的元素时才替换它.

The float nan is neither bigger nor smaller than the integer 1. max starts by choosing the first element, and only replaces it when it finds an element which is strictly larger.

In [31]: max(1,float('nan'))
Out[31]: 1

由于nan不大于1，返回1.

Since nan is not larger than 1, 1 is returned.

In [32]: max(float('nan'),1)
Out[32]: nan

由于1不大于nan，所以返回nan.

Since 1 is not larger than nan, nan is returned.

附注.请注意，np.max 对 float('nan') 的处理方式不同:

PS. Note that np.max treats float('nan') differently:

In [36]: import numpy as np
In [91]: np.max([1,float('nan')])
Out[91]: nan

In [92]: np.max([float('nan'),1])
Out[92]: nan

但是如果你想忽略np.nans，你可以使用np.nanmax:

but if you wish to ignore np.nans, you can use np.nanmax:

In [93]: np.nanmax([1,float('nan')])
Out[93]: 1.0

In [94]: np.nanmax([float('nan'),1])
Out[94]: 1.0

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