Python:绕过零除 [英] python: getting around division by zero

问题描述

我有一个很大的浮点数数据集.我遍历它们并为每个评估np.log(x). 我知道了

I have a big data set of floating point numbers. I iterate through them and evaluate np.log(x) for each of them. I get

RuntimeWarning: divide by zero encountered in log

我想解决这个问题，如果发生此错误，则返回0.

I would like to get around this and return 0 if this error occurs.

我正在考虑定义一个新功能:

I am thinking of defining a new function:

def safe_ln(x):
    #returns: ln(x) but replaces -inf with 0
    l = np.log(x)
    #if l = -inf:
    l = 0
    return l

基本上，我需要一种测试输出是否为-inf的方法，但我不知道如何进行. 谢谢您的帮助！

Basically,I need a way of testing that the output is -inf but I don't know how to proceed. Thank you for your help!

推荐答案

由于x=0的log为负无穷大，因此我只需要检查输入值是否为零并返回您想要的值即可:

Since the log for x=0 is minus infinite, I'd simply check if the input value is zero and return whatever you want there:

def safe_ln(x):
    if x <= 0:
        return 0
    return math.log(x)

编辑:小修改:您应检查所有小于或等于0的值.

EDIT: small edit: you should check for all values smaller than or equal to 0.

编辑2 :np.log当然是在numpy数组上计算的函数，对于单个值，应使用math.log.这就是上面的函数与numpy的外观:

EDIT 2: np.log is of course a function to calculate on a numpy array, for single values you should use math.log. This is how the above function looks with numpy:

def safe_ln(x, minval=0.0000000001):
    return np.log(x.clip(min=minval))

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