取整** 0.5和math.sqrt [英] Rounding ** 0.5 and math.sqrt

问题描述

在Python中，是

n**0.5  # or
math.sqrt(n) 

识别数字是一个完美的平方?具体来说，我应该担心使用

recognized when a number is a perfect square? Specifically, should I worry that when I use

int(n**0.5)  # instead of
int(n**0.5 + 0.000000001)

由于精度误差，我可能会意外地得出比实际平方根小1的数字吗?

I might accidentally end up with the number one less than the actual square root due to precision error?

推荐答案

是的，您应该担心:

In [11]: int((100000000000000000000000000000000000**2) ** 0.5)
Out[11]: 99999999999999996863366107917975552L

In [12]: int(math.sqrt(100000000000000000000000000000000000**2))
Out[12]: 99999999999999996863366107917975552L

显然添加0.000000001在这里也无济于事...

obviously adding the 0.000000001 doesn't help here either...

@DSM指出，您可以使用十进制库:

As @DSM points out, you can use the decimal library:

In [21]: from decimal import Decimal

In [22]: x = Decimal('100000000000000000000000000000000000')

In [23]: (x ** 2).sqrt() == x
Out[23]: True

对于超过10**999999999的数字，只要您检查精度(可配置)，它将抛出错误而不是错误的答案...

for numbers over 10**999999999, provided you keep a check on the precision (configurable), it'll throw an error rather than an incorrect answer...

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