在 Python 中将字符串转换为带小数的整数

问题描述

我在 Python 中有一个格式为nn.nnnnn"的字符串，我想将其转换为整数.

I have a string in the format: 'nn.nnnnn' in Python, and I'd like to convert it to an integer.

直接转换失败:

>>> s = '23.45678'
>>> i = int(s)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: invalid literal for int() with base 10: '23.45678'

我可以使用以下方法将其转换为小数:

I can convert it to a decimal by using:

>>> from decimal import *
>>> d = Decimal(s)
>>> print d
23.45678

我也可以在."上进行拆分，然后从零中减去小数点，然后将其添加到整数中......呸.

I could also split on '.', then subtract the decimal from zero, then add that to the whole number ... yuck.

但我更喜欢将它作为 int 来使用，而不需要进行不必要的类型转换或操作.

But I'd prefer to have it as an int, without unnecessary type conversions or maneuvering.

推荐答案

这个怎么样?

>>> s = '23.45678'
>>> int(float(s))
23

或者……

>>> int(Decimal(s))
23

或者……

>>> int(s.split('.')[0])
23

恐怕它会变得比这简单得多.接受它并继续前进.

I doubt it's going to get much simpler than that, I'm afraid. Just accept it and move on.

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