Python：十进制数字的范围函数 [英] Python: range function for decimal numbers

问题描述

例如，对于浮点数
，Python中是否有任何range（）函数

Is there any range() function in python for float numbers for example

a=0.6

if a in range(0,1):
    a=3

我该如何实现呢？

推荐答案

如果我没看错，你想测试一个数字是否在另外两个之间数字，因此使用：

If I'm reading correctly, you want to test if a number is between two other numbers, so use:

a = 0.6
if 0 <= a < 1: # change to `<= 1` to be inclusive
   a = 3

您无需生成范围并进行成员资格测试-除非您有 a 应该匹配的离散值集-内置的 range < Python 3.x中的/ code>可以对 int s执行高效的查找，因为它可以优化成员资格测试。如果您在较大范围内有大量离散值，则最好还是在数学上进行计算。

You don't need to generate a range and do membership testing - unless you have a discrete set of values that your a should match - the builtin range in Python 3.x can do efficient lookups for ints as it can optimise membership testing. If you have a large amount of discrete values in a large range, then you'd be better of doing it mathematically anyway.

这篇关于Python：十进制数字的范围函数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！