Python 中的模运算符

Modulo operator in Python

提问by KodeWarrior

What does modulo in the following piece of code do?

下面这段代码中的模有什么作用？

from math import *
3.14 % 2 * pi

How do we calculate modulo on a floating point number?

我们如何计算浮点数的模数？

回答by Blender

When you have the expression:

当你有表达式时：

a % b = c

It really means there exists an integer nthat makes cas small as possible, but non-negative.

这实际上意味着存在一个正整数n，使c尽可能小，但非负。

a - n*b = c

By hand, you can just subtract 2(or add 2if your number is negative) over and over until the end result is the smallest positive number possible:

手动，您可以一遍又一遍地减去2（2如果您的数字为负，则添加），直到最终结果是可能的最小正数：

  3.14 % 2
= 3.14 - 1 * 2
= 1.14

Also, 3.14 % 2 * piis interpreted as (3.14 % 2) * pi. I'm not sure if you meant to write 3.14 % (2 * pi)(in either case, the algorithm is the same. Just subtract/add until the number is as small as possible).

此外，3.14 % 2 * pi被解释为(3.14 % 2) * pi。我不确定您是否打算编写3.14 % (2 * pi)（在任何一种情况下，算法都是相同的。只需减/加，直到数字尽可能小）。

回答by Joran Beasley

same as a normal modulo 3.14 % 6.28 = 3.14, just like 3.14%4 =3.143.14%2 = 1.14(the remainder...)

与普通 modulo 相同 3.14 % 6.28 = 3.14，就像3.14%4 =3.143.14%2 = 1.14（余数...）

回答by Thomas

In addition to the other answers, the fmoddocumentationhas some interesting things to say on the subject:

除了其他答案之外，fmod文档中还有一些关于这个主题的有趣内容：

math.fmod(x, y)

Return fmod(x, y), as defined by the platform C library. Note that the Python expression x % ymay not return the same result. The intent of the C standard is that fmod(x, y)be exactly (mathematically; to infinite precision) equal to x - n*yfor some integer n such that the result has the same sign as xand magnitude less than abs(y). Python's x % yreturns a result with the sign of yinstead, and may not be exactly computable for float arguments. For example, fmod(-1e-100, 1e100)is -1e-100, but the result of Python's -1e-100 % 1e100is 1e100-1e-100, which cannot be represented exactly as a float, and rounds to the surprising 1e100. For this reason, function fmod()is generally preferred when working with floats, while Python's x % yis preferred when working with integers.

math.fmod(x, y)

返回fmod(x, y)，由平台 C 库定义。请注意，Python 表达式x % y可能不会返回相同的结果。C 标准的意图是对于某个整数 nfmod(x, y)精确地（数学上；无限精度）等于x - n*y，使得结果具有相同的符号x和小于 的幅度abs(y)。Pythonx % y返回的结果带有y相反的符号，并且对于浮点参数可能无法完全计算。例如，fmod(-1e-100, 1e100)is -1e-100，但 Python 的-1e-100 % 1e100is的结果 1e100-1e-100不能完全表示为浮点数，而是四舍五入到令人惊讶的1e100。出于这个原因，函数fmod()在使用浮点数时通常是首选，而 Python 的x % y处理整数时首选。

回答by Xorlev

Same thing you'd expect from normal modulo .. e.g. 7 % 4 = 3, 7.3 % 4.0 = 3.3

您对正常模数的期望相同.. 例如7 % 4 = 3，7.3 % 4.0 = 3.3

Beware of floating point accuracy issues.

注意浮点精度问题。

回答by praveen kansara

you should use fmod(a,b)

你应该使用 fmod(a,b)

While abs(x%y) < abs(y) is truemathematically, for floatsit may not be true numerically due to roundoff.

While abs(x%y) < abs(y) is true在数学上，由于 ，floats它在数值上可能不是真的roundoff。

For example, and assuming a platform on which a Python floatis an IEEE 754double-precision number, in order that -1e-100 % 1e100have the same sign as 1e100, the computed result is -1e-100 + 1e100, which is numerically exactly equal to 1e100.

例如，假设一个平台上的 aPython float是一个IEEE 754双精度数，为了与-1e-100 % 1e100具有相同的符号1e100，计算结果是-1e-100 + 1e100，在数值上完全等于1e100。

Function fmod()in the math module returns a result whose sign matches the sign of the first argument instead, and so returns -1e-100in this case. Which approach is more appropriate depends on the application.

fmod()math 模块中的函数返回一个结果，其符号与第一个参数的符号匹配，因此-1e-100在这种情况下返回。哪种方法更合适取决于应用程序。

where x = a%bis used for integer modulo

where x = a%b用于整数模