Python 3.7 math.remainder和％(模运算符)之间的区别

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从What’s New In Python 3.7起
 我们可以看到有新的math.remainder.它说

Return the IEEE 754-style remainder of x with respect to y. For finite x and finite nonzero y, this is the difference x - n*y, where n is the closest integer to the exact value of the quotient x / y. If x / y is exactly halfway between two consecutive integers, the nearest even integer is used for n. The remainder r = remainder(x, y) thus always satisfies abs(r) <= 0.5 * abs(y).

Special cases follow IEEE 754: in particular, remainder(x, math.inf) is x for any finite x, and remainder(x, 0) and remainder(math.inf, x) raise ValueError for any non-NaN x. If the result of the remainder operation is zero, that zero will have the same sign as x.

On platforms using IEEE 754 binary floating-point, the result of this operation is always exactly representable: no rounding error is introduced.

但我们也记得有%符号

remainder of x / y

我们还看到操作符有一条说明：

Not for complex numbers. Instead convert to floats using abs() if appropriate.

如果可能的话,我还没有尝试过运行Python 3.7.

但我试过了

Python 3.6.1 (v3.6.1:69c0db5050, Mar 21 2017, 01:21:04)
[GCC 4.2.1 (Apple Inc. build 5666) (dot 3)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> import math
>>> 100 % math.inf
100.0
>>> math.inf % 100
nan
>>> 100 % 0
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ZeroDivisionError: integer division or modulo by zero

所以不同的是,我们会在文档中得到ValueError,而不是nan和ZeroDivisionError.

那么问题是％和math.remainder之间有什么区别？ math.remainder是否也适用于复数(％缺少它)？主要优势是什么？

这是官方CPython github repo的source of math.remainder.

解决方法:

Return the IEEE 754-style remainder of x with respect to y. For finite
x and finite nonzero y, this is the difference x - n*y, where n is the
closest integer to the exact value of the quotient x / y. If x / y is
exactly halfway between two consecutive integers, the nearest even
integer is used for n. The remainder r = remainder(x, y) thus always
satisfies abs(r) <= 0.5 * abs(y).

对于模数,这是m = x-n * y,其中n是floor(x / y),所以0 <= m <y.对于余数,y代替abs(r)< *="" 0.5="" <="" =="" abs(y).="" p="" 所以="">

modulo(2.7, 1) = 0.7
remainder(2.7, 1) = -0.3

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