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

Question

来自 What’s New In Python 3.7 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.

但是我们还记得有一个%符号是

remainder of x / y

我们还看到有一条给操作员的注释：

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

我还没试过 运行 Python 3.7 如果可能的话。

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

所以区别在于，我们会得到 ValueError 而不是 nan 和 ZeroDivisionError，正如文档中所说的那样。

所以问题是%和math.remainder有什么区别？ math.remainder 是否也适用于复数（% 缺少复数）？主要优势是什么？

这是来自官方 CPython github repo 的 source of math.remainder。

Answer 1

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)，因此对于余数，0 <= m < y 而不是 abs(r) <= 0.5 * abs(y)。

所以

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

Answer 2

感谢@MaartenFabré，我没有注意细节：

math.remainder() is the difference x - n*y, where n is the closest integer to the exact value of the quotient x / y

我构建了 Python 3.7:

Python 3.7.0a0 (heads/master:f34c685020, May  8 2017, 15:35:30)
[GCC 4.2.1 Compatible Apple LLVM 8.0.0 (clang-800.0.42.1)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> import math

区别如下：

零作为除数：

>>> math.remainder(1, 0)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: math domain error
>>> 1 % 0
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ZeroDivisionError: integer division or modulo by zero

基本数字，其中math.remainder(x, y) < x % y

>>> math.remainder(5, 3)
-1.0
>>> 5 % 3
2

复数：

>>> math.remainder(3j + 2, 4)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: can't convert complex to float
>>> (3j + 2) % 4
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: can't mod complex numbers.

无穷大(math.inf)

>>> math.remainder(3, math.inf)
3.0
>>> 3 % math.inf
3.0
>>> math.remainder(math.inf, 3)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: math domain error
>>> math.inf % 3
nan