Python3.7的区别数学.余数and%(模运算符)

2024-05-03 08:41:17 发布

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来自What’s New In Python 3.7 我们可以看到新的^{}。上面写着

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

remainder of x / y

我们还可以看到,操作员注意:

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

如果可能的话,我还没有尝试运行Python3.7。在

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,我们将得到nan和{},而不是{},就像它在文档中所说的那样。在

所以问题是%和{}之间有什么区别?math.remainder是否也可以处理复数(%缺少它)?主要优势是什么?在

这是来自官方cpythongithub回购的source of ^{}。在


Tags: andofthetoforismathinteger
2条回答

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,这是m = x - n*y,其中nfloor(x/y),所以剩下的0 <= m < y而不是{}。在

所以

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

多亏了@MaartenFabré,我没有注意到细节:

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

不同之处在于:

零作为除数:

^{pr2}$

基本数,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

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