这是Python中余数的整数版本，它的结果应该与C的“%”操作符相同:

def remainder(n, d):
    return (-1 if n < 0 else 1) * (abs(n) % abs(d))

预期结果:

remainder(123, 10)   ==  3
remainder(123, -10)  ==  3
remainder(-123, 10)  == -3
remainder(-123, -10) == -3

从Python 3.7开始，有一个新的math.remainder()函数:

from math import remainder
print(remainder(26,7))

输出:

-2.0  # not 5

注意，如上所述，它与%不同。

引用文档:

math.remainder(x, y) 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.

Issue29962描述了创建新函数的基本原理。

你正在寻找模运算符:

a % b

例如:

>>> 26 % 7
5

当然，也许他们希望您自己实现它，这也不会太难。

如果你想求除法问题的余数，就用实际的余数法则，就像数学一样。当然，这不会给你一个十进制输出。

valone = 8
valtwo = 3
x = valone / valtwo
r = valone - (valtwo * x)
print "Answer: %s with a remainder of %s" % (x, r)

如果你想用计算器的形式，只要代入valone = 8 用valone = int(输入(“值一”))。对valtwo = 3做同样的处理，但显然变量不同。

除法的余数可以使用运算符%来发现:

>>> 26%7
5

如果你同时需要商和模，有一个内置的divmod函数:

>>> seconds= 137
>>> minutes, seconds= divmod(seconds, 60)

26 % 7(你将得到余数)

26 / 7(你会得到除数，可以是浮点值)

26 // 7(你将得到除数，只有整数值)