的值加上0.5，可以“欺骗”int()使其舍入而不是舍入 传递给int()的数字。

Use:

>>> def round_to_nearest(n, m):
        r = n % m
        return n + m - r if r + r >= m else n - r

它不使用乘法，也不会从/转换为浮点数。

四舍五入到最接近10的倍数:

>>> for n in range(-21, 30, 3): print('{:3d}  =>  {:3d}'.format(n, round_to_nearest(n, 10)))
-21  =>  -20
-18  =>  -20
-15  =>  -10
-12  =>  -10
 -9  =>  -10
 -6  =>  -10
 -3  =>    0
  0  =>    0
  3  =>    0
  6  =>   10
  9  =>   10
 12  =>   10
 15  =>   20
 18  =>   20
 21  =>   20
 24  =>   20
 27  =>   30

如你所见，它对负数和正数都适用。平局(例如-15和15)总是向上四舍五入。

一个类似的例子，四舍五入到5的最接近倍数，证明它也表现为不同的“基数”:

>>> for n in range(-21, 30, 3): print('{:3d}  =>  {:3d}'.format(n, round_to_nearest(n, 5)))
-21  =>  -20
-18  =>  -20
-15  =>  -15
-12  =>  -10
 -9  =>  -10
 -6  =>   -5
 -3  =>   -5
  0  =>    0
  3  =>    5
  6  =>    5
  9  =>   10
 12  =>   10
 15  =>   15
 18  =>   20
 21  =>   20
 24  =>   25
 27  =>   25

去掉“rest”会起作用:

rounded = int(val) - int(val) % 5

如果该值是一个整数:

rounded = val - val % 5

作为函数:

def roundint(value, base=5):
    return int(value) - int(value) % int(base)

def round_to_next5(n):
    return n + (5 - n) % 5

对于整数和Python 3:

def divround_down(value, step):
    return value//step*step


def divround_up(value, step):
    return (value+step-1)//step*step

生产:

>>> [divround_down(x,5) for x in range(20)]
[0, 0, 0, 0, 0, 5, 5, 5, 5, 5, 10, 10, 10, 10, 10, 15, 15, 15, 15, 15]
>>> [divround_up(x,5) for x in range(20)]
[0, 5, 5, 5, 5, 5, 10, 10, 10, 10, 10, 15, 15, 15, 15, 15, 20, 20, 20, 20]

对已接受答案的补充，用于指定四舍五入到最接近的5或其他值

import math

def my_round(x, base, down = True):
    return base * math.floor(x/base) + (not down) * base