是否有一个内置函数可以像下面这样舍入?

10 -> 10
12 -> 10
13 -> 15
14 -> 15
16 -> 15
18 -> 20

当前回答

这是我的C代码。如果我理解正确的话,应该是这样的;

#include <stdio.h>

int main(){
int number;

printf("Enter number: \n");
scanf("%d" , &number);

if(number%5 == 0)
    printf("It is multiple of 5\n");
else{
    while(number%5 != 0)
        number++;
  printf("%d\n",number);
  }
}

这也是四舍五入到最接近5的倍数而不是四舍五入;

#include <stdio.h>

int main(){
int number;

printf("Enter number: \n");
scanf("%d" , &number);

if(number%5 == 0)
    printf("It is multiple of 5\n");
else{
    while(number%5 != 0)
        if (number%5 < 3)
            number--;
        else
        number++;
  printf("nearest multiple of 5 is: %d\n",number);
  }
}

其他回答

另一种方法(不需要显式的乘法或除法运算符):

def rnd(x, b=5):
    return round(x + min(-(x % b), b - (x % b), key=abs))
def round_to_next5(n):
    return n + (5 - n) % 5

如果有人需要“财务四舍五入”(0.5位总是向上):

def myround(x, base=5):
    roundcontext = decimal.Context(rounding=decimal.ROUND_HALF_UP)
    decimal.setcontext(roundcontext)
    return int(base *float(decimal.Decimal(x/base).quantize(decimal.Decimal('0'))))

根据文档,其他舍入选项是:

ROUND_CEILING(朝向无限) ROUND_DOWN(趋近于零) ROUND_FLOOR(朝向-∞) ROUND_HALF_DOWN(当平局趋于0时最接近) ROUND_HALF_EVEN(最接近偶数) ROUND_HALF_UP(到最接近的平局从0开始) ROUND_UP(远离零) ROUND_05UP(如果四舍五入后的最后一位为0或5,则远离0;否则趋于零)

默认情况下,Python使用ROUND_HALF_EVEN,因为它有一些统计优势(四舍五入的结果没有偏见)。

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

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]