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
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
def round_up_to_base(x, base=10):
return x + (base - x) % base
def round_down_to_base(x, base=10):
return x - (x % base)
这给了
基础= 5:
>>> [i for i in range(20)]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
>>> [round_down_to_base(x=i, base=5) for i in range(20)]
[0, 0, 0, 0, 0, 5, 5, 5, 5, 5, 10, 10, 10, 10, 10, 15, 15, 15, 15, 15]
>>> [round_up_to_base(x=i, base=5) for i in range(20)]
[0, 5, 5, 5, 5, 5, 10, 10, 10, 10, 10, 15, 15, 15, 15, 15, 20, 20, 20, 20]
基础= 10:
>>> [i for i in range(20)]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
>>> [round_down_to_base(x=i, base=10) for i in range(20)]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10]
>>> [round_up_to_base(x=i, base=10) for i in range(20)]
[0, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 20, 20, 20, 20, 20, 20, 20, 20, 20]
在Python 3.7.9中测试
其他回答
那么这个呢:
def divround(value, step):
return divmod(value, step)[0] * step
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);
}
}
这只是一个比例的问题
>>> a=[10,11,12,13,14,15,16,17,18,19,20]
>>> for b in a:
... int(round(b/5.0)*5.0)
...
10
10
10
15
15
15
15
15
20
20
20
舍入到非整数值,例如0.05:
def myround(x, prec=2, base=.05):
return round(base * round(float(x)/base),prec)
我发现这很有用,因为我只需要在代码中进行搜索和替换,就可以将“round(”更改为“myround(”,而不必更改参数值。
def round_to_next5(n):
return n + (5 - n) % 5