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
我想还没有人写过这个,但你可以这样做:
round(12, -1) --> 10
round(18, -1) --> 20
其他回答
divround的修改版本:-)
def divround(value, step, barrage):
result, rest = divmod(value, step)
return result*step if rest < barrage else (result+1)*step
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
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))
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);
}
}
下一个5的倍数
考虑51需要转换为55:
code here
mark = 51;
r = 100 - mark;
a = r%5;
new_mark = mark + a;