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
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
其他回答
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
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
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]
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;
round(x[, n]):数值四舍五入到10的负n次方的最接近倍数。所以如果n是负的…
def round5(x):
return int(round(x*2, -1)) / 2
由于10 = 5 * 2,您可以对2使用整数除法和乘法,而不是对5.0使用浮点除法和乘法。这并不重要,除非你喜欢位移位
def round5(x):
return int(round(x << 1, -1)) >> 1