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。

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

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

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

我需要四舍五入到前面的5。

举个例子，16发到15发或者19发到15发

下面是使用的代码

    def myround(x,segment):
        preRound = x / segment
        roundNum = int(preRound)
        segVal = segment * roundNum
        return segVal

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