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)。

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

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。

去掉“rest”会起作用:

rounded = int(val) - int(val) % 5

如果该值是一个整数:

rounded = val - val % 5

作为函数:

def roundint(value, base=5):
    return int(value) - int(value) % int(base)

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))

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