在python中舍入浮点到负小数点

Question

我正在寻找在python中的1,2，-1小数点的集合和舍入的东西，我已经通过它编写了一些逻辑--我能够得到+ve点的正确值，但仍然可以向下查看-ve值。

我的代码为-：

import decimal

outcome = '2156.24611496733'
rounding="Alwayes Down" 
decimal_place = -1

#   for positive values
if rounding == "Alwayes Up":
    decimal.getcontext().rounding = decimal.ROUND_UP

elif rounding == "Alwayes Down":
    decimal.getcontext().rounding = decimal.ROUND_DOWN

#   for negative values
if rounding == "Alwayes Down":
    print(round(float(outcome), decimal_place))

rounding_string = "1." + "0"*int(decimal_place)
decimal_outcome = decimal.Decimal(str(outcome)).quantize(decimal.Decimal(rounding_string))

print(decimal_outcome)

表值就像-

​

我期待这样的价值观

​

​

Answer 1

这是符合我矩阵的完整解决方案。

import decimal

outcome = '622222.545454545'
rounding="Alwayes Up" 
decimal_place = -1

#   for positive values
if rounding == "Alwayes Up":
    decimal.getcontext().rounding = decimal.ROUND_UP

elif rounding == "Alwayes Down":
    decimal.getcontext().rounding = decimal.ROUND_DOWN
    
rounding_string = "1." + "0"*int(decimal_place)
decimal_outcome = decimal.Decimal(str(outcome)).quantize(decimal.Decimal(rounding_string))

print(decimal_outcome)

# for negative values
if rounding == "Alwayes Up":
    print(round(float(outcome), decimal_place))

elif rounding == "Alwayes Up":
    decimal.getcontext().rounding = decimal.ROUND_UP
    rounding_string = "1." + "0"*int(decimal_place)
    decimal_outcome = decimal.Decimal(str(outcome)).quantize(decimal.Decimal(rounding_string))
    print(float(decimal_outcome))