基于python中所需精度的unround float

当我在写这个问题时 - 我已经设法使用一些定制的函数来解决它。也许它会对某人有用。

解决方案

def split_value_into_range(number,needed_precision,current_precision = -1):
    # you can decide if you want to pass the current precision from outside or not..
    # probably doesn't really matter perfoRMANce wise
    if current_precision == -1:
        # convert to String and remove Trailing zeros from float
        numberStr = f'{number:g}'

        # get current precision
        scale = numberStr[::-1].find('.')
        current_precision = scale

    # get precision difference
    precision_diff = needed_precision - current_precision

    if precision_diff < 1:
        raise Exception("Operation not supported. NOTE: modify the function to suit your needs")

    # get number of 'digits' to substract and to add
    partial_substractor = 5 * (10 ** (precision_diff - 1))
    partial_adder = partial_substractor - 1

    # count how many `digits` are in range
    diff = partial_substractor + partial_adder

    denominator = 10 ** needed_precision
    substractor = partial_substractor / denominator
    adder = partial_adder / denominator

    min_range = round(number - substractor,needed_precision)
    max_range = round(number + adder,needed_precision)

    return min_range,max_range,diff

number = 5.86

min,max,diff = split_value_into_range(number,3,2)
print(f"min={min},max={max},diff={diff}")

min,4)
print(f"min={min},5)
print(f"min={min},6)
print(f"min={min},diff={diff}")

输出

@H_660_3@min=5.855,max=5.864,diff=9
min=5.855,max=5.8649,diff=99
min=5.855,max=5.86499,diff=999
min=5.855,max=5.864999,diff=9999

已知限制

如果您想将尾随零保留在浮点数中，那么您必须使用 decimal。不要使用float。 :) 在上面的示例中，实际上没有必要对 @H_660_3@min_range 进行舍入，因为无论如何，python 都会处理尾随零。不过，我希望代码在输出结果中保持一致，因此我将其保留在代码中。 . 另外，据我所知，当您从 floats 中添加或减去值时，可能会有一些最小偏差，这是由于系统在 CPU 指令级别（低级内容）中的浮点处理......所以请记住这一点使用浮点数时 :)

附言如果您正在阅读此答案并知道如何解决问题的另一种方法 - 请尽可能附上性能比较报告:)