在自定义函数的上下文中，在Python中舍入十进制数的最简单方法是什么？

Question

我很难处理浮点数，而且我什至发现Python中的内置十进制类有时也很令人困惑。作为解决方案，我自己编写了一个方法，该方法只接受一个值和一些数字，这些数字应在浮点数之后显示，并以一种不错的格式返回字符串。看起来像这样：

def format_to_decimal_string(value, decimal_places):
    """
    # Given the `value` and `decimal_places`, this method returns a string
      in the format of a decimal number. Eg: `format_to_decimal_string('99.99', 1)`
      returns '99.9', while `format_to_decimal_string('99.99', 0)` return '99'.

    # The arguments are validated, to make sure `value` can be turned into
      a decimal number, and `decimal_places` is an integer greater than 0.
      The `value` is first turned into a string to accomodate for edge-cases,
      such as `format_to_decimal_string(99.99, 2)`.

    # If `decimal_number` is zero, we return a formatted zero string.

    # The start index is calculated, and using that, we append numbers
      to the string. Once again, we make sure to add zeroes to the start 
      and/or the end if needed.

    # Finally, we append the minus sign if needed, and return the result.
    """

    # Validating arguments
    try:
        decimal_number = decimal.Decimal(str(value)).as_tuple()
    except:
        raise ValueError('The value is not a valid decimal number.')

    if not isinstance(decimal_places, int):
        raise ValueError('The given decimal places is not an integer.')
    if not decimal_places >= 0:
        raise ValueError('The given decimal places must be greater than or equal to zero.')

    # Check if `decimal_number` is zero
    if decimal_number == 0:
        result = '0'
        if not decimal_places == 0:
            result += '.'
        for i in range(decimal_places):
            result += '0'
        return result

    # Finding the start index
    exponent_start_index = len(decimal_number.digits) + decimal_number.exponent

    # Appending the first digit
    if exponent_start_index == 0:
        result = '0'
    else:
        result = ''
        for digit in decimal_number.digits[0:exponent_start_index]:
            result += str(digit)

    # Appending the exponents
    exponent = ''
    if not decimal_places == 0:
        result += '.'
        for digit in decimal_number.digits[exponent_start_index:(decimal_places + exponent_start_index)]:
            exponent += str(digit)
        # Adding extra zeroes to make the number have a valid precision
        if decimal_places > len(exponent):
            for i in range(decimal_places - len(exponent)):
                exponent += '0'

    # Combining the first digit and the exponents
    result = result + exponent

    # Appending the minus sign if needed
    if decimal_number.sign == 1:
        if not decimal.Decimal(result) == 0:
            result = '-' + result

    return result

不言自明。它也考虑了零，只是为了使输出看起来不错。但是，我完全忘记的一件事是对数字进行取整。例如，当我希望它返回format_to_decimal_string(19.2189, 3)时，'19.218'返回'19.219'。有了这么大的方法，我觉得没有简单的方法可以对此进行添加？我是否应该尝试用另一种解决方法重写新方法，或者将其合并到现有方法中？谢谢。

Answer 1

如果要对此使用小数模块，则可以执行以下操作：

decimal.Decimal(19.2189).quantize(decimal.Decimal(10) ** -3)

quantize方法会将小数点四舍五入到其参数中的小数位数（因此您可以传递.001而不是将小数点提高到-3）。

Answer 2

使用{this.props.posts.slice(...

您可以使用round()完成此操作。

round()

如果要将此数字转换为字符串：

>>> round(19.2189, 3)
19.219

使用>>> str(round(19.2189, 3)) '19.219'

您也可以使用format()完成此操作：

format()