在 Python 中确定特定数字的精度和小数位数

问题描述

我在 Python 中有一个包含浮点数的变量(例如 num = 24654.123)，我想确定数字的精度和比例值(在 Oracle 意义上)，所以 123.45678应该给我 (8,5)，12.76 应该给我 (4,2)，等等.

I have a variable in Python containing a floating point number (e.g. num = 24654.123), and I'd like to determine the number's precision and scale values (in the Oracle sense), so 123.45678 should give me (8,5), 12.76 should give me (4,2), etc.

我首先考虑使用字符串表示(通过 str 或 repr)，但是对于大数来说这些都失败了(虽然我现在明白这是浮点的限制表示这是这里的问题):

I was first thinking about using the string representation (via str or repr), but those fail for large numbers (although I understand now it's the limitations of floating point representation that's the issue here):

>>> num = 1234567890.0987654321
>>> str(num) = 1234567890.1
>>> repr(num) = 1234567890.0987654

下面的好点.我应该澄清一下.该数字已经是一个浮点数，并且正在通过 cx_Oracle 推送到数据库.我试图在 Python 中尽我所能来处理对于相应数据库类型来说太大的浮点数，而不是执行 INSERT 和处理 Oracle 错误(因为我想处理字段中的数字，而不是记录，在一次).我猜 map(len, repr(num).split('.')) 是最接近浮点数的精度和比例的?

Good points below. I should clarify. The number is already a float and is being pushed to a database via cx_Oracle. I'm trying to do the best I can in Python to handle floats that are too large for the corresponding database type short of executing the INSERT and handling Oracle errors (because I want to deal with the numbers a field, not a record, at a time). I guess map(len, repr(num).split('.')) is the closest I'll get to the precision and scale of the float?

推荐答案

获取小数点左边的位数很简单:

Getting the number of digits to the left of the decimal point is easy:

int(log10(x))+1

小数点右边的位数比较棘手，因为浮点值固有的不准确性.我还需要几分钟才能弄清楚这一点.

The number of digits to the right of the decimal point is trickier, because of the inherent inaccuracy of floating point values. I'll need a few more minutes to figure that one out.

基于这个原则，这里是完整的代码.

Based on that principle, here's the complete code.

import math

def precision_and_scale(x):
    max_digits = 14
    int_part = int(abs(x))
    magnitude = 1 if int_part == 0 else int(math.log10(int_part)) + 1
    if magnitude >= max_digits:
        return (magnitude, 0)
    frac_part = abs(x) - int_part
    multiplier = 10 ** (max_digits - magnitude)
    frac_digits = multiplier + int(multiplier * frac_part + 0.5)
    while frac_digits % 10 == 0:
        frac_digits /= 10
    scale = int(math.log10(frac_digits))
    return (magnitude + scale, scale)

这篇关于在 Python 中确定特定数字的精度和小数位数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持html5模板网！