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)