Determine precision and scale of particular number

问题:

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.

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

Edit:

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?

回答1:

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

int(log10(x))+1

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.

Edit: 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)

回答2:

Not possible with floating point variables. For example, typing

>>> 10.2345

gives:

10.234500000000001

So, to get 6,4 out of this, you will have to find a way to distinguish between a user entering 10.2345 and 10.234500000000001, which is impossible using floats. This has to do with the way floating point numbers are stored. Use decimal.

import decimal
a = decimal.Decimal('10.234539048538495')
>>> str(a)
'10.234539048538495'
>>>  (len(str(a))-1, len(str(a).split('.')[1]))
(17,15)

回答3:

I think you should consider using the decimal type instead of a float. The float type will give rounding errors because the numbers are represented internally in binary but many decimal numbers don't have an exact binary representation.

回答4:

seems like str is better choice than repr:

>>> r=10.2345678
>>> r
10.234567800000001
>>> repr(r)
'10.234567800000001'
>>> str(r)
'10.2345678'

回答5:

(0) Please confirm or deny: You are given floats to use, this is unavoidable, you can't get your data as decimal, the Oracle datatypes include decimal-based types, and this fundamental mismatch is unavoidable. Please explain any full or partial denial.

(1) Your "fail for large numbers" remark is misleading/irrelevant/wrong -- you say that your starting point is a float, but 1234567890.0987654321 can't be represented as a float, as shown by the result of repr().

(2) Perhaps you could use the NEW repr (Python 2.7 and 3.1) which provides the minimum possible precision of repr(x) that still satisfies float(repr(x)) == x

E.g. old repr(1.1) produces "1.1000000000000001", new repr(1.1) produces "1.1"

About "I guess map(len, repr(num).split('.')) is the closest I'll get to the precision and scale of the float?": You need a strategy to handle (a) negative and zero numbers (b) numbers like 1.1e20

Digging in Objects/floatobject.c should turn up the C code for the new repr() of a float object, should you need to use Python 2.6 or earlier.

(3) Perhaps if you told us the specs for the relevant Oracle data types, we could help you devise checks for choosing which type can contain a given float value.

回答6:

Basically, you can't with floating point numbers. Using the decimal type would help and if you want really large precision, consider using gmpy, the GNU Multiple Precision library's port to Python.

回答7:

I found another solution that seems to be simpler, but I'm not sure exactly if it will work for all cases.

   import math
   x = 1.2345678

   def flip(string):
       result = ""
       for ch in string:
           result = ch + result
      return result

   prec = int(math.log10(float(flip(str(x)))) + 1   # precision as int