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)
Answer from Mark Ransom on Stack OverflowWhen the matter is precision - like, really precise - floats are crap. Even double precision can be a gamble when working with scientific data - very small and very large numbers.
What are the best options when I need to work with numbers on the -15th and +20th orders of magnitude? (at the same time, mind you)
Is the decimal.py module precise enough for those sorts of calculations? If not, is it possible to get precise results with python or will I have to write some matlab modules to crunch my numbers?
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)
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)
You are running into the old problem with floating point numbers that not all numbers can be represented exactly. The command line is just showing you the full floating point form from memory.
With floating point representation, your rounded version is the same number. Since computers are binary, they store floating point numbers as an integer and then divide it by a power of two so 13.95 will be represented in a similar fashion to 125650429603636838/(2**53).
Double precision numbers have 53 bits (16 digits) of precision and regular floats have 24 bits (8 digits) of precision. The floating point type in Python uses double precision to store the values.
For example,
>>> 125650429603636838/(2**53)
13.949999999999999
>>> 234042163/(2**24)
13.949999988079071
>>> a = 13.946
>>> print(a)
13.946
>>> print("%.2f" % a)
13.95
>>> round(a,2)
13.949999999999999
>>> print("%.2f" % round(a, 2))
13.95
>>> print("{:.2f}".format(a))
13.95
>>> print("{:.2f}".format(round(a, 2)))
13.95
>>> print("{:.15f}".format(round(a, 2)))
13.949999999999999
If you are after only two decimal places (to display a currency value, for example), then you have a couple of better choices:
- Use integers and store values in cents, not dollars and then divide by 100 to convert to dollars.
- Or use a fixed point number like decimal.
There are new format specifications, String Format Specification Mini-Language:
You can do the same as:
"{:.2f}".format(13.949999999999999)
Note 1: the above returns a string. In order to get as float, simply wrap with float(...):
float("{:.2f}".format(13.949999999999999))
Note 2: wrapping with float() doesn't change anything:
>>> x = 13.949999999999999999
>>> x
13.95
>>> g = float("{:.2f}".format(x))
>>> g
13.95
>>> x == g
True
>>> h = round(x, 2)
>>> h
13.95
>>> x == h
True