Clarification on the Decimal type in Python

Question

Everybody know, or at least, every programmers should know, that using the float type could lead to precision errors. However, in some cases, an exact solution would be great and there are cases where comparing using an epsilon value is not enough. Anyway, that's not really the point.

I knew about the Decimal type in Python but never tried to use it. It states that "Decimal numbers can be represented exactly" and I thought that it meant a clever implementation that allows to represent any real number. My first try was:

>>> from decimal import Decimal
>>> d = Decimal(1) / Decimal(3)
>>> d3 = d * Decimal(3)
>>> d3 < Decimal(1)
True

Quite disappointed, I went back to the documentation and kept reading:

The context for arithmetic is an environment specifying precision [...]

Ok, so there is actually a precision. And the classic issues can be reproduced:

>>> dd = d * 10**20
>>> dd
Decimal('33333333333333333333.33333333')
>>> for i in range(10000):
...    dd += 1 / Decimal(10**10)
>>> dd
Decimal('33333333333333333333.33333333')

So, my question is: is there a way to have a Decimal type with an infinite precision? If not, what's the more elegant way of comparing 2 decimal numbers (e.g. d3 < 1 should return False if the delta is less than the precision).

Currently, when I only do divisions and multiplications, I use the Fraction type:

>>> from fractions import Fraction
>>> f = Fraction(1) / Fraction(3)
>>> f
Fraction(1, 3)
>>> f * 3 < 1
False
>>> f * 3 == 1
True

Is it the best approach? What could be the other options?

question from:https://stackoverflow.com/questions/20354423/clarification-on-the-decimal-type-in-python

Answer 1

The Decimal class is best for financial type addition, subtraction multiplication, division type problems:

>>> (1.1+2.2-3.3)*10000000000000000000
4440.892098500626                            # relevant for government invoices...
>>> import decimal
>>> D=decimal.Decimal
>>> (D('1.1')+D('2.2')-D('3.3'))*10000000000000000000
Decimal('0.0')

The Fraction module works well with the rational number problem domain you describe:

>>> from fractions import Fraction
>>> f = Fraction(1) / Fraction(3)
>>> f
Fraction(1, 3)
>>> f * 3 < 1
False
>>> f * 3 == 1
True

For pure multi precision floating point for scientific work, consider mpmath.

If your problem can be held to the symbolic realm, consider sympy. Here is how you would handle the 1/3 issue:

>>> sympy.sympify('1/3')*3
1
>>> (sympy.sympify('1/3')*3) == 1
True

Sympy uses mpmath for arbitrary precision floating point, includes the ability to handle rational numbers and irrational numbers symbolically.

Consider the pure floating point representation of the irrational value of √2:

>>> math.sqrt(2)
1.4142135623730951
>>> math.sqrt(2)*math.sqrt(2)
2.0000000000000004
>>> math.sqrt(2)*math.sqrt(2)==2
False

Compare to sympy:

>>> sympy.sqrt(2)
sqrt(2)                              # treated symbolically
>>> sympy.sqrt(2)*sympy.sqrt(2)==2
True

You can also reduce values:

>>> import sympy
>>> sympy.sqrt(8)
2*sqrt(2)                            # √8 == √(4 x 2) == 2*√2...

However, you can see issues with Sympy similar to straight floating point if not careful:

>>> 1.1+2.2-3.3
4.440892098500626e-16
>>> sympy.sympify('1.1+2.2-3.3')
4.44089209850063e-16                   # :-(

This is better done with Decimal:

>>> D('1.1')+D('2.2')-D('3.3')
Decimal('0.0')

Or using Fractions or Sympy and keeping values such as 1.1 as ratios:

>>> sympy.sympify('11/10+22/10-33/10')==0
True
>>> Fraction('1.1')+Fraction('2.2')-Fraction('3.3')==0
True

Or use Rational in sympy:

>>> frac=sympy.Rational
>>> frac('1.1')+frac('2.2')-frac('3.3')==0
True
>>> frac('1/3')*3
1

You can play with sympy live.