Clarification on the Decimal type in Python

Question:

Everybody knows, or at least, every programmer 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?

Asked By: Maxime Chéramy

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Answers:

So, my question is: is there a way to have a Decimal type with an infinite precision?

No, since storing an irrational number would require infinite memory.

Where Decimal is useful is representing things like monetary amounts, where the values need to be exact and the precision is known a priori.

From the question, it is not entirely clear that Decimal is more appropriate for your use case than float.

Answered By: NPE

is there a way to have a Decimal type with an infinite precision?

No; for any non-empty interval on the real line, you cannot represent all the numbers in the set with infinite precision using a finite number of bits. This is why Fraction is useful, as it stores the numerator and denominator as integers, which can be represented precisely:

>>> Fraction("1.25")
Fraction(5, 4)
Answered By: jonrsharpe

If you are new to Decimal, this post is relevant: Python floating point arbitrary precision available?

The essential idea from the answers and comments is that for computationally tough problems where precision is needed, you should use the mpmath module https://code.google.com/p/mpmath/. An important observation is that,

The problem with using Decimal numbers is that you can’t do much in the way of math functions on Decimal objects

Answered By: William Denman

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.

Answered By: dawg

Just to point out something that might not be immediately obvious to everyone:

The documentation for the decimal module says

… The exactness carries over into arithmetic. In decimal floating point, 0.1 + 0.1 + 0.1 – 0.3 is exactly equal to zero.

(Also see the classic: Is floating point math broken?)

However, if we use decimal.Decimal naively, we get the same "unexpected" result

>>> Decimal(0.1) + Decimal(0.1) + Decimal(0.1) == Decimal(0.3)
False

The problem in the naive example above is the use of float arguments, which are "losslessly converted to [their] exact decimal equivalent," as explained in the docs.

The trick (implicit in the accepted answer) is to construct the Decimal instances using e.g. strings, instead of floats

>>> Decimal('0.1') + Decimal('0.1') + Decimal('0.1') == Decimal('0.3')
True

or, perhaps more convenient in some cases, using tuples (<sign>, <digits>, <exponent>)

>>> Decimal((0, (1,), -1)) + Decimal((0, (1,), -1)) + Decimal((0, (1,), -1)) == Decimal((0, (3,), -1))
True

Note: this does not answer the original question, but it is closely related, and may be of help to people who end up here based on the question title.

Answered By: djvg