Extreme floating point precision number 1.0-(0.1^200) in python-CodePudding

I was creating a threshold with extreme floating point precision. I am in need of a value something like 0.999999999... with 200 9s. Is that possible in python?

One solution I thought of was to create it using 1.-1e-200, but that just gives me the value 1.

CodePudding user response：

It is not possible with normal variables. Python uses the processor's double precision floating point numbers, which hold about 17 decimal digits.

There are add-on packages like bigfloat that can do arbitrary floating point arithmetic, but you'd need a pretty good reason to use them. Most people overestimate the need for precision in floating point math.

CodePudding user response：

The standard fractions.Fraction class provides unlimited-precision exact rational arithmetic.

>>> from fractions import Fraction
>>> x = Fraction(1) - Fraction(1, 10**200)

decimal.Decimal might also suit your needs if you're looking for high-precision rather than exact arithmetic, but you'll have to tweak the context settings to allow more than the default 28 significant digits.

>>> from decimal import Decimal, Context
>>> decimal.setcontext(Context(prec=1000))
>>> x = Decimal(1) - 1 / Decimal(10**200)