How can I store float numbers larger than 1*10^310 in python?

Question

I am working on a program that outputs the condition number of a big matrix, so I used the Power Method to get the Largest EigenValue, but the values are large numbers (float) larger than 1*10^310, and in the end the values become "Infinity", I tried the decimal module but it's the same. How can I store those large float values? Or maybe another method that uses shorter values? (I'm not allowed to use any module that helps explicitly the proccess as Numpy)

Answer 1

Don't work with floating point values if you can help it; they are very difficult to reason about and will bite you!

Whenever you are trying to work with floats, especially ones with lots of digits, you should consider how you can shift it into an integer range and if you have invalid or needless accuracy beyond the floating part of your value

• perhaps into a bigger int such as 10**400 or 10**100000, which should provide plenty of room for your floating point digits, while allowing you to work in the integer space
• directly convert or scale down, discarding digits beyond the decimal point (consider how accurate the measurement really is)

>>> int(1.0 * 10) * 10**999  # divide off 10**690 later or note in units
10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
>>> int(1.0 * 10**10)  # multiply by 10**300 later or note in units
10000000000

Practically, this is why you would want scientific notation - don't store the data with all its digits if you don't need them, keep the smallest amount you need and a second multiplier for the size factor (scientific notation does use a floating-point, but the idea is the same for integers)

Then, rather than working with floating points, you can recall the multiplier(s) at the end when you're done with your math (even multiplying them out separately)

It may even be sufficient to remove a significant portion of the digits entirely in some regular manner, and display the factor in the post-calculation units for whom or whatever is consuming the data

---

While this question is about large numbers, even decimal.Decimal unfortunately does not handle the small bits of floating points the way one might expect, as they're subject to some aliasing from how they're stored

https://en.wikipedia.org/wiki/Floating-point_arithmetic#IEEE_754:_floating_point_in_modern_computers

This is problematic with normal python floats, and so extends to Decimals, even of a size you may expect to see in normal use!

>>> 9007199254740993.0
9007199254740992.0
>>> Decimal(9007199254740993.0)  # NOTE converted to float before Decimal
Decimal('9007199254740992')

Adapted from Which is the first integer that an IEEE 754 float is incapable of representing exactly?

Example to the original question

>>> a = Decimal(10**310) * Decimal(1.0)
>>> b = Decimal(1)
>>> a + b - a
Decimal('0E+283')

Further examples

>>> a = Decimal(10**310)
>>> b = Decimal(0.1)
>>> a + b - a
Decimal('0')
>>> a
Decimal('10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000')
>>> b
Decimal('0.1000000000000000055511151231257827021181583404541015625')

>>> 10**-100
1e-100
>>> Decimal(10**-100)
Decimal('1.00000000000000001999189980260288361964776078853415942018260300593659569925554346761767628861329298958274607481091185079852827053974965402226843604196126360835628314127871794272492894246908066589163059300043457860230145025079449986855914338755579873208034769049845635890960693359375E-100')
>>> 10**-1000
0.0
>>> Decimal(10**-1000)
Decimal('0')

Answer 2

You want to use the decimal module:

from decimal import Decimal

x = Decimal('1.345e1310')
y = Decimal('1.0e1310')
print(x + y)

Result:

2.345E+1310