long int computations

[Duncan Booth]

Peter Otten <__peter__ at web.de> wrote:

> Victor Eijkhout wrote:
> 
>> I have two long ints, both too long to convert to float, but their ratio
>> is something reasonable. How can I compute that? The obvious "(1.*x)/y"
>> does not work.
> 
>>>> import fractions
>>>> x = 12345 * 10**1000
>>>> y = 765 * 10**1000
>>>> float(x)
> Traceback (most recent call last):
>   File "<stdin>", line 1, in <module>
> OverflowError: long int too large to convert to float
>>>> fractions.Fraction(x, y)
> Fraction(823, 51)
>>>> float(_)
> 16.137254901960784
> 

Another way of doing this is simply to scale the values down until they do 
fit in a float:

>>> x = 12345 * 10**1000
>>> y = 765 * 10**1000
>>> maxfloat = 1e308 # or whatever it actually is...
>>> d = max(max(abs(x),abs(y))//long(maxfloat), 1)
>>> float(x//d)/float(y//d)
16.137254901960784

I think that should work for any cases where the ratio fits in a float (if 
it doesn't then you might get a ZeroDivisionError or inf or -inf).

-- 
Duncan Booth http://kupuguy.blogspot.com