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From: Jussi Piitulainen on 3 May 2010 12:23
Peter Otten writes:
> 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

Logarithms agree to 13 decimal digits or so:

>>> math.exp(math.log(x) - math.log(y))
16.13725490196353
From: Peter Otten on 3 May 2010 12:39
Peter Pearson wrote:

> On Mon, 03 May 2010 17:30:03 +0200, Peter Otten <__peter__(a)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
>
> Does this still work if y = 765 * 10**1000 + 1 ? It looks
> as if it might rely on x and y having a large common divisor.

You mean something like this?

>>> import fractions
>>> x = 765 * 10**1000
>>> fractions.gcd(x, x+1)
1L
>>> float(fractions.Fraction(x, x+1))
1.0

But my D'oh!* moment really was seeing Jerry Hill doing it with

>>> from __future__ import division
>>> x/(x+1)
1.0

Peter

(*) I'm having a lot of these lately. Looks like I need a new brain.
From: Victor Eijkhout on 3 May 2010 16:49
Jerry Hill <malaclypse2(a)gmail.com> wrote:

> >>> from __future__ import division
> >>> long1/long2
> 0.5

Beautiful. Thanks so much guys.

Victor.
--
Victor Eijkhout -- eijkhout at tacc utexas edu
From: Mensanator on 4 May 2010 12:50
On May 3, 10:17 am, s...(a)sig.for.address (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.

You could try using the gmpy module. It supports arbitrary precision
floats, so converting long to float is no problem.

YY = 3**10684
float(YY)
Traceback (most recent call last)
File "pyshell#78>", line 1, in <module>
float(YY)
OverflowError: Python int too large to convert to C double

import gmpy
gmpy.mpf(YY)
mpf('3.6600365709...0197681e5097',16936)



>
> Victor.
>
> --
> Victor Eijkhout -- eijkhout at tacc utexas edu

From: Victor Eijkhout on 4 May 2010 13:35
Mensanator <mensanator(a)aol.com> wrote:

> You could try using the gmpy module. It supports arbitrary precision
> floats, so converting long to float is no problem.

I fear I may actually have to go symbolic. I'm now having to use the
12th root of 2, and I would like the twelfth power of that to be exactly
2.

Victor.
--
Victor Eijkhout -- eijkhout at tacc utexas edu
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