From: Paul Rubin on 6 May 2010 02:55 see(a)sig.for.address (Victor Eijkhout) writes: > 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. The math.log function has a special hack for long ints, that might help: Python 2.6.2 (r262:71600, Jan 25 2010, 18:46:47) >>> from math import * >>> a = log(3**100000) >>> a 109861.22886681097 >>> b = log(3**100001) >>> exp(ba) 2.9999999999994813
From: Mark Dickinson on 6 May 2010 04:22 On May 3, 9:49 pm, s...(a)sig.for.address (Victor Eijkhout) wrote: > Jerry Hill <malaclyp...(a)gmail.com> wrote: > > >>> from __future__ import division > > >>> long1/long2 > > 0.5 > > Beautiful. Thanks so much guys. And if for some reason you don't want to use the 'from __future__' import, then you can do long1.__truediv__(long2): >>> n = 765*10**1000 + 1 >>> n.__truediv__(n+1) 1.0 If you care about speed at all, I'd avoid the Fractions solution; it does an expensive and slow gcd computation.  Mark
First

Prev

Pages: 1 2 3 Prev: strange interaction between open and cwd Next: Django as exemplary design 