Bruce Southey wrote: > Alan McIntyre wrote: >> On Thu, Jun 5, 2008 at 10:16 PM, Keith Goodman <[EMAIL PROTECTED]> >> wrote: >> >>> How can that lead to instability? If the last half-million values are >>> small then they won't have a big impact on the mean even if they are >>> ignored. The variance is a mean too (of the squares), so it should be >>> stable too. Or am I, once again, missing the point? >>> >> >> No, I just didn't think about it long enough, and I shouldn't have >> tried to make an example off the cuff. ;) After thinking about it >> again, I think some loss of accuracy is probably the worst that can >> happen. >> _______________________________________________ >> Numpy-discussion mailing list >> Numpy-discussion@scipy.org >> http://projects.scipy.org/mailman/listinfo/numpy-discussion >> >> > Any problems are going to mainly due to the distribution of numbers > especially if there are very small numbers and very large numbers. > This is mitigated by numerical precision and algorithm - my guess is > that it will take a rather extreme case to cause you any problems. > > Python and NumPy are already using high numerical precision (may > depend on architecture) and NumPy defines 32-bit, 64-bit and 128-bit > precision if you want to go higher (or lower). This means that > calculations are rather insensitive to numbers used so typically there > is no reason for any concern (ignoring the old Pentium FDIV bug, > http://en.wikipedia.org/wiki/Pentium_FDIV_bug ). > > The second issue is the algorithm where you need to balance > performance with precision. For simple calculations: > http://en.wikipedia.org/wiki/Algorithms_for_calculating_variance > > Bruce > I forgot to add: import numpy x = 1e305 * numpy.ones(10000000, np.float128) type(x[0]) # gives <type 'numpy.float128'> x.mean() # gives 1.000000000000036542e+305
Bruce _______________________________________________ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
