Raul Miller <[email protected]> writes: > And, of course, Welford's approach does achieve that. But Welford's > algorithm assumes a scalar computing architecture. Meanwhile, it > seems to me that subtracting the computed mean achieves approximately > the same thing that Welford's approach achieves, but modularized to > eliminate the "scalar architecture" aspect.
What I found surprising from all this is that the iteration of the difference helps. e.g., r=:1e12+?1e6#0 %: (+/@:*: % <:@#) (- mean) r 0.573789 %: (+/@:*: % <:@#) (- mean)^:2 r 0.288728 I certainly did not expect that to work, since I expected the second iteration of (- mean) to be a no-op, but it looks like that's where the "numerical weirdness" lies. > That said, if there are generic cases where Welford's approach behaves > better than subtracting the computed mean from the data set, I would > be interested in hearing about them. (But I doubt they exist, except > in hand crafted special cases with no general utility, because > Welford's algorithm is doing essentially the same thing, with a > running average.) I'd also like to see it, but I've not been able to find an example. However, I do find it interesting that Welford's algorithm can do this on a single pass through the data. Just a few months ago, I bought myself the Dover edition of Hamming's numerical methods book. I think I'll use this as an excuse to read more of it. Regards, Johann P.S. On my particular sample, adding 0.5 makes the problem worse, so it appears to depend on the random numbers involved. %: (+/@:*: % <:@#) (- mean) r+0.5 1.02374 ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
