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*Subject*: Re: STANDARD DEVIATON*From*: Struan Gray <struan.gray(at)sljus.lu.se>*Date*: 4 Aug 2000 12:43:09 GMT*Distribution*: world*Newsgroups*: comp.lang.idl-pvwave*Organization*: This line intentionally left bland*References*: <3986B6E9.48105613@met.ed.ac.uk> <8m6ims$jr8$1@news.lth.se> <398AA19C.7B7427B1@met.ed.ac.uk> <398AADC0.92604B5D@met.ed.ac.uk>*Xref*: news.doit.wisc.edu comp.lang.idl-pvwave:20667

Ben Marriage, ben@met.ed.ac.uk writes: > I believe that this gives different results to > the inbuilt IDL routines. It's possible I'm missing something, but the result of my function should be the same as yours, but multiplied by sqrt(8/9). You and IDL are dividing by (N-1) in the definition of variance, I am dividing by N. Statisticians love to argue about which is appropriate and where, but for an image of the s.d. it doesn't matter much. The box smooth always divides by N, but you can always add a renormalisation step and it'll still be faster than nested loops. I use this as part of a routine to do so-called statistical differencing of images, which brings out fine detail on a varying background (it's like an unsharp mask weighted by the local image statistics). In this case the the s.d. image is multiplied by an arbitrary, user-selectable factor, so the difference between N and N-1 is irrelevant. Struan

**Follow-Ups**:**Re: STANDARD DEVIATON***From:*Struan Gray

**References**:**STANDARD DEVIATON***From:*Ben Marriage

**Re: STANDARD DEVIATON***From:*Struan Gray

**Re: STANDARD DEVIATON***From:*Ben Marriage

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