Thanks. I guess that -"_1 _ means that subtraction is performed over the first dimension of the left argument
cheers Roelof Contact information Dr. Roelof K. Brouwer PEng Professor Department of Computing Science School of Advanced Technologies and Mathematics Thompson Rivers University 900 McGill Road, PO Box 3010 Kamloops, BC, Canada,V2C 5N3 skype RoelofBrouwer telephone numbers 1-250 828 5219 work 1-250-377-6133 fax home page www.tru.ca/advtech/compsci/faculty/rbrouwer email [EMAIL PROTECTED] www.cariboo.tru.ca/advtech/compsci/faculty/rbrouwer/pickup/flighttoNanaimo.w mv www.picasaweb.google.com/rkbrouwer -----Original Message----- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Roger Hui Sent: September 4, 2007 6:23 PM To: General forum Subject: Re: [Jgeneral] normalize data In norm, if you change the -"1 and %"1 to -"_1 _ and %"_1 _ then the function would work on an argument of any rank. Also, the [ can be elided. Thus: norm1=: (-"_1 _ mean) %"_1 _ %:@:variance ----- Original Message ----- From: Raul Miller <[EMAIL PROTECTED]> Date: Tuesday, September 4, 2007 17:40 Subject: Re: [Jgeneral] normalize data To: General forum <[email protected]> > On 9/4/07, Roelof Brouwer <[EMAIL PROTECTED]> wrote: > mean=:+/%# > variance=: (mean@:*:)-(*:@ mean) > normalize_data=: ([-mean)%(%:@: variance) > norm_data=: normalize_data"1&.|: > norm=: ([- " 1 mean)%"1 %:@: variance > > ... > > With the above definitions (which include a couple names supplied > by me), the following holds: > > (norm -: norm_data) ?3 3$0 > 1 > > So... what was the question? ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
