search for: uncenter

Hi, Seeking suggestions to compute uncentered (pearson correlation) correlation efficiently. corr from stats library works on x and y columns. dist from amap library works on x and y rows. My data layout is slightly different such that row(i) of matrix x is compared to row(i) of matrix y. Thanks [[alternative HTML version deleted]]

Dear all: I am comparing the PLS outputs of R and SAS for the following data set: Y x1 x2 x3 3 6 2 2 3 1 5 5 4 7 4 1 5 6 5 6 2 4 3 2 8 5 0 9 where Y is the dependent variable and x1, x2, x3 are the independent variables. I found several PLS algorithms in R (NIPALS,SIMPLS,KERNEL PLS). SAS has SIMPLS and NIPALS. The following are the NIPALS calculations of

...) # center and scale m2 <- scale(m1) # find cholesky decomp c1 <- chol(var(m2)) # force to be independent m3 <- m2 %*% solve(c1) # create new correlation matrix: cor2 <- cbind( rbind( cor1, z=c(.5,.3,.1,.05) ), z=c(.5,.3,.1,.05,1) ) # create new matrix m4 <- m3 %*% chol(cor2) # uncenter and unscale m5 <- sweep( m4, 2, attr(m2, 'scaled:scale'), '*') m5 <- sweep( m5, 2, attr(m2, 'scaled:center'), '+') ##Check they are equal zapsmall(cor(m5))==zapsmall(cor2) Thanks, ben

...omparative (preferrable simple and intuitive) measure(s)? Something that would graphically perhaps give an indication without time-consuming clustering, sampling or whatsoever processing? Even though the above mentoined authors mention some measure for the assymetry of the yielded compoenents ( uncentered -> unipolar, centered -> bipolar) I find the concept a bit hard to understand. Isn't there a quick way (function) to just say (with numbers of plots of course) "well, it seems that the data are heterogenous looking at between- axes" or the other way around "it looks l...

...the options of the prcomp function. But as far as I know and as I understand the help text it should have no influence on the eigenvalue equation whether the data are centered or not. I know about the advantages of centered date but I want to understand how the prcomp function works in the case of uncentered data. Thank you very much for your efforts. -- View this message in context: http://r.789695.n4.nabble.com/prcomp-function-tp3036088p3036088.html Sent from the R help mailing list archive at Nabble.com.

...;. I also wondered whether it would be possible to convert to a phylogenetic tree and use the functions in the 'ape' package? Any suggestion on how to split up a PVclust plot would be greatly appreciated (code for the plot below) Thanks Tom result <- pvclust(df.1, method.dist="uncentered", method.hclust="average",nboot=10) par(mar=c(0,0,0,0)) par(oma=c(0,0,0,0)) plot(result, print.pv =FALSE, col.pv=c("red","",""), print.num=FALSE, float = 0.02, font=1, axes=T, cex =0.85, main="", sub="", xlab="", ylab= &...

I am trying to generate a fourth vector,z, given three known and fixed vectors, x1,x2,x3 with corresponding known and fixed correlations with themeselves and with z. That is, all correlations are known and prespecified. How can I do this? Thank you, ben

Hi people, I want to obtain the adjusted r-square given a set of coefficients (without the intercept), and I don't know if there is a function that does it. Exist???????????????? I know that if you make a linear regression, you enter the dataset and have in "summary" the adjusted r-square. But this is calculated using the coefficients that R obtained,and I want other coefficients