similar to: Parzen Windows

Take a look at the manual page for "layout". The final example does pretty much what you're asking about (with a scatterplot and histograms, but the idea should be the same). Hope this helps, Matthew Wiener Applied Computer Science & Mathematics Dept. Merck Research Labs Rahway, NJ 07090 732-594-5303 -----Original Message----- From: Adaikalavan Ramasamy [mailto:ramasamy at

You are using a method that needs to estimate the covariance matrix of all the variables. If you have 80 variables, there are (80+1)*80/2 = 3240 variances and covariances to estimate. How many data points do you think you need to do that? Some people assume the covariance matrix is diagonal (i.e., assuming all the variables are uncorrelated). Even then you still have 80 variances to estimate.

thank you to both Adaikalavan and Patrick. on the basis of Adaikalavan example this is an example that point out my problems. >log_raise=function(num, exp, base){return(log(num^exp,base))} I would like to have optional fields, so some settings to be default parameters; if I declare the function as above the operator must of course input all the variables. Finally I could fix as follow.

Dear all, I am having trouble trying to install R-1.8.1 on a Sun Solaris 8 (Generic_108528-23 version) machine. The configuration was successful but make check fails. I traced the the problem to the plot() function. > 1 + 1 [1] 2 > capabilities() jpeg png tcltk X11 GNOME libz http/ftp sockets TRUE FALSE TRUE TRUE FALSE TRUE TRUE TRUE

I worked this out over the weekend. I appreciate that using temporary variables would be simpler but I think this makes for quite readable code: # in RProfile.site inplace <- function (f, arg=1) eval.parent(call("<-",substitute(f)[[arg+1]], f),2) # examples in code inplace(foo[bar,baz] *2) # or inplace(paste(foo[bar,baz], 1:10)) # or inplace(sub("blah",

I have been struggling with this problem for a while and I hope someone could help me. Or if someone could point me to a section in the manual I would be grateful. x <- "my" plot(1:10, main=expression(paste( x, Delta, "values" ))) Q : How do I get the title to say "my (triangle symbol) values" ? The following trial-and-error produced mainly errors :

Dear all, A few weeks ago I asked about X11() availability using CGI-perl. The general advice was to use bitmap which requires ghostscript (Xvnc and Xvbf were the other alternatives). My system administrator got around to installing Aladin Ghostscript 6.0 and the setting the R_GSCMD environment variable. Other system specification is given below. But now I have a new problem that says that font

Could some kindly tell me if I am supposed to be getting the same test statistic value with var.equal=TRUE and var.equal=FALSE in t.test ? set.seed(1066) x1 <- rnorm(50) x2 <- rnorm(50) t.test(x1, x2, var.equal=FALSE)$statistic # 0.5989774 t.test(x1, x2, var.equal=TRUE)$statistic # 0.5989774 ??? Here are my own calculations that shows that perhaps the result when var.equal=TRUE is

First, many thanks to Frank Harrell for once again helping me out. This actually relates to the next point, which is my contribution to the 'why don't social scientists use R' discussion. I am a hybrid social scientist(child psychiatrist) who trained on SPSS. Many of my difficulties in coming to terms with R have been to do with trying to apply the logic underlying SPSS, with dire

Thanks Adaikalavan, however the problem remains. Considering AIC() as applied to the linear model in AIC() help documentation: > data(swiss) > lm1 <- lm(Fertility ~ . , data = swiss) > AIC(lm1) [1] 326.0716 Clearly this includes the estimation of the residual standard error as an estimated parameter, as this gives the correct score: > -2*logLik(lm1) + 2*(length(coef(lm1))+1)

Dear R-user and -programmer, has one R-package the ability to compute smoothed periodograms of time series using the Tukey-window and/or the Parzen-window? In the ts- and tseries-packages I have found only Daniell-smoothers. With many thanks in advance for any hint Albrecht Kauffmann -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read

I really appreciate all the helpful answers from Richard Müller, Marc Schwartz, Adaikalavan Ramasamy, Vito Ricci. I tried Marc's simple suggestion which was to copy only the data area and paste to a new excel sheet, and that solved my problem. I'll try other suggestions while I'm learning more about R. Unedited answers are provided below followed by my question. Thanks again. Kyong

There was a BioConductor thread today where the poster wanted to find pairwise difference between columns of a matrix. I suggested the slow solution below, hoping that someone might suggest a faster and/or more elegant solution, but no other response. I tried unsuccessfully with the apply() family. Searching the mailing list was not very fruitful either. The closest I got to was a cryptic chunk

[In future, please press reply all as there might be others in the mailing list who can help you] Without a (short) reproducible example, it would be difficult to say. BTW, I think you have mistyped the error/warning message. Does it actually say "cannot compute _correct_ p-value" or "cannot compute _exact_ p-value". See example below. > wilcox.test( 1:3, 1:10 )

I am trying to write a wrapper for the last example in help(cv.glm) that deals with leave-one-out-cross-validation (LOOCV) for a logistic model. This wrapper will be used as part of a bigger program. Here is my wrapper funtion : logistic.LOOCV.err <- function( formu=NULL, data=NULL ){ cost.fn <- function(cl, pred) mean( abs(cl-pred) > 0.5 ) glmfit <- glm(

I now have versions of code that is destined (I believe) for 0.63 which is in a suitable state for comment. The files are at ftp://ftp.stats.ox.ac.uk/pub/R (Our www server is being moved, so may be intermittently down, but this ftp server should be stable.) All are R packages, for the moment for personal use only (no re-distribution). Use with 0.62.3 or 0.63 (although I am aware of some

dear R-helper, I tried to find out a package in which i can have penalized maximum likelihood estimator applying on generalized extreme value distribution with beta function) but could not. would you please help me to know the name of the package. thanks for your help. S.Murshed --- r-help-request at stat.math.ethz.ch wrote: > Send R-help mailing list submissions to > r-help at

Hi list members I have a vector of strings x=c("w","ex","ee") And I want to get a logical vector showing the positions where my search string "e" matches the elements partially, i.e. is at least the left-hand part of the target strings, i.e. I want to get a vector FALSE TRUE TRUE. Any ideas? Thanks Steve Powell proMENTE social research research |

Apologies in advance if my comments don't help, in which case, no need to respond, but I noted in ?ksmooth: "bandwidth the bandwidth. The kernels are scaled so that their quartiles (viewed as probability densities) are at ? 0.25*bandwidth." So, could this be a source of the discrepancies you cited? Given that ?ksmooth explicitly says: "Note: This function was implemented for

Just out of curiosity, what is the difference between the terms for package and library ? Why are we loading a package with the library() command ? If this is a case of RTFM, I would be happy to do so if pointed in the right direction. I have searched the FAQ and mail archives and only came up with http://tolstoy.newcastle.edu.au/R/help/05/02/12162.html but this still does not explain what is