similar to: Attaching marginal summary plots to the main matrix plot

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.

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.

I suspect Prof. Ripley's response suffices. However, there *is* a Parzen kernel for kernel smoothing: Parzen K(z) = { 4/3 - 8z^2 + 8|z|^3 if |z| <= 1/2 8(1 -|z|)^3/3 if 1/2 < |z| <= 1 0 otherwise If I'm not mistaken, this appeared in Parzen's original 1962 paper on kernel density estimation. I also seem to recall

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",

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

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)

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

Hi. I did write those functions, and sent them (I thought) to one of the R maintainers to see whether they would be appropriate for inclusion (because I'd seen some requests on the mailing lists). However, I'm happy to post them -- I should have thought of it before. WARNING: I've tested these functions on some data arising in my work and also on the USArrests data that comes with

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

[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 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 :

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

Hi, I have huge matrices in which the response variable is in the first column and the regressors are in the other columns. What I wanted to do now is something like this: #this is just to get an example-matrix DataMatrix <- rep(1,1000); Disturbance <- rnorm(900); DataMatrix[101:1000] <- DataMatrix[101:1000]+Disturbance; DataMatrix <- matrix(DataMatrix,ncol=10,nrow=100); #estimate

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

In datamatrix[, "y"] == datamatrix[, "y"][-1] : longer object length is not a multiple of shorter object length out = c(FALSE,datamatrix[,'y'] == datamatrix[,'y'][-1]) and I do not know why I get that error, the resulting out matrix is somehow one row larger than datamatrix... all I try to do is filter matrix by dropping rows where [,'y'][-1] ==

Hi. I have been looking at the PCAproj function in package pcaPP (R 2.4.1) for robust principal components, and I'm trying to interpret the results. I started with a data matrix of dimensions RxC (R is the number of rows / observations, C the number of columns / variables). PCAproj returns a list of class princomp, similar to the output of the function princomp. In a case where I can

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

Hi, I need to permute the rows of a matrix, where each row is independently rearranged. A simple solution is this: shuffled <- datamatrix <- matrix(1:24, ncol = 4) for (i in 1:nrow(datamatrix)) { shuffled[i, ] <- sample(datamatrix[i, ]) } > datamatrix [,1] [,2] [,3] [,4] [1,] 1 7 13 19 [2,] 2 8 14 20 [3,] 3 9 15 21 [4,] 4 10 16 22 [5,]

Hello R users, I'm wondering whether it is possible to manage big data set in R? I have a data set with 3 million rows and 3 columns (X,Y,Z), where X is the group id. For each X, I need to run 2 regression on the submatrix. I used the function "split": datamatrix<-read.csv("datas.csv", header=F, sep=",") dim(datamatrix) # [1] 2980523 3

> From: Duncan Murdoch <murdoch at stats.uwo.ca> > To: ramasamy at cancer.org.uk > Date: Fri, 10 Mar 2006 08:33:09 -0500 > Subject: Re: [R] To improve my understanding of workspaces > Other than Emacs, I use the same work habits as Adai. An advantage of > this workflow is that almost everything is stored in text format, so it > is easy to compare different versions to