similar to: Upgrade R?

Hello, I am going to install the new version of R 2.14.1. After the installation, I want to copy my installed packages to the new library. But since over time I forgot which ones I installed I want to get a list of all the packages I installed among the packages installed initially by the R-installer. Is this possible? Cem Cem Girit, PhD Biopticon Corporation 182

Hello, I need some results from the survival analysis of my data that I do not know whether exist in Survival Package or how to obtain if they do: 1. The Mean survival time 2. The standard error of the mean 3. Point and 95% Lower & Upper Confidence Intervals estimates Any help will be greatly appreciated. Cem [[alternative HTML version

Hello, I am having hard time obtaining a value from a function. "fit" is a survival function that produces some results, such as "median", "confidence intervals" etc. But str() function does not list these values. How can I extract these to be able use them? For example, I need "median" value for the group DrugA which is 48. "Print" function does

Hello, Is there a web version of this R-Help user group (such as the ones under Google Groups) such that 1. I can do a search on any topic over thousands of posts on R easily and effectively 2. My mailbox do not overflow with emails so that I do not need to edit it every day 3. I can arrange to receive only the responses to my posts automatically 4.

Hello, I am interested in parametric multi comparison tests such as Dunnett, Duncan, Tukey, Newman-Keuls, Bonferonni, Scheffe, and non-parametric tests such as Kruskal-Wallis, and Mann-Whitney U. Are there packages that include most of these tests in each category? Many packages exist for an individual test but their outputs vary in great detail (test statistics, p-values, etc.)

I noticed that when I fit a linear model using 'lm' there is an attribute called "factors" that is added to the "term". It doesn't seem to appear for 'model.matrix', just 'lm'. I have been unable to find where it gets constructed or what it means? It looks like a two dimensional array that I may be able to use so I would just like to get some

Can someone enlighten me on why the following doesn't work? setwd('C:/Temp/R') d <- rep(1:53,2) (s <- ts(d, frequency=53, start=c(2000,10))) n <- length(s) k <- n%/%3 for(i in (n-k):n) { st <- c(start(s)[1] + (start(s)[2] + i)%/%frequency(s), (start(s)[2] + i) %% frequency(s)) ed <- c(start(s)[1] +

I noticed that the text() command adds text to a plot. Is there a way to either make the plot blank or add text to a "blank sheet". I would like to "plot" a page that contains just text, no plot lines, labels, etc. Suggestions? Kevin [[alternative HTML version deleted]]

I can get multiple plots on a page like: op <- par(mfcol = c(3, 1)) What I was wondering is if there is a way to have a title for the whole page? I can specify the title for each individual plot like: plot(xxx, main=".") But I would like a 'title' for the group of plots. Is this possible? Thank you. Kevin [[alternative HTML version deleted]]

In browsing the source I see the following construct: res <- switch(type, working = , response = r, deviance = , pearson = if (is.null(object$weights)) r else r * sqrt(object$weights), partial = r) I understand that 'switch' will execute the code that is matched by its corresponding string value (in this case 'type'). What I don't

I have two time series a <- ts(1:10, start=c(1,6), end=c(2,5), frequency=10) b <- ts(1:5, start=c(2,1), end=c(2,5), frequency=10) Obviously 'b' is a subset of 'a'. I want a single index value indicating where that start of 'b' lines up with the start of 'a'. So in this simple example I would expect an index of 5. I was playing with 'merge'.

What is wrong with the following? x <- 1:2 if(x[1] > 0) { if(x[2] > 0) { print("1 & 2 > 0") } else { print("1 > 0") } } else { if(x[2] > 0) {

I have some data where the frequency is heavily weighted on the lower end. So I have lots of low values with very few higher values. I would like to find breakpoints that cover the data with as much detail as possible. I find that if I use hist() to automatically find the breaks for me it finds breaks that are too coarse for the low values. I have tried the other algorithms (like 'Scott'

Hello, I cannot access the r-help website although after registration I am getting all the posts sent to the side. Each time I click on the "Visit Subscriber List" on the https://stat.ethz.ch/mailman/listinfo/r-help site, I get "R-help roster authentication failed." error. Any ideas? Cem [[alternative HTML version deleted]]

I can put a legend on a plot with something like: legend('bottom', leg.txt, horiz = TRUE, fill = colors) But what if the arrays leg.txt and colors are too big? I would still like to provide a legend but to save space I would like to just show small boxes with the color filled in so it will still fit on the plot. If I could adjust the size of the boxes (at least in one dimension)

Has anyone come across the right combinations to print a limited number of digits? My trial and error approach is taking too much time. Here is what I have tried: > op <- options() > a <- c(1e-10,1,2,3,.5,.25) > names(a) <- c("A", "B", "C", "D", "E", "F") > # default > a A B C D

I started to check what I thought I knew with autocovariance and it doesn’t jive with the the calculations given by ‘R’. I was wondering if there is some scaling or something that I am not aware of. Take the example Ø d <- 1:10 Ø (a <- acf(d, type="covariance", demean=FALSE, plot=FALSE)) Autocovariances of series ‘d’, by lag 0 1 2 3 4 5 6

I can get an array of strings for the data that I want using 'paste()' as follows: paste('ma', 1:am$arma[2], '=', coef(am)[1:am$arma[2] + am$arma[1]], sep='') This results in a vector of strings like: [1] "ma1=1.17760133668255" "ma2=0.649795570407939" "ma3=0.329456750858276" What I would like is fixed.pars <-

I was wondering what the best approach is for missing data in a time series. I give an example using xts but I would like to know what seems to be the "best" method. Say I have library(xts) xts.ts <- xts(1:4,as.Date(c("1970-01-01", "1970-1-3", "1980-10-10", "2007-8-19")), frequency=52) I would like to turn this into a time series (still

I would like the text plotted with 'mtext' to be alighned like it is for printing on the console. Here is what I have: > print(emt) ME RMSE MAE MPE MAPE MASE original -1.034568e+07 1.097695e+08 2.433160e+07 -31.30554 37.47713 1.5100050 xreg 1.561235e+01 2.008599e+03 9.089473e+02 267.05490 280.66734