similar to: Converting characters to numbers in data frames

Moi! A student here has been getting a bit irritated with some side effects of scale() (OS is Windows XP, the behaviour occurs in R 2.0.0, but not 1.7.1). The problem is that she scales a variable in a data frame, then does a regression, and tries to get some predictions for some new data. However, at this point she gets an error (see the example below). This seems to be because the

This is slightly off-topic, but there was a discussion about MCMC a couple of weeks ago. OpenBUGS can operate from R, at least in Windows (there are some problems in Linux at the moment). The interface with R is one are that there are plans to work on: it's a bit basic at the moment. Bob -- Bob O'Hara Dept. of Mathematics and Statistics P.O. Box 68 (Gustaf H??llstr??min katu 2b)

Hi List, A friend of mine recently asked the same question as Heinz T?chler. Since I've already written the code I'd like to share with the list. # x is an object returned by "survfit"; # "smed" returns a matrix of 5 columns of # n, events, median, 0.95LCL, 0.95UCL. # The matrix returned has rownames as the # group labels (eg., treatment arms) if any. smed <-

I can only assume I'm betraying my ignorance here, but this is not what I would expect. I'm getting the following from a stepwise selection (with both step and stepAIC): > step(lm(sqrt(Grids)~ SE + Edge + NH), scope=~ (Edge + SE + NH)^2) Start: AIC= 593.56 sqrt(Grids) ~ SE + Edge + NH Df Sum of Sq RSS AIC <none> 2147.0 593.6 + Edge:NH 1

Hi, I apologize for this often/old question. I found some hints but couldn't solve the problem so far. I have C functions (incl. the header files) as well as the R wrapper functions which I want to use for faster calculations. These functions are included in a R package. The installation process seems to be ok (no errors). I also can load the package without errors. But when I call the

The background: I'm trying to fit a Poisson-lognormal distrbutuion to some data. This is a way of modelling species abundances: N ~ Pois(lam) log(lam) ~ N(mu, sigma2) The number of individuals are Poisson distributed with an abundance drawn from a log-normal distrbution. To fit this to data, I need to integrate out lam. In principle, I can do it this way: PLN1 <- function(lam, Count,

I have a simple (!) problem. What I want to do is to create a legend with Greek letters, and substituting numbers into the legend. Like this: mu=1:3 Mean=rep(mu, each=20) plot(runif(60), rnorm(60,Mean,0.1), pch=Mean) # create a plot legend(0.6,1.7, paste("mu =", 1:3), pch=mu) but with a Greek letter mu. I think the solution has something to do with substitute() possibly along with

I'm using the RIP Linux mini rescue distribution. http://www.tux.org/pub/people/kent-robotti/looplinux/rip/ I have modified v. 7.0 for usage on our 32 node Tyan Tiger MPX Linux cluster with a serial console. Because of a bug with Grub on this motherboard when running headless and using a serial console I use the optional Isolinux version 2.08 also included with RIP Linux to boot from a

Please consider a data frame A B C D 1 4 5 0 2 3 2 75 3 4 1 84 4 5 1 90 5 3 0 100 Is there a way to plot column B and C as barplot *and* D as line on the same graph?? R-2.0.1 powered by Mandrake 10.1 ------------------------------------------------------------------------------------------------------------------------- Landini dr. Massimiliano Tel. mob. (+39) 347 140 11 94 Tel./Fax. (+39)

Dear All, I have a data frame of vectors of publication names such as 'pub': pub1 <- c('Brown DK, Santos R, Rome DF, Don Juan X') pub2 <- c('Benigni D') pub3 <- c('Arstra SD, Van den Hoops DD, lamarque D') pub <- rbind(pub1, pub2, pub3) I would like to construct a dataframe with only author's last name and each last name in columns and the

Hello, I am currently using R to run an external program and then read the results the external program sends to the stdout which are tsv data. When R reads the results in it converts it to to a list of strings which I then have to maniuplate with a whole slew of commands (which, figuring out how to do was a reall challenge for a newbie like myself)--see below. Here's the code I'm

4000:6000 gives you 4000, 4001, ..., 6000. I suspect you want population= c(seq(4000, 6000, length=5), seq(3500, 4300, length=5), seq(3000, 3200, length=5)) Bob On 20 September 2017 at 17:07, Shivi Bhatia <shivipmp82 at gmail.com> wrote: > Hi Team, > > I using the syntax as: > > data.df<- data.frame( > city= c(rep(c("Delhi",

Dear friends, Over the past weeks, I have been asking a lot of questions about how to use R in Bayesian analysis. I am brand new to R, but I am very pleased with it. I started with winbugs but I found winbugs to be a limited software, not bad but has several limitations. By contrast, R allows the analyst to tackle any problem with a huge set of tools for any kind of analysis. I love R. In

Hello, can anyone tell me if R has any special function for simulating the structure of human populations? Something like the genetic algorithm? I need to simulate a sample of a population with a specific structure. Is there something on R that can help me? Thanks to everyone. __________________________________________________________

Hello, I am working on extracting data from sound recording (fee bee song of the black-capped chickadee). I obtained a two column matrix with x=time(s) and y=frequency(khz). The part of the recording that is interesting me is when the frequency is stable. Does somebody ever used a fonction that extract only constant data among a huge list of data (sample= 180000)? Thanks Thibault Grava

The loop never assigns anything to d0, only t. The first line makes t a character string "d0$V1" (or "d0$V2" etc.). The second line assigns either 0 or 1 to t. Looking at this, I don't think you've got into the R psychology (bad news if you want to use R, good news in many other ways). I assume d0 is a list, so could you put the V's into a vector, and then just use

It might help if you provided the code you used. It's possible that you didn't use direction="backward" in stepAIC(). Or if you did, it was still running, so whatever else you try will still be slow. The statement "R provides only the pvalues for each level" is wrong: look at the anova() function. Bob On 29 June 2017 at 11:13, Beno?t PELE <benoit.pele at

Dear R-help I am using R version 2.6.2. I am trying to subtract specific values from a larger data frame. I feel this should be straightforward, but I am struggling. I have a dataframe "Bk" as follows: DateTime cumPrecip 01/01/2008 00:00 348 01/01/2008 01:00 348 01/01/2008 02:00 348 01/01/2008 03:00 347 01/01/2008 04:00 348 01/01/2008 05:00 348 01/01/2008 06:00 349

Dear all, I have several questions regarding fisher.test() in R, and I'd highly appreciate any help with it. I have a group of observations, each having people's income, and an indicator of whether selected in or out a program. I want to test the difference between income of people who are in and out. Because the distribution is far from normal, I decide to use the fisher's exact

Dear all, I have several questions regarding fisher.test() in R, and I'd highly appreciate any help with it. I have a group of observations, each having people's income, and an indicator of whether selected in or out a program. I want to test the difference between income of people who are in and out. Because the distribution is far from normal, I decide to use the fisher's exact