similar to: Selecting first 7 elements

Hi, How to rbind these vectors from a list?: > l <- list(a = c(1, 2), b = c(1, 2, 3)) > l $a [1] 1 2 $b [1] 1 2 3 > do.call(rbind, l) [,1] [,2] [,3] a 1 2 1 b 1 2 3 Warning message: In function (..., deparse.level = 1) : number of columns of result is not a multiple of vector length (arg 1) > -J

Hello, Can anyone think of a non-iterative way to generate a decreasing geometric sequence in R? For example, for a hypothetical function dg, I would like: > dg(20) [1] 20 10 5 2 1 where I am using integer division by 2 to get each subsequent value in the sequence. There is of course: dg <- function(x) { res <- integer() while(x >= 1) { res <- c(res, x) x

Hi R Users, I have produced a simulated scatter plot of y versus x tightly clustered around the 45 degree line through the origin with the following code: x <- seq(1,100) y <- x+rnorm(100,0,10) plot(x,y,col="blue") abline(0,1) Is there some way to generate a 3-dimensional analogue of this? Can I get a similar simulated scatter plot of points in 3 dimensions where the points

dear useRs, i'm working with a mean reverting model of the following specification: y = mu + beta(x - mu) + errorterm, where mu is a constant currently I estimate just y = x (with lm()) to get beta and then calculate mu = estimated intercept / (1-beta). but I'd like to estimate mu and beta together in one regression-step and also get the test-statistics (including parameter

Hi, I have a matrix of results of multiple 2x2 chi^2 tests, non- significant tests are marked as TRUE. Is there a function for grouping those variables in a similar way LSD.test from agricolae library does? I reviewed LSD.test's source but it's not helpful for me. This is my matrix: [,1] [,2] [,3] [,4] [,5] [,6] [,7] 1 TRUE FALSE TRUE TRUE TRUE FALSE TRUE 2 FALSE TRUE

Hi folks, Debian 504 64-bit What is the correct syntax to check the available of a package on R repo? > available.packages("emacs", "OS_type=linux") Warning: unable to access index for repository emacs Package Version Priority Depends Imports LinkingTo Suggests Enhances OS_type License File Repository > available.packages("emacs") Warning: unable

Dear all, I have a list like this: l <- list(list(a=1,b=NULL), list(a=2,b=2)) I want to find out the elements with value of NULL and replace them with NA. The actual case has a very long list, so manually find out and replace them is not an option. I can use for loop to do this, but I want to know if there is vectorized way (or other ways) to do it? Thanks -- Wincent Rong-gui HUANG Doctoral

Dear R-users, I have installed the latest version of Tinn R (Version 2.2.0.2). With the older versions it was possible to run Tinn R next to R and then select code that one could send to R. Now it seems that one has to start the R gui in Tinn R and then it is possible to select en run code of R. But if I start R within Tinn R it does NOT start the R-Gui but only the R console. When I start R

Dear list, I have a question and I'm going to give an example of my problem f<- function(d1, d2, d3) { d<- d1*d2/d3 return(d) } v1<- 1 v2<- 2 If I try f(v1, v2, v3) Error in f(v1, v2, v3) : object "v3" not found I obviously got the above error message. I would like to add something to my function to allow me to get a certain value (say zero) if one of the arguments is

I've always found the chron library to be useful for tasks like this: > x <- round(runif(10)*100000, digits=0) > y <- as.Date(x, origin="1970-01-01") > library(chron) > days(y) [1] 7 25 26 25 10 24 1 31 12 8 31 Levels: 1 < 2 < 3 < 4 < 5 < 6 < 7 < 8 < 9 < 10 < 11 < 12 < 13 < ... < 31 Notice that it returns the days as

Hello! I have something like this: test1 <- data.frame(intx=c(4,3,1,1,2,2,3), status=c(1,1,1,0,1,1,0), x1=c(0,2,1,1,1,0,0), x2=c(1,1,0,0,2,2,0), sex=c(0,0,0,0,1,1,1)) and I can easily fit a cox model: library(survival) coxph(Surv(intx,status) ~ x1 + x2 + strata(sex),test1) However, I want to

Hi, I am looking for a R package to draw 3d plot. But not in a static way like scatterplot3d or stuff like that. I would like to make rotate the plot, to zoom/unzoom etc..., actually to render the graph 'dynamic'... (for the biologist and bioinformatician, a R package which permitts to reproduce what pdb viewer does). Does a Way exist to do something like that in R ? Thanks,

Hi R helpers, If I have a dataset looks like: ID record 1 20 . 30 . 25 2 26 . 15 3 21 4..................... And I want it becomes ID record 1 20 1 30 1 25 2 26 2 15 3 21 4..................... That is, I have to duplicate IDs for those with multiple records. I am wondering it is possible to be

Hello, How would you go about handling the following situation? This is on R 2.12.0 on Ubuntu 32-bit. I have a wrapper function to lm. I want to pass in a subset argument. First, I just thought I'd use "...". ## make example reproducible set.seed(123) df1 <- data.frame(age = rnorm(100, 50, 10), bmi = rnorm(100, 30, sd = 2)) ## create a wrapper using

All, I installed R-2.10.1 with Readline=no. Now for some reason R does not recognize some key strokes like the directional arrows. I am not sure if Readline is the problem or not. I have tried .Cofigure with Readline = yes but it doesn't fix the problem nor do I really know if readline is the problem to start with. Has anybody else run into similar problems? Thanks, Steve [[alternative

Hello R-Users! I am looking for an editor to be able to execute commands into R in Linux ubuntu. Is there any suggestion? Thanks. [[alternative HTML version deleted]]

Hi, Suppose a write a function a_fn<-function(arg1) { return(table(arg1)); } I have a column called AGE. Now I call the function c = a_fn(AGE); When a_fn is called, AGE is received in arg1. My question is, how do I access the actual name of the argument arg1? i.e, inside the function, i need to know that the actual name of arg1 is "AGE" in this case. Thanks in advance,

Hi, when I have made a decision tree with rpart, is it possible to "apply" this tree to a new set of data in order to find out the distribution of observations? Ideally I would like to plot my original tree, with the counts (at each node) of the new data. Reagards, Jay

Is there a less cryptic way to compare three or more values? allTheSame<-c("red","red","red","red") notAllTheSame<-c(132,132,132,999) all.identical <- function(vectorToTest){ cIdentical=sum(vectorToTest %in% vectorToTest[1]) return(cIdentical==length(vectorToTest)) } all.identical(allTheSame) all.identical(notAllTheSame) Thanks in

Hi, dear R users I am a newbie in R and I wantto use the method of meximum likelihood to fit a Weibull distribution to my survival data. I use "optim" as follows: optim(c(1, 0.25),weibull.like,mydata=mydata,method="L-BFGS-B",hessian = TRUE) My question is: how do I setup the constraints so that the two parametrs of Weibull to be pisotive? Or should I use other function