similar to: SAPPLY function XXXX

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 everyone, What is the best way to have R print a title and then skip a line (both without the [1] line preffix) when writing a function that prints an object (to the console)? Simplified example: fn3<-function(x,y){ c1<-cov(x,y) print(c1)} data1<-data.frame(age=rnorm(50),weight=rnorm(50)) attach(data1) fn3(age,weight) ===== How do I have R print a nicely formatted title

Hello everyone, I am attempting to use the %in% operator with the ! to produce a NOT IN type of operation. Why does this not work? Suggestions? > data2[data1$char1 %in% c("string1","string2"),1]<-min(data1$x1) > data2[data1$char1 ! %in% c("string1","string2"),1]<-max(data1$x1)+1000 Error: unexpected '!' in "data2[data1$char1

Hello, The application ChairGun2 can't be started up. When starting it shows the mistake: Code: Run-time error ?429? ActiveX conponent can?t create object Wine version 1.2.3 for Cent OS 6.2. is used. This application also didn't work on older versions. What should I do? How to solve the problem? Download ChairGun2: http://turbobit.net/sj18namk8947.html

Hello everyone, I have the following call to sapply() and error message. Is the most efficient way to deal with this to make sum(!is.na(x)) a function in a separate line prior to this call? If not, please advise. N.Valid=sapply(x,sum(!is.na(x))) Error in match.fun(FUN) : 'sum(!is.na(x))' is not a function, character or symbol Thanks! Dan [[alternative HTML version deleted]]

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 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

Hi, I apologize if this has previously been posted. I've just subscribed to the R-help digest. I'm writing a plotting function that uses layout() to plot several different plots on the same device. This function uses plot(), image(), and a custom function that uses text(). Each cell of the layout needs different par() parameters, so what I'd like to do is pass them as lists:

Hi all I'm trying to use the 'by' function to extract the co-efficients from a mixed model which is performed on multiple individuals. I basically have a group of individuals and for each individual I want the co-efficient for there change in 'pots_hauled' in response to a change in 'vpue' with my random variable being 'ns_a_vpue'. The problem I am having is

Dear R-users, I have 1 remark and 1 question on the inclusion of interactions in the gam function from the gam package. I need to fit quantitative predictors in interactions with factors. You can see an example of what I need in fig 9.13 p265 from Hastie and Tibshirani book (1990). It's clearly stated that in ?gam "Interactions with nonparametric smooth terms are not fully

Hello R-helpers, I have 10 dataframes (named data1, data2, ... data10) and I would like to add 5 new columns to each dataframe using the following code: data1$LogDepth<-log10(data1[,2]/data1[,4]) data1$LogArea<-log10(data1[,3]/data1[,5]) data1$p<-2*data1[,6]/data1[,7] data1$Exp<-data1[,2]^(2/data1[,8]) data1$s<-data1[,3]/data1[,9] ...but I would prefer not to repeat this chunk of

hello folks, Im trying to clean out a large file with data i dont need. The column im manipulating in the file is called "legal_status" There are three kinds of rows i want to remove. Those that have "Private", "Private (Op", or "Unknown" in the legal_status column. I wrote this code but i get errors and it says im missing a TRUE/ False thingy...im

I am trying to run my first app on the Solaris server, in a production environment. I get a network unreachable. Why? Does it belongs to database.yml config? Processing CategoriesController#index (for 10.3.70.129 at 2010-01-05 14:00:47) [GET] Errno::ENETUNREACH (Network is unreachable - connect(2)): /usr/ruby-enterprise/lib/ruby/1.8/net/http.rb:560:in `initialize''

Hi This question is far less simple than the title suggests, please read carefully, thanks. I have 2 sets of data, both read into R >data1<-read.table ("1.txt", header=T, sep="\t") >data2<-read.table ("2.txt", header=T, sep="\t") >data1 Taxon stage1 stage2 stage3 stage4 T1 0 0 1 1 T2 0

Hi there, data1<-matrix(data=c(1,1.2,1.3,"3/23/2004",1,1.5,2.3,"3/22/2004",2,0.2,3.3,"4/23/2004",3,1.5,1.3,"5/22/2004"),nrow=4,ncol=4,byrow=TRUE) data1<-data.frame(data1) names(data1)<-c("areaid","x","y","date") data1 areaid x y date 1 1 1.2 1.3 3/23/2004 2 1 1.5 2.3 3/22/2004 3 2

I am trying to find methods for testing and visualizing calibration to Cox models with time-depended coefficients. I have read this nice article <http://journals.sagepub.com/doi/10.1177/0962280213497434>. In this paper, we can fit three models: fit0 <- coxph(Surv(futime, status) ~ x1 + x2 + x3, data = data0) p <- log(predict(fit0, newdata = data1, type = "expected")) lp

I need help in subseting a dataframe: data1<-data.frame(year=c(2001,2002,2003,2004,2001,2002,2003,2004, 2001,2002,2003,2004,2001,2002,2003,2004), firm=c(1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4),x=c(11,22,-32,25,-26,47,85,98, 101,14,87,56,12,43,67,54), y=c(110,220,302,250,260,470,850,980,1010,140,870,560,120,430,670,540)) data1 I want to keep the firms where all x>0 (where there are

Hi everybody, I'm using the method described here to make a linear regression: http://www.apsnet.org/education/advancedplantpath/topics/Rmodules/Doc1/05_Nonlinear_regression.html > ## Input the data that include the variables time, plant ID, and severity > time <- c(seq(0,10),seq(0,10),seq(0,10)) > plant <- c(rep(1,11),rep(2,11),rep(3,11)) > > ## Severity

Here is the example. > age=18:29 > height=c(76.1,77,78.1,78.2,78.8,79.7,79.9,81.1,81.2,81.8,82.8,83.5) > type=c("A", "B", "C", "D","A", "B", "C", "D","A", "B", "C", "D") >

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