similar to: duplicate 'row.names' are not allowed

Dear All I would need to generate differences between variates such as nam1<-nam2-nam3 in the following loop: for(i in c("13","26","38")) { for (j in c("HR","PQ","QRS","QT")){ nam1<-paste("d",j,i,sep="") nam2<-paste(j,i,sep=".")

Hi, I have gote the following data x1 <- c(rep(1,6),rep(4,7),rep(6,10)) x2 <- rnorm(length(x1),6,1) data <- data.frame(x1,x2) and I would like to compute the mean of the x2 for each individual of x1, i. e. x1=1,4 and 6? Thank you very much in advance, Amori [[alternative HTML version deleted]]

Here is a small reproducible example: data <- structure(list(V1 = structure(1:3, .Label = c("Name1", "Name2", "Name3"), class = "factor"), V2 = structure(c(1L, 3L, 2L), .Label = c("nam1", "name-1", "name_12"), class = "factor"), V3 = structure(1:3, .Label = c("nam2", "nam_34",

Hi all, I am trying to compare two distance matrices with R. I would like to create a XY plot of these matrices and do some linear regression on it. But, I am a bit new to R, so i have a few questions (I searched in the documentation with no success). The first problem is loading a distance matrix into R. This matrix is the output of a the Phylip program Protdist and lookes like this: 5

Hi everyone,   Is there any counter function in R for the following purpose: x <- matrix(c(1,1,0,2,1,0,0,2,0,1,2,1,2,1,0,1),nrow=4) As I would like to know how many zeros, ones, and twos in each row of x?   Many thank in advance, Amor     [[alternative HTML version deleted]]

Hello everbody, I want to generate different subsets of my data-set and safe this subsets with names listet in a vector. Because in reality I have got about 70 subsets I want to realize this in a loop Somehow like this: names=c("nam1",

Hi gues,   Is there any function in R for boxplot with different time points? t1 <- c(rep(1,20),rep(2,20)) t2 <- c(rep(1,10),rep(2,10),rep(1,10),rep(2,10)) x <- rnorm(40,5,1) dat <- data.frame(t1,t2,x) boxplot(x~t1,t2) Many thanks, Amor [[alternative HTML version deleted]]

Hi,   I have the following data and I would like to delete douple names, it is almost similar to SAS PROC SORT nodupkey! Is there any function in R does this?   x1 <- rnorm(11,5,1) x2 <- runif(11,0,1) nam <-paste("A", c(1:4,4,5:9,9), sep=".") mydata <- data.frame(x1,x2) crownames(mydata) <- nam   Many thanks in advance, Amor [[alternative HTML version

Hello everybody,   I wonder if there is any swap function in R that does the following: x <- seq(1,10,12) x1 <- swap(x) x1 1 8 3 4 5 6 7 2 10 Thank you very much!   Amor __________________________________________________ Schutz gegen Massenmails. [[alternative HTML version deleted]]

Untitled Document Áö¿ªÀ» °ü¸®ÇÏ½Ç ÁÖºÎÆÀÀåÀ» ÃʺùÇÕ´Ï´Ù ! ´ç»ç´Â ±èÄ¡³ÃÀå°í Àü¹®À¯Åë¾÷ü·Î¼­ °¢ Áö¿ªÀ» °ü¸®ÇÏ½Ç ÁÖºÎÆÀÀå´ÔÀ» ¸ð½Ê´Ï´Ù ÇöÀç ±èÄ¡³ÃÀå°í º¸±ÞÀ²Àº ´ëµµ½Ã°¡ 25%,Áö¹æÀº 15% ¼öÁØ¿¡ ºÒ°ú ÇÏÁö¸¸ À̹ø³âµµ ºÎÅÍ ¸¹Àº ÆǸÅÁõ°¡·Î ÀÎÇØ ³â°£ ÆǸŴë¼ö°¡ ÀÌ¹Ì TV³ª ³ÃÀå°í¿¡ À°¹ÚÇÏ°í ÀÖÀ¸¸ç ¾ÕÀ¸·Î ´õ¿í ´õ ¸¹Àº ÁֺεéÀÌ ±èÄ¡³ÃÀå°í¸¦ ±¸ÀÔ ÇÒ °ÍÀ¸·Î ¿¹»óµÇ°í ÀÖ½À´Ï´Ù.

I am trying to call a funtion within another function and I clearly am misunderstanding what I should do. Below is a simple example. I know lstfun works on its own but I cannot seem to figure out how to get it to work within ukn. Basically I need to create the variable "nts". I have probably missed something simple in the Intro or FAQ. Any help would be much appreciated. EXAMPLE

Hello,   I am using mod1 <- lrm(y~x1+x2,na.action=na.pass,method="lrm.fit") summary(mod1) and I've got the following error: Error in summary.Design(mod1) : adjustment values not defined here or with datadist for x1 x2   Many thank, Amor [[alternative HTML version deleted]]

I am sorry, I'd like to split my column ("names") such that all the beginning of a string ("X..") is gone and only the rest of the text is left. x<-data.frame(names=c("X..aba","X..abb","X..abc","X..abd")) x$names<-as.character(x$names) (x) str(x) Can't figure out how to apply strsplit in this situation - without using a

Dear Elisa, Try this: el<- matrix(1:100,ncol=20) ?set.seed(25) ?el1<- matrix(sample(1:100,20,replace=TRUE),ncol=1) In the example you showed, there were no column names.? ?list(el[,sort(el1)[1:3]],sort(el1,index.return=TRUE)$ix[1:3]) #[[1]] ?# ?? [,1] [,2] [,3] #[1,]?? 31?? 61?? 71 #[2,]?? 32?? 62?? 72 #[3,]?? 33?? 63?? 73 #[4,]?? 34?? 64?? 74 #[5,]?? 35?? 65?? 75 # #[[2]] #[1] 9 5 3 A.K.

List, here''s how I got 4.1 up on suse 11.4 There is a small bug with the xencommons init script preventing xenstored from starting. Add this repo for 4.1 RPM''s ------------- zypper ar http://download.opensuse.org/repositories/Virtualization/openSUSE_11.4 Virt zypper ref zypper in xen-4.1.0_01-106.1 Fix the xen inits  (observe the order here) ------------ chkconfig -s

Is this the solution? > d1<- as.data.frame(lapply(data,as.character),stringsAsFactors=FALSE) > str(d1) 'data.frame': 3 obs. of 5 variables: $ V1: chr "Name1" "Name2" "Name3" $ V2: chr "nam1" "name_12" "name-1" $ V3: chr "nam2" "nam_34" "name-2" $ V4: chr "nam3"

Hi everyone- I have a dataset with multiple "pre" and "post" variables I want to compare. The variables are named "apple_pre" or "pre_banana" with the corresponding post variables named "apple_post" or "post_banana". The variables are in no particular order. apple_pre orange_pre orange_post pre_banana apple_post post_banana person_1

Dear R users,Sorry for such a basic question. I really need to know that how can i pick the indices of 5 lowest values from each row of a matrix with dimensions 12*12??Thank you very much in advance Elisa [[alternative HTML version deleted]]

I am not really sure what the warning means but I think your underlying problem is that all your variables are factors. Did you intend the values in each variable to be character? data.frame':??? 3 obs. of? 5 variables: ?$ V1: Factor w/ 3 levels "Name1","Name2",..: 1 2 3 ?$ V2: Factor w/ 3 levels "nam1","name-1",..: 1 3 2 ?$ V3: Factor w/ 3 levels

Untitled Document Áö¿ªÀ» °ü¸®ÇÏ½Ç ÁÖºÎÆÀÀåÀ» ÃʺùÇÕ´Ï´Ù ! ´ç»ç´Â ±èÄ¡³ÃÀå°í Àü¹®À¯Åë¾÷ü·Î¼­ °¢ Áö¿ªÀ» °ü¸®ÇÏ½Ç ÁÖºÎÆÀÀå´ÔÀ» ¸ð½Ê´Ï´Ù ÇöÀç ±èÄ¡³ÃÀå°í º¸±ÞÀ²Àº ´ëµµ½Ã°¡ 25%,Áö¹æÀº 15% ¼öÁØ¿¡ ºÒ°ú ÇÏÁö¸¸ À̹ø³âµµ ºÎÅÍ ¸¹Àº ÆǸÅÁõ°¡·Î ÀÎÇØ ³â°£ ÆǸŴë¼ö°¡ ÀÌ¹Ì TV³ª ³ÃÀå°í¿¡ À°¹ÚÇÏ°í ÀÖÀ¸¸ç ¾ÕÀ¸·Î ´õ¿í ´õ ¸¹Àº ÁֺεéÀÌ ±èÄ¡³ÃÀå°í¸¦ ±¸ÀÔ ÇÒ °ÍÀ¸·Î ¿¹»óµÇ°í ÀÖ½À´Ï´Ù.