similar to: how to add new rows in a dataframe?

Hi, Your question is still not clear. May be this helps: dat2<- read.table(text=" patient_id????? t???????? scores 1????????????????????? 0??????????????? 1.6 1????????????????????? 1??????????????? 2.6 1????????????????????? 2???????????????? 2.2 1????????????????????? 3???????????????? 1.8 2????????????????????? 0????????????????? 2.3 2?????????????????????? 2???????????????? 2.5

Hi, It would be better if you provided the output of dput(dataset).? I am not sure about the structure of your dataset. Just from reading the data as is shown. dat1<- read.table(text=" separator,tissID >,>,2 ,2,1 ,6,5 ,11,13 >,>,4 ,4,9 ,6,2 ,7,3 ,21,1 ,23,58 ,25,9 ,26,4 >,>,11 ,1,12 >,>,21 ,4,1 ,11,3

Hi Elisa, I hope this is what you wanted. dat1<-read.csv("peaks.csv",sep=",") #Subset dat2<-dat1[1:5,] res1<-do.call(cbind,lapply(seq_len(nrow(dat2)),function(i) do.call(rbind,lapply(split(rbind(dat2[i,],dat2[-i,]),1:nrow(rbind(dat2[i,],dat2[-i,]))), function(x) {x1<-rbind(dat2[i,],x);

Hi there, I have some data which are convenient to enter as lists. For example: t1<-list(fname="animal1",testname="hyla",dspkr="left",res1=39.7,res2=15.0) t2<-list(fname="animal1",testname="bufo",dspkr="left",res1=14.4,res2=56.1)

Hi R helpers! I have a question. I'm trying to create a function for an exercise. Here are the arguments I should include: x and y are numeric z is a name ("plus","minus","multiply","divide") and swap is logical. Here is what the function should do: When z="plus", then x+y is performed and so on for the other z names. It should give a NA

Dear R users, I have a data structure as follows: id two res1 res2 c1 c2 inter 1 -0.786093166 1 0 1 2 6 3 -0.308495749 1 0 0 1 2 5 -0.738033048 1 0 0 0 1 7 -0.52176252 1 0

I am using R 1.1 with Redhat 6.2 and RW 1.001 with Win98 (the upkey doesn't work on my IBM either as has been previously reported by others). The function aov doesn't return either the residuals or the residual degrees of freedom for split-plot designs. If you use the following code from Baron and Li's "Notes on the use of R for psycology experiments and questionnaires"

Thank you very much for your reply. Your code work well with this example. I modified a little to fit my real data, I got an error massage. Error in split.default(x = seq_len(nrow(x)), f = f, drop = drop, ...) : Group length is 0 but data length > 0 On Thu, Jan 31, 2013 at 12:21 PM, arun kirshna [via R] < ml-node+s789695n4657196h87@n4.nabble.com> wrote: > Hi, > Try this: >

Dear R-users I have a problem choosing initial points for the function arms() in the package HI I intend to implement a Gibbs sampler and one of my conditional distributions is nonstandard and not logconcave. Therefore I'd like to use arms. But there seem to be a strong influence of the initial point y.start. To show the effect I constructed a demonstration example. It is reproducible

Hi, I use lme to fit models like R> res1 <- lme(y~A+B, data=mydata, random=~1|subject) R> res2 <- lme(y~B+A, data=mydata, random=~1|subject) (only difference between these two models are the sequence in which the indep variables are written in formula) where y is continuous and A, B, and subject are factors. To get ANOVA table I used R> anova(res1) R> anova(res2) and found

Hi, I am new to clustering and was wondering why pvclust using "maximum" as distance measure nearly always results in p-values above 95%. I wrote an example programme which demonstrates this effect. I uploaded a PDF showing the results Here is the code which produces the PDF file: ------------------------------------------------------------------------------------- s <-

Dear R-users, I would like to use optim( ) to minimize a function which depends on 4 parameters: 2 vectors, a scalar, and a matrix. And I have a hard to define the parameters at the beginning of the function, and then to call optim. Indeed, all the examples I have seen dont treat cases where parameters are not all real. Here is my code, it doesnt work but its just to show you where is exactly my

Hi, Suppose, if I create 15 files in my working directory. set.seed(48) lapply(1:15,function(i) {m1 <- matrix(sample(1:20,1686*2,replace=TRUE),nrow=1686,ncol=2); write.table(m1,paste0("file_",i,".txt"),row.names=FALSE,quote=FALSE)}) ?D <-dir() D1 <- D[order(as.numeric(gsub("\\D+","",D)))] D1 ?res <- t(sapply(D1,function(x) {x1<-

Dear R-Users, I want to use mtable from package "memisc" to produce Latex-style estimation output. However, mtable() only gives me a p-value and not the corresponding test-statistic. Does anyone know how to extract it, either from a glm/anova object or mtable? Here is a short example: # Run this #################### install.packages("memisc") library(memisc) set.seed(1)

I initially thought, this should better be posted to r-devel but alas! no response. so I try it here. sory for the lengthy explanation but it seems unavoidable. to quickly see the problem simply copy the litte example below and execute f(n=5) which crashes. called with n != 5 (and of course n>3 since there are 3 parameters in the model...) everything is as it should be. in detail:

HI, Using c11<- 0.01 c12<- 0.01 c1<- 0.10 c2<- 0.10 One possible problem is that: dim(res5) #[1] 513? 20 res6<-aggregate(.~m1+n1+m+n,data=res5[,c(1:6,9:12,21:24)] ,max) #Error in `[.data.frame`(res5, , c(1:6, 9:12, 21:24)) : ?# undefined columns selected A.K. ________________________________ From: Joanna Zhang <zjoanna2013 at gmail.com> To: arun <smartpink111 at

Deal all, as MCMClogit does not allow for the specification of several chains, I have run my model 3 times with different random number seeds and differently dispersed multivariate normal priors. For example: res1 = MCMClogit(y~x,b0=0,B0=0.001,data=mydat, burnin=500, mcmc=5500, seed=1234, thin=5) res2 = MCMClogit(y~x,b0=1,B0=0.01,data=mydat, burnin=500, mcmc=5500, seed=5678, thin=5) res3 =

I noticed Doug only circulated the following on S-news, but it may be of interest to R users who don't follow S-news. His findings are certainly consistent with my own, where my problems often force me into the element by element type of situation. Paul Gilbert _______ At the risk of beating this example to death, I went back and compared the execution time for the element-by-element method

I am trying to run separate regressions for different groups of observations using the lapply function. It works fine when I write the formula inside the lm() function. But I would like to pass formulae into lm(), so I can do multiple models more easily. I got an error message when I tried to do that. Here is my sample code: #generating data x1 <- rnorm(100,1) x2 <- rnorm(100,1) y <-

What is going on here? In the lines ending in #### the inputs and outputs are identical yet one gives a warning and the other does not. a1 <- `rownames<-`(anscombe[1:3, ], NULL) a2 <- anscombe[1:3, ] ix <- 5:8 # input arguments to #### are identical in both cases identical(stack(a1[ix]), stack(a2[ix])) ## [1] TRUE identical(a1[-ix], a2[-ix]) ## [1] TRUE res1 <-