similar to: Transforming results of the summary function

Hi Im pretty new to R and I have run in to a problem. How do I name all the levels in this list. Lev1 <- c("A1","A2") Lev2 <- c("B1","B2") Lev3 <- c("C1","C2") MyList <- lapply(Lev1,function(x){ lapply(Lev2,function(y){ lapply(Lev3,function(z){ paste(unlist(x),unlist(y),unlist(z)) })})}) I would like to name the different

Hi, I have a data.frame like this: y <- rnorm(60) lev <- gl(3,20, labels=paste("lev", 1:3, sep="")) date1 <- as.Date(seq(ISOdate(2007,9,1), ISOdate(2007,11,5), by=60*60*24)) date1 <- date1[-c(3,4,15,34,38,40)] df <- data.frame(lev=lev, date1=date1, y=y) I would like to produce a new data.frame with missing days in df$date1 in each df$lev, like this: lev

#Hi R-users, #I have an example DF like this: y1 <- rnorm(10) + 6.8 y2 <- rnorm(10) + (1:10*1.7 + 1) y3 <- rnorm(10) + (1:10*6.7 + 3.7) y <- c(y1,y2,y3) x <- rep(1:3,10) f <- gl(2,15, labels=paste("lev", 1:2, sep="")) g <- seq(as.Date("2000/1/1"), by="day", length=30) DF <- data.frame(x=x,y=y, f=f, g=g) DF$wdays <- weekdays(DF$g)

Dear list members, I have problems to interpret the coefficients from a lm model involving the interaction of a numeric and factor variable compared to separate lm models for each level of the factor variable. ## data: y1 <- rnorm(20) + 6.8 y2 <- rnorm(20) + (1:20*1.7 + 1) y3 <- rnorm(20) + (1:20*6.7 + 3.7) y <- c(y1,y2,y3) x <- rep(1:20,3) f <- gl(3,20,

Hi R-users, I have a data.frame like this (modificated from https://stat.ethz.ch/pipermail/r-help/2007-August/138124.html). y1 <- rnorm(20) + 6.8 y2 <- rnorm(20) + (1:20*1.7 + 1) y3 <- rnorm(20) + (1:20*6.7 + 3.7) y <- c(y1,y2,y3) x <- rep(1:5,12) f <- gl(3,20, labels=paste("lev", 1:3, sep="")) d <- data.frame(x=x,y=y, f=f) and this is how I can

R-users E-mail: r-help@r-project.org Hi! R-users. I am just wondering what the definition of "dffits" in R language is. Let me show you an simple example. function() { library(MASS) xx <- c(1,2,3,4,5) yy <- c(1,3,4,2,4) data1 <- data.frame(x=xx, y=yy) lm.out <- lm(y~., data=data1, x=T) lev1 <- lm.influence(lm.out)$hat sig1 <-

Dear R Gurus, I am having a little difficulty with nlm. I've searched the archives and found nothing that tells me why this is occuring -- though there are some slightly similar issues. A simple example: lev2<-function(aaa,bbb,ccc,ddd,eee){ res<-aaa+bbb+ccc+ddd+eee res } nlm(lev2,p=c(32,4,5),ddd=45,eee=23) Error in f(x, ...) : argument "bbb" is missing, with no default

#Hi R-users, #Suppose that I have a data.frame like this: y1 <- rnorm(10) + 6.8 y2 <- rnorm(10) + (1:10*1.7 + 1) y3 <- rnorm(10) + (1:10*6.7 + 3.7) y <- c(y1,y2,y3) x <- rep(1:3,10) f <- gl(2,15, labels=paste("lev", 1:2, sep="")) g <- seq(as.Date("2000/1/1"), by="day", length=30) DF <- data.frame(x=x,y=y, f=f, g=g) DF$g[DF$x == 1]

i am trying to fit a linear model with both continuous covariates and factors. When fitted with the intercept term the first level of the factor is treated by R as intercept and the estimate of the effects of remaining levels(say i th level) are given as true estimate of i th level - estimate of 1st level.can any please help me? thanks in advance..... -- View this message in context:

Hi folks, I would like to trim the trailing spaces in my factor variables using lapply (described in this post by Marc Schwartz: http://tolstoy.newcastle.edu.au/R/e2/help/07/08/22826.html) but the code is not functioning (in this example there is only one factor with trailing spaces): y1 <- rnorm(20) + 6.8 y2 <- rnorm(20) + (1:20*1.7 + 1) y3 <- rnorm(20) + (1:20*6.7 + 3.7) y <-

Hi I have a data.frame with a grouping variable having the levels C, mild AD, mod AD, O and S since I want to compute a lda only for the two groups 'C' and 'mod AD' I call lda with data=subset(mydata.pca,GROUP == 'mod AD' | GROUP == 'C') my.lda <- lda(GROUP ~ Comp.1 + Comp.2 + Comp.3 + Comp.4+ Comp.5 + Comp.6 + Comp.7 + Comp.8 ,

Hi all, I didn't get a response to my post of this issue a week ago, so I've tried to clarify: When I use lme to analyze a model of nested random effects, the variance estimates of levels higher in the hierarchy appear to have much more variance than they should. In the example below with 4 levels, I simulate variance in level 2 (sd=1.0) and level 4 (sd=0.1), but levels 1 and 3 do

I have a data.frame object like: > data.frame(x=c('A','A','B','B'), y=c('Ab','Ac','Ba','Bd')) x y 1 A Ab 2 A Ac 3 B Ba 4 B Bd how could I create a tree structure object like this: |---Ab A---| _| |---Ac | | |---Ba B---| |---Bb Thanks, Zech [[alternative HTML version deleted]]

URL: http://www.tek-tips.com/gviewthread.cfm/lev2/3/lev3/19/pid/865/qid/336795 I have the same problem described in the above posting: using WinXP (he used Win2K, presumably WinNT does the same thing), you print something, but the print queue display shows an error message and doesn't show your job(s). The described solution worked for me too: add to /etc/samba/smb.conf [global] disable

Folks, Thanks to Ricardo Stella and "Chris" for quick assistance, I had sent about 15 help requests to listproc and given up, seeing no sign of a samba-digest list which is what my majordomo training has taught me. What a strange default setting . . . So onto my question about Samba. I've installed the latest RPM from the samba FTP site (samba-2.0.6-19991110) on a RedHat Linux

Dear all, I've got a problem in fitting multilevel model in lme. I don't know to much about that but suspect that something is wrong with my model. I'm trying to fit: m1<-lme(X~Y,~1|group,data=data,na.action=na.exclude,method="ML") m2<-lme(X~Y+Z,~1|group,data=data,na.action=na.exclude,method="ML") where: X - dependent var. measured on a scale ranging from

https://bugzilla.samba.org/show_bug.cgi?id=2947 wayned@samba.org changed: What |Removed |Added ---------------------------------------------------------------------------- Status|NEW |ASSIGNED ------- Additional Comments From wayned@samba.org 2005-08-03 16:45 ------- Keep in mind that --link-dest only hard-links regular

###Dear R users ###I have been using SensoMineR package from CRAN for most of my work in sensory data analysis and from my usage experience, I encountered some areas for improvement and considered ###modifying the function in SensoMineR package for my personal use. I felt that it could be useful to share this to the community for enabling adoption by other users where they might require a

Hello, assume I have an "unstructured" text line from a connection. Unfortunately, it is in string format: R> x [1] "\talpha0\t-0.638\t0.4043\t0.4043\t-2.215\t-0.5765\t-0.137\t501\t2000" How can I extract the data included in this string object "x" in order to get the elements for the parameter vector called "alpha0", i.e. -0.638 0.4043 0.0467

Hola a todos, Me gustaria saber como puedo evaluar / obtener el resultado guardado en un objeto cuando este se ejecuta. Quiero decir, el resultado de la siguiente expresion: > paste("res_UP_sep_dec_","16_17_18",sep="") [1] "res_UP_sep_dec_16_17_18" es un objeto (una lista) cuyo nombre es precisamente res_UP_sep_dec_16_17_18 >