similar to: Regression with nominal data

### example:start v <- sample(rnorm(200), 100, replace=T) k <- rep.int(c("locA", "locB", "locC", "locD"), 25) tapply(v, k, summary) ### example:end ... (hopefully) produces 4 summaries of v according to k group membership. How can I transform the output into a nice table with the croups as columns and the interesting statistics as lines? Thx,

Hello, how can I return the name of a variable, say "a$b", from a function? fun <- function(x){ return(substitute(x)); } a <- data.frame(b=1:10); fun(a$b) ... returns a$b, but this is a type language, thus I can't use it as a character string, can I? How? Thanks for help, S?ren

Hello, after the creation of a data.frame I like to add NAs as follows: n <- 743; x <- runif(n, 1, 7); Y <- runif(n, 1, 7); Ag6 <- runif(n, 1, 7); df <- data.frame(x, Y, Ag6); # a list with positions: v <- apply(df, 2, function(x) sample(n, sample(1:ceiling(5*n/100), 1), repl=F)); # a loop too much? for (i in 1:length(df)){ df[unlist(v[i]), i] <- NA; } summary(df); This

I have created a function to do something: i <- factor(sample(c("A", "B", "C", NA), 793, rep=T, prob=c(8, 7, 5, 1))) k <- factor(sample(c("X", "Y", "Z", NA), 793, rep=T, prob=c(12, 7, 9, 1))) mytable <- function(x){ xtb <- x btx <- x # do more with x, not relevant here cat("The table has been created,

A Likert scale may have produced counts of answers per category. According to theory I may expect equality over the categories. A statistical test shall reveal the actual equality in my sample. When applying a chi square test with increasing number of repetitions (simulate.p.value) over a fixed sample, the p-value decreases dramatically (looks as if converge to zero). (1) Why? (2) (If

How to I "recode" a factor into a binary data frame according to the factor levels: ### example:start set.seed(20) l <- sample(rep.int(c("locA", "locB", "locC", "locD"), 100), 10, replace=T) # [1] "locD" "locD" "locD" "locD" "locB" "locA" "locA" "locA"

A list contains several matrices. Over all matrices (list elements) I'd like to access one matrix cell: m <- matrix(1:9, nrow=3, dimnames=list(LETTERS[1:3], letters[1:3])) l <- list(m1=m, m2=m*2, m3=m*3) l[[3]] # works l[[3]][1:2, ] # works l[[1:3]][1, 1] # does not work How can I slice all C-c combinations in the list? S?ren -- S?ren Vogel, Dipl.-Psych. (Univ.), PhD-Student, Eawag,

Hello, I read the help as well as the examples, but I can not figure out why the following code does not produce the *given* row names, "x" and "y": x <- 1:20 y <- 21:40 rbind( x=cbind(N=length(x), M=mean(x), SD=sd(x)), y=cbind(N=length(y), M=mean(y), SD=sd(y)) ) Could you please help? Thank you S?ren

A dataframe holds 3 vars, each checked true or false (1, 0). Another var holds the grouping, r and s: ### start:example set.seed(20) d <- data.frame(sample(c(0, 1), 20, replace=T), sample(c(0, 1), 20, replace=T), sample(c(0, 1), 20, replace=T)) names(d) <- c("A", "B", "C") e <- rep(c("r", "s"), 10) ### end:example How do I get the

Hello, I am writing as an administrator, not as an R user, so forgive me if I am not completely knowledgeable about R. I have a user who is creating an R package for windows from a Linux environment using the crossbuild environment by Jun Yan and A.J. Rossini. The packages she generated worked fine until she tried to install in R 1.9.1 for Windows. Now when she installs with

Hello, r-help at r-project.orgbarplot(twcons.area, beside=T, col=c("green4", "blue", "red3", "gray"), xlab="estate", ylab="number of persons", ylim=c(0, 110), legend.text=c("treated", "mix", "untreated", "NA")) produces a barplot very fine. In addition, I'd like to get the

r11 -- r16 are variables showing a reason for usage of a product in 6 different situations. Each variable is a factor with 4 levels imported from a SPSS sav file with labels ranging from "not important" to "very important", and NA's for a sample of N = 276. (1) I need a chi square test of independence showing that the reason does not differ depending on the

Dear R-user: After the fitting the Tobit model using zelig, if I use the following command then I can get the regression coefficents: beta=coefficients(il6.out) > beta (Intercept) apache 4.7826 0.9655 How may I extract the "Naive SE" from the following output please? > summary(il6w.out) Call: zelig(formula = il6.data$il6 ~ il6.data$apache, model =

Could you help me with a problem? I should put non-linear variables into zelig-model, how can that be done? I'm dealing with air pollution data, trying to find out daily associations between mortality and air pollutants. Weather variables used as confounders are in some cases non-linear. Since smoothing is not an option I don't know how to proceed. Thanks, Jaana

Hello a <- c("Mama", "Papa", "Papa; Mama", "", "Sammy; Mama; Papa") a <- strsplit(a, "; ") mama <- rep(F, length(a)) mama[sapply(a, function(x) { sum(x=="Mama") }, simplify=T) > 0] <- T papa <- rep(F, length(a)) papa[sapply(a, function(x) { sum(x=="Papa") }, simplify=T) > 0] <- T # ... more

Greetings all, please consider the following data: #Build Data frame Slope<-c(1.291370, 12.208500, 2.110930, 0.578990, 5.019520, 0.807444, 0.554079 , 1.257080, 0.241504 , 0.184337 , 0.383044 , 0.342021) Exposure<-c(790.54, 1167.79 , 845.58 , 1082.47 , 1189.61 , 677.17 , 2058.56 , 469.09 , 112.02 , 803.31 , 254.14 ,1336.16) FwyDist<-c(11809.4222 ,10623.0458, 12279.6271,

In https://stat.ethz.ch/pipermail/r-help/2008-March/156868.html I found what linear separability means. But what can I do if I find such a situation in my data? Field (2005) suggest to reduce the number of predictors or increase the number of cases. But I am not sure whether I can, as an alternative, take the findings from my analysis and report them. And if so, how can I find the linear

Dear Group I have few errors while installing package rattle from CRAN i do the installing from the local zip files... I am using R 2.4.0 do i have to upgrade to R2.4.1 ? ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ utils:::menuInstallLocal() package 'rattle' successfully unpacked and MD5 sums checked updating HTML package descriptions > help(rattle) No

our group at Duke is currently attempting to connect two data sets fed through a web-based system, one containing research questions (i.e., variables positioned with certain roles) and a database of statistical methods. this connection is done informally in a number of R packages and related software such as Rcmdr, R4calc, and Rkward where a variable of a certain type is matched to a certain

Pls advise how I can use R in conjoint analysis?? regds Faisal Afzal Siddiqui Karachi, Pakistan ____________________________________________________________________________________ Looking for last minute shopping deals?