similar to: Return a variable name

### 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,

Hi, y is nominal (3 categories), x1 to 3 is scale. What I want is a regression, showing the probability to fall in one of the three categories of y according to the x. How can I perform such a regression in R? Thanks for your 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,

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 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

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

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

Hi, how can I attach variable labels originally read by read.spss() to the resulting variables? <pre> X <- read.spss('data.sav', use.value.labels = TRUE, to.data.frame = TRUE, trim.factor.names = TRUE, trim_values = TRUE, reencode = "UTF-8") names(X) <- tolower(names(X)) attach(X) </pre> Thank you S?ren

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

Can't make sense of calculated results and hope I'll find help here. I've collected answers from about 600 persons concerning three variables. I hypothesise those three variables to be components (or indicators) of one latent factor. In order to reduce data (vars), I had the following idea: Calculate the factor underlying these three vars. Use the loadings and the original var

Dear all, I have 100 files which are used as input.and I have to input the name of my files again and again.the name of the files are 1.out, 2.out......100.out. I want to know if there is anything like perl so that i can use something like this- for($f = 1; $f <= 100; $f++) { $file = $f.".out"; I have tried this thing in R but it does not work.Can somebody please help me.

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:

Some help with how to re-label the vertical axis in a histogram would be appreciated. qplot(off.sc,weight=rel.freq,binwidth=.29,main="test Figure"+ylab("New from inside"))+ylab("New from outside")+ xlab("off.sc\nAggregated frequency plots for 17 equal intervals.") The code

Hi helpeRs, I have inherited a set of data files that use the file system as a sort of poor man's database, i.e., the data files are nested in directories that indicate which city they come from. For example: dir.create("deleteme") for(i in paste("deleteme", c("New York", "Los Angeles"), sep="/")) { dir.create(i) for(j in

Hi guys, Are there any feasible methods in searching & finding non-ASCII characters in R? For example, from the following object, x <- mia. SzaÌmitaÌó The desired output is, x.out <- mia. SzaImitaIA Your help in resolving this would be greatly appreciated. [[alternative HTML version deleted]]