similar to: Define return values of a function

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

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

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"

Hi, keine ahnung. Das liegt jetzt bei Hr. Feld. Frag mal bei ihm nach. Gr??e Thushyanthan r-help-request at r-project.org wrote: > Send R-help mailing list submissions to > r-help at r-project.org > > To subscribe or unsubscribe via the World Wide Web, visit > https://stat.ethz.ch/mailman/listinfo/r-help > or, via email, send a message with subject or body 'help'

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 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

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

### 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, 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, 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

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

I have no real idea what you are trying to do, but if a table is what you want, you can probably get it using the table() function. Or, more likely, the xtabs() function. Using your example from an earlier post (adjusted to make it comprehensible to the human mind): set.seed(1000) time <- factor(rep(c("Pre","Post"),each=200)) treatment <-

Estimate region of highest probabilty density Dear R-community I have data consisting of x and y. To each pair (x,y) a z value (weight) is assigned. With kde2d I can estimate the densities on a regular grid and based on this make a contour plot (not considering the z-values). According to an earlier post in the list I adjusted the kde2d to kde2d.weighted (see code below) to estimate the

Hi Rolf, No it is not. I don't know to which question did you want to respond ? I desribed everything in my first email and attached links from SO with pictures included, which are quite understandable. Cheers, Jacek wt., 27 lut 2024 o 02:29 Rolf Turner <rolfturner at posteo.net> napisa?(a): > > I have no real idea what you are trying to do, but if a table is > what you

Hello The "Craddock-Flood Test" is recommended for large tables with small degrees of freedom and low-frequency cells. Is there an R procedure and/or package which does the test? Thank you for your help! S?ren Vogel -- S?ren Vogel, Dipl.-Psych. (Univ.), PhD-Student, Eawag, Dept. SIAM http://www.eawag.ch, http://sozmod.eawag.ch

Hello I found some small postings dated to 22 Oct 2008 on the message subject. Recently, I have been working with binary logistic regressions. I didn't use the design package. Yet, I needed the "fit" indices. Therefore, I wrote a small function to output the Nagelkerke's R, and the Cox-&-Snell R from a fitted model. I am no professional programmer by far, yet, I hope, that

Hello, for my data preparation and administration (data, labels, etc.) I use OpenOffice.org ODS spreadsheet files with several sheets in one file. However, I find it inconvenient to export every single sheet to a csv file whenever I apply changes to the labelling or so, and I haven't found a plugin or an application which does this batch in OOorg. Is there new development on a direct import