similar to: sunflowerplot error

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,

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

### 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,

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"

Hello, R-help, does anybody have already a work-around for the problem that the formula version of sunflowerplot() throws an error when provided with a value for xlab (or ylab) different from NULL: > sunflowerplot( Sepal.Length ~ Sepal.Width, data = iris, xlab = "A") Error in model.frame.default(formula = Sepal.Length ~ Sepal.Width, data = iris, : variable lengths differ

Hi, I use panel.points to add points to a xyplot graphic. But I like to use the sunflowerplot to plot my points because this is very superimposed. It is possible to use this? I try but it dont work directly. It may be need to put this function inside a panel.??? Thanks Ronaldo -- Where there's a will, there's a relative. -- > Prof. Ronaldo Reis J?nior | .''`.

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

Dear fellow R users, I would like to draw a "sunflowerplot" because I have data (decade by month) that plots multiple times on the same x-y co-ordinates. Further I would like to colour each of the points/sunflower leaves on the plot according to the group they belong to (i.e. which type of event each represents within that decade and month). I thought that this would be relatively

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

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

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

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

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

x <- c(1, 1, 2, 2, 2, 3) y <- c(1, 2, 1, 3, NA, 3) > str(xyTable(x,y)) List of 3 $ x : num [1:6] 1 1 2 2 NA 3 $ y : num [1:6] 1 2 1 3 NA 3 $ number: int [1:6] 1 1 1 NA NA 1 How many (2,3)s do we have? At least one, the third entry, but the fourth entry, (2,NA), is possibly a (2,3) so we don't know and make the count NA. I suspect this is not the intended logic, but a

Le 25/04/2023 ? 10:24, Viechtbauer, Wolfgang (NP) a ?crit?: > Hi all, > > Posted this many years ago (https://stat.ethz.ch/pipermail/r-devel/2017-December/075224.html), but either this slipped under the radar or my feeble mind is unable to understand what xyTable() is doing here and nobody bothered to correct me. I now stumbled again across this issue. > > x <- c(1, 1, 2, 2, 2,