similar to: add absolute value to bars in barplot

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"

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

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

On Mon, 2007-05-14 at 23:41 +0200, vax9000 at gmail.com wrote: > Full_Name: vax, 9000 > Version: 2.4.0, 2.2.1 > OS: 2.4.0: Mac OS X; 2.2.1: Linux > Submission from: (NULL) (192.35.79.70) > > > To reproduce this bug, first go to the website "http://llmpp.nih.gov/DLBCL/" and > download the 14.8M data set "Web Figure 1 Data file". The direct link is >

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

Hallo, What is the right way to get vertical bars in a barchart? For instance barchart(VADeaths, key=simpleKey(colnames(VADeaths),points=F,rectangles=T)) gives what I need, only I would like the bars to be vertical. But barchart(VADeaths,horizontal=F, key=simpleKey(colnames(VADeaths),points=F,rectangles=T)) does not give what I need, and I do not understand how to change the

Hello I have data of body length and body weight of people of different skin colors. I tried to write a code to plot body length and body weight according to the skin colors. (Thanks for Petr's advice so far.) A loop is used but an error shows up in the following code. It says: unexpected '}' in "

Hi Dieter and R community: I tried both of these three versions with ylim as suggested, none work: I am getting only single (pch = 16) not overlayed (pch =3) everytime. *vs 1* require(lattice) xyplot(Sepal.Length ~ Sepal.Width | Species , data= iris, panel= function(x, y, subscripts) { panel.xyplot(x, y, pch=16, col = "green4", ylim = c(0, 10)) panel.lmline(x, y, lty=4, col =

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

Dear list, I have done a scatter plot of multiple variables in the same graph, with different col and pch. I managed to do it with the following code but not know how to make a function of these so that next time if I want to do similar graph but with new variables, I dont have to copy the code and then change the old variables with the new ones but just call a function with the new

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

Given this example mean.values<-colMeans(VADeaths) mean.values<-apply(VADeaths, 2, mean) median.values<-apply(VADeaths, 2, median) dotchart(VADeaths, gdata=mean.values) dotchart(VADeaths, gdata=median.values) is it possible to ?combine? a single dotchart showing both the mean and the median for each single group (with different plotting symbols)? ?is it that possible with the use of

Dear R users, using dotplot (R1.2.0, WinNT4.0), I am trying to change the character size of the labels of the points: > # example > data(VADeaths) > dotplot(VADeaths, main = "Death Rates in Virginia - 1940") > # I'd like to have smaller character size of the labels (for age and population groups) > ?dotplot > # for argument "cex", this says: