similar to: chisq.test: decreasing p-value

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 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"

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,

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

Dear All Where can I lookup good methods to compute p from q=bin(m,n)p^n*(1-p)^(m-n) such that q<=alfa, alfa small. Are there such libs, code and source in R? Best Regards -- Jan Burse SIAM, EAWAG Scheuchzerstr. 67 ?berlandstr. 133 8006 Z?rich 8600 D?bendorf tel: +41-1-364 17 66 tel: +41-1-823 55 34

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

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

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

heya, i am fitting linear mixed effect model to a response Y. Y shows an s-shaped distribution when using QQ-plots (some zero values and some very high values). hence, which transformation should i apply that Y follows a normal distribution? any r-function/package available to do this? thanks for any hint, regards, lukas ??? Lukas Indermaur, PhD student eawag / Swiss Federal Institute of

Hello Instead of testing against 0 i would like to test regression slopes against -1. Any idea if there's an R script (package?) available. Thanks for any hint. Cheers Lukas ??? Lukas Indermaur, PhD student eawag / Swiss Federal Institute of Aquatic Science and Technology ECO - Department of Aquatic Ecology ?berlandstrasse 133 CH-8600 D?bendorf Switzerland Phone: +41 (0) 71 220