similar to: getting R2.4 (Win)

i was searching for the source of bca.ci, a function of the package boot. I tried require(boot, keep.source=TRUE) but again the source was not viewable. How should i do? Best regards Meinhard Ploner ---------------- PS > version _ platform i386-apple-darwin8.8.1 arch i386 os darwin8.8.1 system i386, darwin8.8.1 status major 2

i have a data set with repeated measures on same people, structure like below: id x1 x2 ... 001 10 20 ... 001 8 45 ... 001 4 2 ... 002 .... 002 ... 002 .... 002 .... 003 .... ....... what is the easist way to show how many observations for each subject id? [[alternative HTML version deleted]]

Hello, How ls() only functions or only data objects (basically anything other than functions) such as data.frame, numeric ...? John [[alternative HTML version deleted]]

Hi, I am trying to create a power curve to show how the power of a t-test varies depending on the mean. Any ideas how I should go about this? Louisa _________________________________________________________________ [[alternative HTML version deleted]]

Dear list, I have a vector (array, table row, whatever is best) of frequency values for categories (or bins), and I need to find the median category. Trivial to do by hand, but I was wondering if there is a means to do it in R in an elegant way. The obvious medioan(vector) returns the median frequency for the binns, and that is not what I want. i.e,: freq cat1 1 cat2 10

Hello I have a dataset in which I would like to replace 0s with NAs. There is a lot of information on how to replace NAs with 0, but I have struggled to find anything with regards to doing the reverse. Any recommendations would be great. Cheers Christine

I have a data set where I want the correlations between 2 variables conditional on a students grade level. This code works just fine. by(tmp[,c('mtsc07', 'DCBASmathscoreSPRING')], tmp$Grade, cor, use='complete', method='pearson') However, this generates an error by(tmp[,c('mtsc07', 'DCBASmathscoreSPRING')], tmp$Grade, cor, use='complete',

Hi Everyone, I am trying to use lmFit function; however, i cannot find it function anywhere. I have been trying to find the function in Bioconductor and elsewhere. I re-install bioconductor source, update package and update R as well. no luck Is there a command in R where i can just type, and it will download it for me? -- View this message in context:

Hello, I'm new to R, I've been working with it for the last 2 weeks. I am plotting some data and not getting the labels on the x axis I am expecting on my plot. my code reads #hours in the day h <- c(0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23) #hp is a data frame with a pivot table of 25 columns (label and data for 24 hours) plot(h, as.matrix(hp[1,2:25]),

Hello, A made a xyplot using the lattice library in R (latest version). The publisher of our paper has requested: 1. all tick marks should point inwards instead of outwards. 2. All lines should be thicker (lines, axes, boxes, etc. Everything). Lines is easy...I used: lwd=1.5 but what about the lines of the axes, and the lines that build up the plot itself?....? Any

Dear R users I have a very basic question. I tried but could not find the required result. using dat <- pima f <- table(dat[,9]) > f 0 1 500 268 i want to find that class say "0" having maximum frequency i.e 500. I used >which.max(f) which provide 0 1 How can i get only the "0". Thanks and best regards Muhammad Azam Ph.D. Student Department of

Hello! I?m fitting a model with glm(family binomial). The best model counts 9 Variables and includes an interaction term that was generated by the product of to continuous variables (a*b). All variables are correlated under a value of 0.7 (Spearman rank order) While the estimates of both main effects are negativ, the resulting interaction term is positiv. This change of sign makes it difficult to

Good afternoon Running R2.10.0 on Windows I have a data frame that includes (among much else) a factor (In_2006) and a continuous variable (math_3_4). I would like to find the 2 cases for In_2006 = 0 that are closest to each case where In_2006 = 1. My data looks like In_2006 math_3_4 0 55.1 1 51.6 1 18.1 1 26.6 1 14.1

Hi folk, I tried this and it works just perfectly tapply(iris[,1],iris[5],mean) but, how to obtain a single table from multiple variables? In tapply x is an atomic object so this code doesn't work tapply(iris[,1:4],iris[5],mean) Thanx and great summer holidays Gianandrea -- View this message in context: http://www.nabble.com/multiple-tapply-tp18868063p18868063.html Sent from the R help

Hello: I recall something like this being discuss recently, but I can't seem to locate an example in the archives. I have data like the following: df <- expand.grid(1:4, 1992:2002) names(df) <- c("MSA", "YEAR") df$IDUPREV <- runif(44) df$VALIDAT <- rnorm(44) I want to create an xyplot() with separate loess lines for each series (IDUPREV and VALIDAT) in

Hi R, I have a vector V1 of unknown length, say n. I need to convert this into a matrix C of row size=5, and accordingly the column should be updated. I tried with: C=as.matrix(V1,5,n/5) But it is not working...Could somebody help me on this? Thanks in advance... [[alternative HTML version deleted]]

I have two points of collection across 20 subjects (pre and post for each), so 20 pairs of data points. I would like to plot the actual raw data points for each subject for both pre and post and connect lines between these two points (20 in all) to depict real change between the two timepoints. I have tried using stripchart which adequately plots the two lines of subject data points. Attempting

Hi All: I have a seemingly typical SPSS data file with 219 rows and 486 variables. When I attempt to read this file into R with read.spss() in the foreign package, I consistently get a "crash". This is the sequence of events: > library(foreign) > sessionInfo() R version 2.7.1 Patched (2008-07-24 r46120) i386-pc-mingw32 locale: LC_COLLATE=English_United

Hello! I have a data frame with a factor and a numeric variable: x<-data.frame(factor=c("b","b","d","d","e","e"),values=c(1,2,10,20,100,200)) For each level of "factor" - I would like to divide each value of "values" by the mean of "values" that corresponds to the level of "factor" In other

What is the right way to assign a new variable into each a of list of data frames? Here is my failed attempt: mylist <- list(df1 = data.frame(A = runif(5), B = runif(5)), df2 = data.frame(A = runif(5), B= runif(5))) lapply(mylist, function(x){x$Y <- x$A * x$B}) $df1 [1] 0.25589928 0.03446026 0.94992362 0.21388326 0.08668821 $df2 [1] 0.08771839 0.05643553 0.09036894