similar to: how to handle a longitudinal data

i have a large longitudinal data set. The number of observations for each subject is not the same across the sample. The largest number of a subject is 5 and the smallest number is 1. now i want to make each subject to have the same number of observations by filling zero, e.g., my original sample is id x 001 10 001 30 001 20 002 10 002 20 002 40 002 80 002 70 003 20 003 40 004 ...... now i wish

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',

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

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

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

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

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

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

... and is there also such a nice tool (like spss.get) for exporting data frames to SPSS? write.table does not keep the data frame labels - neither did the other exporting tools that I found. Thanks! Michael [[alternative HTML version deleted]]

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]),

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]]

A colleague was asking me if R does multi-level modelling as opposed to multiple regression. Since I have no knowledge of multi-level modelling (except 5 minutes googling ) I thought that I would as here. Does are offer any multi-level modeling packages? It looked like arm might be one but I was not sure. Thanks

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

Running R 1.8.1 on a Windows machine In dotchart, I would like to shrink the labels on the tick marks (that is, the numbers) without shrinking anything else. I could not find this in either the Rhelp archives or in ?dotchart, which recmmends cex to avoid 'label overlap', but cex shrinks all the characters in the plot. Is there a way to do this? Thanks Peter Peter L. Flom, PhD

Dear R Users, If it is possible, i would like to learn whether a R package that can be used for Taguchi design is available. Sincerely -- Ýbrahim Mutlay [[alternative HTML version deleted]]

Dear All, Need some help in polynomials transformation to get the coefficients. I have tried "poly.transform" as applied in S-plus but it does not work. Thanks in advanced for any helps. Regards. Abd. Rahman Kassim (PhD) Head Forest Ecology Branch Forest Management & Ecology Program Forestry and Conservation Division Forest Research Institute Malaysia Kepong 52109 Selangor,

Dear latexRs, I tried to make a latex printout of a simple categorial ftable. It should look like the output of print.ftable. Any ideas how to get the syntax of summary.formula right. Or some alternative? As far I see, xtable does not have method for ftable. Dieter library(Hmisc) n=500 sex <- factor(sample(c("m","f"), n, rep=TRUE)) treatment <-

I am having a hard time understanding how to perform a "repeated measures" type of ANOVA with R. When reading the document found here: http://cran.r-project.org/doc/contrib/Lemon-kickstart/kr_repms.html I find that there is a reference to a function make.rm () that is supposed to rearrange a "one row per person" type of frame to a "one row per observation" type

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

R-experts: A quick question, please. >From a lab exp, I got 12 positives out of 50. To get 90% CI for this , I think binom.test might be the one to be used. Is there a better way or function to calculate this? > binom.test(x=12, n=50, p=12/50, conf.level = 0.90) Exact binomial test data: 12 and 50 number of successes = 12, number of trials = 50, p-value = 1 alternative