similar to: trying to produce an array of ranks

I predicted that y would increase as x increased. However, I only made the prediction on the ranks of the scores. The ranks don't correlate with predicted. And, I don't think a regression on the ranks is warranted. However, the actual scores do yield a significant slope for b, and a significant R^2 using a linear regression (y is the value and x is the predicted rank). What

How can I add an extra column containing the rank to a ragged array indexed by more than one grouping factors? E.g. with the barley dataset: How can I to add an additional column ``rank'' containing the rank of the ``yield'' of the different varieties in relation to the indices ``year'' and ``site'' to the barley dataframe? I achieved to calculate the ranks with:

Dear r-helpers, Spencer Graves and Manual Morales proposed the following methods to simulate p-values in lme4: ************preliminary************ require(lme4) require(MASS) summary(glm(y ~ lbase*trt + lage + V4, family = poisson, data = epil), cor = FALSE) epil2 <- epil[epil$period == 1, ] epil2["period"] <- rep(0, 59); epil2["y"] <- epil2["base"]

Hi , I am new to R. I am trying to run a simple R script as shown below: aov.R ------ data1<-c(49,47,46,47,48,47,41,46,43,47,46,45,48,46,47,45,49,44,44,45,42,45,45,40 ,49,46,47,45,49,45,41,43,44,46,45,40,45,43,44,45,48,46,40,45,40,45,47,40) matrix(data1, ncol= 4, dimnames = list(paste("subj", 1:12), c("Shape1.Color1", "Shape2.Color1", "Shape1.Color2",

I am extremely puzzled by this behavior in R. I have a data frame called Trials in which I have results from an experiment. I am trying to do a subjects analysis, but getting weird results. Each row has 1 trial in it, which includes a column for the subject number I get the list of subject numbers like so: > Subj=unique(sort(Trials$Subj)) Then I loop over them. But I get strange results. As

How does one go about doing a repeated measure regression? The documentation I have on it (Lorch & Myers 1990) says to use linear / (subj x linear) to get your F. However, if I put subject into glm or lm I can't get back a straight error term because it assumes (rightly) that subject is a nominal predictor of some sort. In looking at LME it seems like it just does the right thing

Dear all, In lme, models in which a factor is fully "contained" in another lead to an error. This is not the case when using lm/aov. I understand that these factors are aliased, but believe that such models make sense when the factors are fitted sequentially. For example, I sometimes fit a factor first as linear term (continuous variable with discrete levels, e.g. 1,2,4,6), and

Hello, I am having a problem where code that plots lines using a different data frame plots bars with the current data frame (I am intended to plot lines). The code specifies lines (see below), so I can't figure out why the results are bars. I suspect that it may have something to do with the fact that in the data frame where the code worked as intended, the both variables specifying

All, How does one extract the level-1 variance from a model fit via lmer()? In the code below the level-2 variance component may be obtained via subscripting, but what about the level-1 variance, viz., the 3.215072 term? (actually this term squared) Didn't see anything in the archives on this. Cheers, David > fm <- lmer( dv ~ time.num*drug + (1 | Patient.new), data=dat.new )

Hello: I am new R user stumped why the R code after this paragraph generates "Error: syntax error" messages after each of the last 2 lines. I have tried searching the manuals, Hmisc documentation, contributed manuals, help archives, and Internet. I am running R 1.7.1 under Windows 2000 (I will upgrade when my imminent OS upgrade happens). My data was successfully entered and

Dear R-help, I'm facing a problem with defining a repeated measures anova with weighted data. Here's the code to reproduce the problem: # generate some data seed=11 rtrep <- data.frame(rt=rnorm(100),ti=rep(1:5,20),subj=gl (20,5,100),we=runif(100)) # model with within factor for subjects/repeated measurements, no problem aov(rt~ti + Error(subj/ti),data=rtrep) #model with weights

All, I'm trying to plot an all.effects() object, as shown in the help for all.effects and also Crawley's R book (p.178, 2007). The data has a repeated measures structure, but I'm using all.effects for the simple lm() fit here. Below is a reproducible example that yields the error message. fm.ex = lm(dv ~ time.num*drug*X, data = dat.new) fm.effects = all.effects(fm.ex, xlevels =

Dear all: I m a newer on R.? I have some problem when I use?anova function.? I use anova function to get Type I SS results, but I also need to get Type III SS results.? However, in my code, there is some different between the result of Type I SS and Type III SS.? I don?t know why the ?seqe? factor disappeared in the result of Type III SS.? How can I do?? Here is my example and result.

Hi all, I have a strange problem and rigth now I can't figure out a solution. Trying to calculate an ANOVA with one between subject factor (group) and one within (hemisphere). My dependent variable is source localization (data). My N = 25. My data.frame looks like this: > ML.dist.stack subj group hemisphere data 1 1 tin left 0.7460840 2 2 tin left

Hi All. I'm trying to run a simple model from Baayan, Davidson, & Bates and getting a confusing error message. Any ideas what I'm doing wrong here? # Here's the data..... Subj <- factor(rep(1:3,each=6)) Item <- factor(rep(1:3,6)) SOA <- factor(rep(0:1,3,each=3)) RT <- c(466,520,502,475,494,490,516,566,577,491,544,526,484,529,539,470,511,528) priming

I'm implementing a custom bootstrap resampling procedure in R. This procedure resamples clusters of data points obtained by different subjects in an experiment. Since the bootstrap samples need to have the same size as the original dataset, `target.set.size`, I select speakers compute their data point contributions to make sure I have a set of the right size. set.seed(1)

Hi, I have a data frame and want to merge adjacent rows if some condition is met. There's an obvious solution using a loop but it is prohibitively slow because my data frame is large. Is there an efficient canonical solution for that? > head(d) rt dur tid mood roi x 55 5523 200 4 subj 9 5 56 5523 52 4 subj 7 31 57 5523 209 4 subj 4 9 58 5523 188 4 subj 4 7

#Dear All, #I'm having a bit of a trouble here, please help me... #I have this data set.seed(4) mydata <- data.frame(var = rnorm(100), temp = rnorm(100), subj = as.factor(rep(c(1:10),5)), trt = rep(c("A","B"), 50)) #and this model that fits them lm <- lm(var ~ temp * subj, data = mydata) #i want to

I din't have the patience to go through your missive in detail, but do note that it is not reproducible, as you have not provided a "data" object. You **are** asked to provide a small reproducible example by the posting guide. Of course, others with more patience and/or more smarts may not need the reprex to figure out what's going on. But if not ... Cheers, Bert Bert Gunter

I have a problem where I need to calculate the corresponding cohort percentile ranks for each of several variables. Essentially, what I need is a function that will calculate the distribution-free percentiles from each variable's data vector, returning a corresponding vector of percentiles: e.g.: percentile.my.data<-/function/(my.data) I tried to make ecdf() perform this task but