similar to: comparing two unequal matrices without for loop?

Hello All, Would anyone be able to help me understand how R computes a quantile-quantile plot for comparing two data samples with unequal sample sizes? Normally, the procedure should be to rearrange the larger data sample into n equally-spaced parts using interpolation, where n is the sample size of the smaller sample, and then plot the matching data pairs. I tried using different plotting

Hello- I have run an ANOVA on 4 treatments with unequal sample sizes (n=9,7,10 and 10). I want to determine where my sig. differences are between treatments using a Bonferroni test, and have run the code: pairwise.t.test(Wk16, Treatment, p.adf="bonf") I receive an error message stating that my arguments are of unequal length: Error in tapply(x, g, mean, na.rm = TRUE) :

Dear Douglas and R-help, Does lme assume normal distribution AND equal variance among groups like anova() does? If it does, is there any method like unequal variance T-test (Welch T) in lme when each group has unequal variance in my data? Thanks, Shirley

Hi, I am doing multiple regression, and have several X variables that are categorical. I read that I can use dummy or contrast codes for that, but are there any special rules when there're unequal #observations in each groups (4 females vs 7 males in a "gender" variable)? Also, can R generate these codes for me? THanks.

Hello, I have some patient data for my masters thesis with three groups (n=16, 19 & 20) I have completed compiling the results of 7 tests, for which one of these tests the variances are unequal. I wish to perform an ANOVA between the three groups but for the one test with unequal variance (<0.001 by both bartlett and levene's test) I am not sure what to do. I thought i would run

I am using levelplot() from lattice with grids that have unequal cell sizes. This means that the boundary between two cells is not always half-way between nodes, as levelplot() assumes. The result is that some cell sizes are rendered incorrectly, which can be painfully obvious if using relatively large cells. Is there any work-around? I am using the conditioning capability of lattice and

Dear R users, I need to merge two data frames based on both equal and unequal comparisons. The "sqldf" package used to work well , but today, I cannot resolve the following error by reinstallation of the sqldf package. Can anyone suggest a different way to perform this kind of merge function? Thank you, Ding > DMRlog2pbde47DMS <- sqldf("select * from DMR_log2pbde47 as a

Hi, i'm looking for an R function to fit a one-way ANOVA with one factor containing 10 levels. The factor levels have different numbers of observations (varying between 20 to 40). For most of the dependent variables i'm testing there are unequal variances among the factor levels. I see the function oneway.test: oneway.test(variable ~ factor, data=dataset) which by default does not

This is in followup to a thread on R-help with subject "Sampling". I claim that R does the wrong thing by default when sampling with unequal probabilities without replacement - the selection probabilities are not proportional to 'prob', for any draw after the first: I suggest that R do what S-PLUS now does (though you're free to choose a better implementation). What S-PLUS

Hi, I am having problems getting cohen's kappa to work. I have been using the function: ><-ckappa(x,y) from the psy package. I am trying to test for inter-observer reliability, I have 2 observers and 26 categories, however, the two observers might not necessarily have the same range of categories (I have unequal score ranges). However, I thought R could cope with this. Each time I

I am getting a repeated error when I try to run a logistic regression in R 2.8.1 >(glm(prop1~x1,data=glm1,family=binomial("logit"),weights=nt1)) Error in model.frame.default(formula = prop1 ~ x1, data = glm1, weights = nt1, : invalid type (list) for variable 'x1' x1 is multistate categorical (3 categories). 2 of the categories have 12 observation, one has 9. Is this what

Well fellow R users, I throw myself on your mercy. Help me, the unworthy, satisfy my employer, the ungrateful. My feeble ramblings follow... I've searched R-Help, the R Website and done a GOOGLE without success for a one way ANOVA procedure to analyse data that are both non-normal in nature and which exhibit unequal variances and unequal sample sizes across the 4 treatment levels. My

Hello, I am hoping someone could shed some light on power calculations for me. I have two small data sets of unequal sample size after NA removal (m = 5, f = 7). m <- c(2.0863, 2.1340, 2.1008, 1.9565, 2.0413, NA, NA) f <- c(1.8938, 1.9709, 1.8613, 2.0836, 1.9485, 2.0630, 1.9143) In a R help message/reply from Sep 30, 2001, it was noted that the "power.t.test" function

Hi all, I have probably a basic question, but I can't seem to find the answer in the literature or in the R-archives. I would like to do a robust ANCOVA (using either rlm or lmRob of the MASS and robust packages) - my response variable deviates slightly from normal and I have some "outliers". The data consist of 2 factor variables and 3-5 covariates (fdepending on the model).

Hello, I am trying to plot SADIE red-blue plots of cluster indicies using filled.contour. I want a plot which only has three bins for the data: <-1.5, -1.5 - 1.5, >1.5, but I am having trouble getting there. example X1 X2 X3 X4 X5 1 -5 -4.5 1.0 4.5 6 2 -3 -2.0 1.2 -1.0 3 3 0 0.0 0.0 -0.5 -1 4 -2 -3.0 1.0 1.5 3 5 -6 -2.0 0.5 3.0 2 example<-as.matrix(example)

Dear R folks - Sorry to be so dense, but can you help me with two programming problems? 1. Randomizing within factors. I have factored the values in an array according to a series of ranked breakpoints. Now I want to randomize the values that fall within each of the breakpoint boundaries, without replacement. I want to end up with an array in which the values within each factored category

Hi I am trying to run an ANOVA on data with unequal variance. I am new to nlme, but to my understanding I need to use the gls function. I have single response variable (distance which is continuous) and the explanatory variable is individual ID (class variable: individuals differ in the variance in their distance values hence the need to using nlme). So I would create a model

Hi, vec1<-factor(rep(c("A","B","C"),c(19,8,11)),levels=c("A","B","C")) sapply(split(vec1,vec1),length) # A? B? C #19? 8 11 #or you could change gl() function gl.new<-function (n, k, labels = 1:n, ordered = FALSE) { ? out<- numeric() ?? for(i in 1:n){ ?? out<- append(out,rep(i,length.out=k[i])) ?? } ?? out<-factor(out) ??

I'am trying to develop some code if R, which would correspond to what I did in SAS. The data look like: Treatment Replicate group1 GSI Control A 1 0.81301 Control B 1 1.06061 Control C 1 1.26350 Control D 1 0.93284 Low A 2 0.79359 Low B

Hi all, right now i am simply comparing means. obviously this can be done by the simple t.test respectively the welch test, if var.equal is set to FALSE. just like this t.test( Y ~ group) t.test( Y ~ group, var.equal = FALSE) now that i need to compare weighted means i am using the lm function as an adjusted t-test: like lmtest <- ( Y ~ group ) anova(lmtest)