similar to: anova and contratst

pairwise.t.test is returning NAs when one of the samples only has one entry, while TukeyHSD returns results (maybe not trustworthy or believable, but results). I stumbled on this because I did not realize one of my samples only had one entry while most of the others had several hundred, so I realize this is not a desirable situation. I'm really just curious about the difference between how

Hi everybody, I'm having trouble getting an eval-parse combination to work with lme in a for loop. Consider the code below. The call to lme outside the loop works. The call to aov inside the loop works, but the call to to lme inside the loop does not. The latter fails with Error in model.frame.default(formula = ~meanCol + Group + c1 + c3 + Subject, : variable lengths differ (found for

Dear R-list members, I have a problem with translating a mathematica script into R. The whole script is at the end of the email (with initial values for easy reproduction) and can be pasted directly into R. The problematic part (which is included below of course) is <--- Original Mathematica ---> (* p_svbar *) UiA = Ni (Dsi - 2Di A + A^2)/2; UiiA = Nii (Dsii - 2Dii A + A^2)/2; psvbar =

Hi everybody, I'm trying to construct contrasts for an ANOVA to determine if there is a significant trend in the means of my groups. In the following example, based on the type of 2x3 ANOVA I'm trying to perform, does the linear polynomial contrast generated by contr.poly allow me to test for a linear trend across groups? doi=data.frame( Group=c( rep(1, 5), rep(2, 5), rep(3, 5),

HI, You can do this: dat1<- read.csv("dat7.csv",header=TRUE,stringsAsFactors=FALSE,sep="\t") dat.bru<- dat1[!is.na(dat1$evnmt_brutal),] fun2<- function(dat){? ????? lst1<- split(dat,dat$patient_id) ??? lst2<- lapply(lst1,function(x) x[cumsum(x$evnmt_brutal==0)>0,]) ??? lst3<- lapply(lst2,function(x) x[!(all(x$evnmt_brutal==1)|all(x$evnmt_brutal==0)),])