similar to: drop1 with contr.treatment

Hi, I'm currently trying to find/define a relationship between one dependent and several independant variables. The problem is that i cannot use the normal multiple regression/correlation in Spss because the data is not normal distributed. i calculated the spearman roh and Kendalls tau Correlation and also some partial correlations in R. Now i wanna find out the the multiple correlation

Hallo, zur Analyse von Daten zum Artenreichtum von Pflanzen, habe ich ein Glm (glm) und anschlie?end eine Anova (anova) durchgef??hrt. Nun m??chte ich f??r die signifikanten Einflussfaktoren einen Post Hoc Tukey Test durchf??hren, um zu ermitteln in wie weit die einzelnen Faktorstufen sich signifikant voneinander unterscheiden. Mit dem Befehl (TukeyHSD) komme ich nicht

I'm wondering which textbook discussed the various contrast matrices mentioned in the help page of 'contr.helmert'. Could somebody let me know? BTW, in R version 2.9.1, there is a typo on the help page of 'contr.helmert' ('cont.helmert' should be 'contr.helmert').

Hi, I have 6 career types, represented as a factor in R, coded from 1 to 6. I need to use the effect coding (also known as deviation coding) which is normally done by contr.sum, e.g. contrasts(career) <- contr.sum(6) However, this results in the 6th category being the reference, that is being coded as -1: $contrasts [,1] [,2] [,3] [,4] [,5] 1 1 0 0 0 0 2 0 1 0

Hi R-useRs, after having read http://tolstoy.newcastle.edu.au/R/help/05/07/8498.html with the same topic but five years older. the solution for a contr.sum with names for factor levels for R version 2.10.1 will be to comment out the following line #colnames(cont) <- NULL in contr.sum i guess? by the way, with contrasts=FALSE colnames are set, so i don't know what the aim is to avoid

i am getting the following error Error in `contrasts<-`(`*tmp*`, value = contr.funs[1 + isOF[nn]]) : contrasts can be applied only to factors with 2 or more levels can any on e suggest how to rectify [[alternative HTML version deleted]]

Full_Name: David Clifford Version: Version 1.5.1 (2002-06-17) OS: Red Hat 7.3 Submission from: (NULL) (128.135.149.55) For n values above 100 there appears to be a bug in contr.poly(n). The contrast matrix should have rank n-1. Running the code below gives output (ie errors) at n=98, 100 and every value greater than 102. for(n in 2:150) { K <- contr.poly(n) rnk <-

Hi Folks, I'm a bit puzzled by the following (example): N<-factor(sample(c(1,2,3),1000,replace=TRUE)) unique(N) # [1] 3 2 1 # Levels: 1 2 3 So far so good. Now: contrasts(N)<-contr.treatment(3, base=1, contrasts=FALSE) contrasts(N) # 1 2 # 1 1 0 # 2 0 1 # 3 0 0 whereas: contr.treatment(3, base=1, contrasts=FALSE) # 1 2 3 # 1 1 0 0 # 2 0 1 0 # 3 0 0 1 contr.treatment(3, base=1,

I've been experimenting with drop1 for my biostatistics class, to obtain the so-called Type III sums of squares. I am fully aware of the deficiencies of this method, however I feel that the students should be familiar with it. What I find baffling is that when applied to a fully balanced design, you obtain different sums of squares. I've used this for several years in Splus and R and never

A recent question in r-help made me realize that I should add a drop1 method for coxph and survreg. The default does not handle strata() or cluster() properly. However, for coxph the right options for the "test" argument would be likelihood-ratio, score, and Wald; not chisq and F. All of them reference a chi-square distribution. My thought is use these arguments, and add an

My apologies to the list for sending this without adequate research. I have found my answer; please ignore! Thanks. I've been experimenting with drop1 for my biostatistics class, to obtain the so-called Type III sums of squares. I am fully aware of the deficiencies of this method, however I feel that the students should be familiar with it. What I find baffling is that when applied to a fully

Hello R-help I don’t know how to interpret significance from the contr.poly() function . From the example below : how can I tell if data has a significant Linear/quadratic/cubic trend? > contr.poly(4, c(1,2,4,8))               .L         .Q          .C [1,] -0.51287764  0.5296271 -0.45436947 [2,] -0.32637668 -0.1059254  0.79514657 [3,]  0.04662524 -0.7679594 -0.39757328 [4,]  0.79262909 

Good morning, I used in R contr.sum for the contrast in a lme model: > options(contrasts=c("contr.sum","contr.poly")) > Septo5.lme<-lme(Septo~Variete+DateSemi,Data4.Iso,random=~1|LieuDit) > intervals(Septo5.lme)$fixed lower est. upper (Intercept) 17.0644033 23.106110 29.147816 Variete1 9.5819873 17.335324 25.088661 Variete2 -3.3794907 6.816101 17.011692 Variete3

glm(A~B+C+D+E+F,family = binomial(link = "logit"),data=tre,na.action=na.omit) Error in `contrasts<-`(`*tmp*`, value = "contr.treatment") : contrasts can be applied only to factors with 2 or more levels however, glm(A~B+C+D+E,family = binomial(link = "logit"),data=tre,na.action=na.omit) runs fine glm(A~B+C+D+F,family = binomial(link =

I'd like to fit and explore a collection of hierarchical loglinear models that might range from the independence model, ~ 1 + 2 + 3 + 4 to the saturated model, ~ 1 * 2 * 3 * 4 I can use add1 starting with a baseline model or drop1 starting with the saturated model, but I can't see how to get the model formulas or terms in each model as a *list* that I can work with further. Consider

the data is : >table.8.3<-data.frame(expand.grid( marijuana=factor(c("Yes","No"),levels=c("No","Yes")), cigarette=factor(c("Yes","No"),levels=c("No","Yes")), alcohol=factor(c("Yes","No"),levels=c("No","Yes"))), count=c(911,538,44,456,3,43,2,279))

Hi, I have a dataset with a response variable and multiple factors with more than two levels, which I have been fitting using lm() or glm(). In these fits, I am generally more interested in deviations from the global mean than I am in comparing to a "control" group, so I use contr.sum() as the factor contrasts. I think I'm happy to interpret the coefficients in the model summary

Dear List, recently tried to reproduce the results of some custom model selection function after updating R, which unfortunately failed. However, I ultimately found the issue to be that testing with pchisq() in drop1() seems to have changed. In the below example, earlier versions (e.g. R 2.4.1) produce a missing P-value for the variable x, while newer versions (e.g. R 2.7.1) produce 0 (2.2e-16).

With the following example using contr.sum for both factors, > dd <- data.frame(a = gl(3,4), b = gl(4,1,12)) # balanced 2-way > model.matrix(~ a * b, dd, contrasts = list(a="contr.sum", b="contr.sum")) (Intercept) a1 a2 b1 b2 b3 a1:b1 a2:b1 a1:b2 a2:b2 a1:b3 a2:b3 1 1 1 0 1 0 0 1 0 0 0 0 0 2 1 1 0 0 1 0

This is probably simple, but I just can't see it... I want to calculate the R^2s for a series of linear models where each term is dropped in turn. I can get the RSS from drop1(), and the r.squared from summary() for a given model, but don't know how to use the result of drop1() to get the r.squared for each model with one term dropped. Working example: library(vcd) # for