search for: helmert

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, perhaps this is a stupid question, but i need some help about Helmert contrasts in the Cox model. I have a survival data frame with an unordered factor `group' with levels 0 ... 5. Calculating the Cox model with Helmert contrasts, i expected that the first coefficient would be the same as if i had used treatment contrasts, but this is not true. I this a error...

..."A","B","C") of a factor X: Y<-c(rnorm(mean=0,n=12),rnorm(mean=2,n=12),rnorm(mean=4,n=12)) X<-factor(c(rep("A",12),rep("B",12),rep("C",12))) Then I do a summary(lm(Y~X)...) using first "Treatment" contrasts and then "Helmert" contrasts. Here are the coefficient parts of the results in each case: summary(lm(Y~X,contrasts=list(X="contr.treatment"))) Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 0.2303 0.3220 0.715 0.47944 XB 1.3057 0.4554 2.867 0...

Hi, I found that using different contrasts (e.g. contr.helmert vs. contr.treatment) will generate different fitted probabilities from multinomial logistic regression using multinom(); while the fitted probabilities from binary logistic regression seem to be the same. Why is that? and for multinomial logisitc regression, what contrast should be used? I guess it...

...a repeated-measures design, the other a split-plot. In a standard ANOVA I have usually undertaken a multiple-comparison test on a significant factor with e.g TukeyHSD, but as I understand it such a test is inappropriate for repeated measures or split-plot designs. Is it therefore sensible to use Helmert contrasts for either of these designs? Whilst not providing all the pairwise comparisons of TukeyHSD, presumably the P-statistic for each Helmert contrast will indicate clearly whether that contrast is significant and should be retained in the model. (This seems to come with the disadvantage that...

On 6/23/05, RenE J.V. Bertin <rjvbertin at gmail.com> wrote: > Hello, > > I was just having a look at the aov function source code, and see that when the model used does not have an Error term, Helmert contrasts are imposed: > > if (is.null(indError)) { > ... > } > else { > opcons <- options("contrasts") > options(contrasts = c("contr.helmert", "contr.poly")) > on.exit(options(opcons)) > ... > >...

...("6wks", "4wks", "2wks")) dist <- rep(gl(4,n,labels=c("1mi","2mi","3mi","4mi")),3) What I would like to do is test two interaction contrasts that cross linear coefficients of the variable "dist" with the two sets of helmert contrasts of the variable "train". I have tried so many things I think are wrong I won't reproduce them all here. Here is a representative example: My code: contrasts(dist) = c(-3, -1, 1, 3) contrasts(train) = contr.helmert(3) aov1 = aov(fatigue~dist*train) summary(aov1, intercept...

...ted leaving out library(nlme) and replacing > options(contrasts=c(factor="contr.SAS",ordered="contr.poly")) with the following contrasts > options(contrasts=c(factor="contr.treatment",ordered="contr.poly")) > options(contrasts=c(factor="contr.helmert",ordered="contr.poly")) > options(contrasts=c(factor="contr.sum",ordered="contr.poly")) > library(MASS) # needed for the contrast contr.sdif > options(contrasts=c(factor="contr.sdif",ordered="contr.poly")) The results from anova were...

...with 2 independent variable, one (x1) is continuous, the other (x2) is a categorical variable with 2 levels. The dependent variable (y) is continuous. When I run linear regression y~x1*x2, I found that the p value for the continuous independent variable x1 changes when different contrasts was used (helmert vs. treatment), while the p values for the categorical x2 and interaction are independent of the contrasts used. Can anyone explain why? I guess the p value for x1 is testing different hypothesis under different contrasts? If the interaction is NOT significant, what contrast should I use to test th...

Dear R users, When fitting a lme() object (from the nlme library), is it possible to test interactions *before* main effects? As I understand, R conventionally re-orders all terms such that highest-order interactions come last - but I??d like to know if it??s possible (and sensible) to change this ordering of terms. I??ve tried the terms() command (from aov) but I don??t know if something

Several bugs (no solutions, yet). These might be well known. 1) If one does, e.g., mymod <- lm(y ~ x); formula(mymod) then one does not get back the formula (one gets, Error: invalid formula) 2) if x is of mode numeric, then the model formula mymod <- lm(y ~ x + x^2) is not processed as S would do it. The model is fit ignoring the x^2 term, however mymod$call includes the x^2 term.

Several bugs (no solutions, yet). These might be well known. 1) If one does, e.g., mymod <- lm(y ~ x); formula(mymod) then one does not get back the formula (one gets, Error: invalid formula) 2) if x is of mode numeric, then the model formula mymod <- lm(y ~ x + x^2) is not processed as S would do it. The model is fit ignoring the x^2 term, however mymod$call includes the x^2 term.

...t the Error strata working.) Q 2) If the speed is important, can we give up choice of contrasts used to set up the dummy factor variables? ( contr.treatment, contr.sum) Why do I ask? In pursuit of speed (we're talking nanoseconds for a reasonable size data set) I have forced in helmert contrasts because their orthogonality makes it slick to compute. So in this version, whatever option(contrasts) is set will be ignored. I don't like making that choice for the user esp. since it gives wrong answers under the default contrasts, and have thought about converting...

...or variables which have only a few levels and will be used in modeling. Every character variable of every dataset in "Statistical Models in S", which introduced factors, is of this type so auto-transformation made a lot of sense. The "solder" data set there is one for which Helmert contrasts are proper so guess what the default contrast option was? (I think there are only a few data sets in the world for which Helmert makes sense, however, and R eventually changed the default.) For character variables that should not be factors such as a street adress stringsAsFactors ca...

Is there any logical reason why glm prints out the labels of factor levels after variable names when baseline contrasts (contr.treatment) are used but the codes for the levels when mean contrasts (contr.sum) are used? Jim -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info",

Dear R-helpers, I try to perform glm's with negative binomial distributed data. So I use the MASS library and the commands: model_1 = glm.nb(response ~ y1 + y2 + ...+ yi, data = data.frame) and predict(model_1, newdata = data.frame) So far, I think everything should be ok. But when I want to perform a glm with a subset of the data, I run into an error message as soon as I want to predict

...In fact, I get the same answers with the default parametrization and the alternative). Since I'm only interested in the interaction term, is it even necessary to change the structure of the model matrix, or will I get the answer I need using the default contrast matrices (i.e. "contr.helmert", "contr.poly")? ===================== Dr. Marc R. Feldesman Professor and Chairman Anthropology Department Portland State University 1721 SW Broadway Portland, Oregon 97201 email: feldesmanm at pdx.edu phone: 503-725-3081 fax: 503-725-3905 http://odin.cc.pdx.edu/~h1mf =====...

I am running a two-way anova with Type III sums of squares and would like to be able to understand what the different SS mean when I use different contrasts, e.g. treatment contrasts vs helmert contrasts. I have read John Fox's "An R and S-Plus Companion to Applied Regression" approach -p. 140- suggesting that treatment contrasts do not usually result in meaningful results with Type III SS but it's not clear to me why. Any suggestions on a stats text discussing this woul...

..., it follows that an identification constraint applies, namely v'phi = 0. By multiplying both sides by (C'C)^{-1} C', it also follows that *alpha =(C'C)^{-1}C'phi which provides an interpretation for the alpha's in terms of the (constrained) phi's. For example take the Helmert contrasts. > contr.helmert(4) [,1] [,2] [,3] 1 -1 -1 -1 2 1 -1 -1 3 0 2 -1 4 0 0 3 The constraint vector is clearly v= (1,1,1,1), since the columns add to zero. In this case the columns are also mutually orthogonal, so the matrix (C'C^{-l} C' (the generalized inverse of C) has a simila...

.../ running Anova(model,type=c("III")). However summary(model) gives the same results in both cases. Is this how it is set up? > local({pkg <- select.list(sort(.packages(all.available = TRUE))) + if(nchar(pkg)) library(pkg, character.only=TRUE)}) > options(contrasts=c("contr.helmert","contr.poly")) > model2<-aov(tdrate~temp*sex) > summary(model2) Df Sum Sq Mean Sq F value Pr(>F) temp 3 0.110137 0.036712 1005.6947 < 2e-16 *** sex 1 0.000141 0.000141 3.8593 0.05095 . temp:sex 3 0.000154 0.000051 1.407...