similar to: model.matrix for a factor effect with no intercept

Dear all, I do not seem to grasp how contrasts are set for ordered factors. Perhaps someone can elighten me? When I work with ordered factors, I would often like to be able to reduce the used polynomial to a simpler one (where possible). Thus, I would like to explicetly code the polynomial but ideally, the intial model (thus, the full polynomial) would be identical to one with an ordered factor.

For ordered factor the natural contrast coding would be to parametrize by the succsessive differences between levels, which does not assume equal spacing of factor levels as does the polynomial contrasts (implicitly at least). This requires the contr.cum, which could be: contr.cum <- function (n, contrasts = TRUE) { if (is.numeric(n) && length(n) == 1) levs <- 1:n

How can I get the result of, e.g., poly(1:3. degree=2) to give me the unnormalized integer coefficients usually used to explain orthogonal polynomial contrasts, e.g, -1 1 0 -2 1 1 As I understand things, the columns of x^{1:degree} are first centered and then are normalized by 1/sqrt(col sum of squares), but I can't see how to relate this to what is returned by poly(). >

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's helmert? here is an example

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 in reasoning, or is it

Hi all, I have dataset 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

I am using RW 1.2.3. on an IBM PC 300GL. Using the data bp.dat which accompanies Helen Brown and Robin Prescott 1999 Applied Mixed Models in Medicine. Statistics in Practice. John Wiley & Sons, Inc., New York, NY, USA which is also found at www.med.ed.ac.uk/phs/mixed. The data file was opened and initialized with > dat <- read.table("bp.dat") >

Hi all, I'm working with a dataset from 10 treatments, each treatment with 30 subjects, each subject measured 5 times. The plot of the dataset suggests that a 3-parameter logistic could be a reasonable function to describe the data. When I try to fit the model using gnls I got the message 'Step halving factor reduced below minimum in NLS step'. I´m using as the initial values of the

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") >

Hi all, I have been trying to calculate Type III SS in R for an unbalanced two-way anova. However, the Type III SS are lower for the first factor compared to type I but higher for the second factor (see below). I have the impression that Type III are always lower than Type I - is that right? And a clarification about how to fit Type III SS. Fitting model<-aov(y~a*b) in the base package and

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Hola compañeros de la lista, qué tal. Los molesto con la siguiente duda: Tengo un experimento con dos factores A y B, cada uno de los cuales tiene los siguientes niveles (que son concentraciones de dos hormonas vegetales aplicadas a plantas): niveles del factor A: 0, 0.2, 0.5, 1 niveles del factor B: 0, 0.1, 0.2, 0.5, 1 y mi variable de respuesta es continua, todo dentro del set de datos

Dear R-users, I want to fit a linear model with Y as response variable and X a categorical variable (with 4 categories), with the aim of comparing the basal category of X (category=1) with category 4. Unfortunately, there is another categorical variable with 2 categories which interact with x and I have to include it, so my model is s "reg3: Y=x*x3". Using fit.contrast to make the

Hi, I am analyzing a repeated-measures Greco-Latin Square with the aov command. I am using aov to calculate the MSs and then picking by hand the appropriate neumerator and denominator terms for the F tests. The data are the following: responseFinger mapping.code Subject.n index middle ring little ---------------------------------------------------------------------------- 1 1

Hi everyone, Can someone explain me how to calculate SAS type III sums of squares in R? Not that I would like to use them, I know they are problematic. I would like to know how to calculate them in order to demonstrate that strange things happen when you use them (for a course for example). I know you can use drop1(lm(), test="F") but for an lm(y~A+B+A:B), type III SSQs are only

Dear R-users: I can't run the following code from "Mixed Effects Models in S and S-plus". library( nlme ) options( width = 65, digits = 5 ) options( contrasts = c(unordered = "contr.helmert", ordered = "contr.poly") ) # Chapter 5 Extending the Basic Linear Mixed-Effects Models # 5.1 General Formulation of the Extended Model data( Orthodont ) vf1Fixed

Hello R-users, I am trying to obtain Type III SS for an ANOVA with subsampling. My design is slightly unbalanced with either 3 or 4 subsamples per replicate. The basic aov model would be: fit <- aov(y~x+Error(subsample)) But this gives Type I SS and not Type III. But, using the drop() option: drop1(fit, test="F") I get an error message: "Error in

I'm running a series of simple bivariate linear regressions on grouped data. I want to test the slopes to see if they are parallel. I normally use analysis of covariance to do so, looking at interaction between the covariate and the factor to make this determination. VR3 pp.149 - 154 has a very nice example of an ANOCOVA, ending with a discussion of this very operation. My question has

Code: > options("contrasts") $contrasts factor ordered "contr.treatment" "contr.poly" I want to change the first entry ONLY, without retyping "contr.poly". How do I do it? I have tried various possibilities and cannot get anything to work. I found out that the response to options("contrasts") has class

Hi all, I am trying to do a ordered probit regression using polr(), replicating a result from SAS. >polr(y ~ x, dat, method='probit') suppose the model is y ~ x, where y is a factor with 3 levels and x is a factor with 5 levels, To get coefficients, SAS by default use the last level as reference, R by default use the first level (correct me if I was wrong), The result I got is a