similar to: glm prb (Error in `contrasts<-`(`*tmp*`, value = "contr.treatment") : )

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]]

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,

Hi! I would like to suggest to make it possible, in one way or another, to get meaningful contrast names when using contr.sum(). Currently, when using contr.treatment(), one gets factor levels as contrast names; but when using contr.sum(), contrasts are merely numbered, which is not practical and can lead to mistakes (see code at the end of this message). This issue was discussed quickly in 2005

Dear list members, I have a 2-factor ANOVA where the summary.lm output looks like this (using treatment contrasts): Value Std. Error t value Pr(>|t|) (Intercept) 0.0389 0.0220 1.7695 0.0817 as.factor(Block)1 0.0156 0.0066 2.3597 0.0215 as.factor(Block)2 -0.0018 0.0037 -0.4857 0.6289 as.factor(Block)3 -0.0007 0.0026 -0.2812 0.7795

Hello all, (Apologies in advance if my terminology is incorrect, I'm relatively new to R and statistics). I have data from a factorial design with two treatments (CRF-23), and I'm trying to compute treatment-contrast interactions through analysis of variance. I can't figure out how to do contrasts properly, despite reading the help for "C" and "contrasts"

Hi, I am trying to find out a collinearity in explanatory variables with alias(). I creat a dataframe: dat <- ds[,sapply(ds,nlevels)>=2] dat$Y <- Response Explanatory variables are factor and response is continuous random variable. When I run a regression, I have the following error: fit <- aov( Y ~ . , data = dat) Error in "contrasts<-"(`*tmp*`, value =

Dear R Core Team I've a proposal to improve drop1(). The function should change the contrast from the default ("treatment") to "sum". If you fit a model with an interaction (which ist not signifikant) and you display the main effect with drop1( , scope = .~., test = "F") If you remove the interaction, then everything's okay. There is no way to fit a

I have a rather large data set (about 30 predictor variables) I need to preform a logistic regression on this data. My response variable is binary. My code looks like this: mylogit <- glm(Enrolled ~ A + B + C + ... + EE, data = data, family = binomial(link="logit")) with A,B,C, ... as my predictor variables. Some categorical, some continuous, some binary. I run the code and get

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

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 

I need to use contr.sum and observe that some levels are not statistically different from the overall mean of zero. What is the proper way of forcing the zero estimate? It seems the column corresponding to that level should become a column of zeros. Is there a way to achieve that without me constructing the design matrix? Thank you. Stephen Bond [[alternative HTML version deleted]]

Dear list, I want to perform a logistic regression analysis with multiple categorical predictors (i.e., a logit) on some data where there is a very definite relationship between one predicator and the response/independent variable. The problem I have is that in such a case the p value goes very high (while I as a naive newbie would expect it to crash towards 0). I'll illustrate my problem

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

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, I need to build a function to generate one column for each level of a factor in the model matrix created on an arbitrary formula (instead of using the available contrasts options such as contr.treatment, contr.SAS, etc). My approach to this was first to use the built-in function for contr.treatment but changing the default value of the contrasts argument to FALSE (I named this function

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

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

Hi Folks, I'm comparing some output from R with output from SPSS. The coefficients of the independent variables (which are all factors, each at 2 levels) are identical. However, R's Intercept (using default contr.treatment) differs from SPSS's 'constant'. It seems that the contrasts were set in SPSS using /CONTRAST (varname)=Simple(1) I can get R's Intercept to match

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

A common point made in discussion of contrasts, type I, II, III SS etc is that for sensible comparisons one should use contrasts that are 'orthogonal in the row-basis of the model matrix' (to quote from http://finzi.psych.upenn.edu/R/Rhelp02/archive/111550.html) Question: How would one check, in R, that this is so for a particular fitted linear model object? Steve Ellison