similar to: contr.sum() and contrast names

Hi all, I'm doing nonlinear regressions on data with several factors. I want to fit say a logistic curve with different parameter values for each factor level. So I'm doing something like: tmp <- by( myData, list(myFactor1, myFactor2), function(x) nls(...) ) It works fine. However, I could not find an easy way to retrieve fitted values. I can use fitted() on each element of tmp, but

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

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

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 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

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]]

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 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,

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 

Dear R-devel list members, I'd like to suggest a more flexible procedure for generating contrast names. I apologise for a relatively long message -- I want my proposal to be clear. I've never liked the current approach. For example, the names generated by contr.treatment paste factor to level names with no separation between the two; contr.sum simply numbers contrasts (I recall an

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

OK, my last question didn't get any replies so I am going to try and ask a different way. When I generate contrasts with contr.sum() for a 3 level categorical variable I get the 2 orthogonal contrasts: > contr.sum( c(1,2,3) ) [,1] [,2] 1 1 0 2 0 1 3 -1 -1 This provides the contrasts <1-3> and <2-3> as expected. But I also want it to create <1-2> (i.e.

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

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').

Hello, I've got what I'd expect to be a pretty simple issue: I fit an aov object using multiple error strata, and would like some significance tests for the contrasts I specified. In this contrived example, I model some test score as the interaction of a subject's gender and two emotion variables (angry, happy, neutral), measured at entry to the experiment (entry) and later

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, 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

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

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