similar to: contr.sum, model summaries and `missing' information

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,

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 =

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

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

On 23/11/2020 17:37, MORILLO Jordi via samba wrote: > Hi Rowland, > Sorry to inform that none of thus packages solve my problem. > > But today, with some Tranquil.it helps, I have some news: > > - Upgrade from 4.11.14 -> 4.12.9 is OK > - Upgrade from 4.12.9 -> 4.13.2 : problem is present with Tranquil.it AND Louis package > - Fresh install + member join with 4.13.2 is

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

Hello ! I have just upgraded 40 x Samba domain member file server from 4.11.14 to 4.13.2 - No problem with 20 x domain member that are in a unique Samba domain (only samba DC) - But for my other domain (with composed of Windows 2016 DC), all of 20 x Samba domain member failed to desserve file after this upgrade :-/ I have triple check /etc/hosts, hostname, krb5 etc .... And

Hi Rowland, Sorry to inform that none of thus packages solve my problem. But today, with some Tranquil.it helps, I have some news: - Upgrade from 4.11.14 -> 4.12.9 is OK - Upgrade from 4.12.9 -> 4.13.2 : problem is present with Tranquil.it AND Louis package - Fresh install + member join with 4.13.2 is OK (Centos AND Buster packages) Problem only occur when upgrading member to 4.13.2 with

I'm running a linear model in R using the car package. I have a variable education, which i have recoded and regrouped to my wishes. However, R seems to place each element of that variable in alphabetical order. When I am running the model, don't I need the model order from lowest to highest to make an inference that a one unit change in one variable produced a one unit change in

Steve Frampton [mailto:frampton@j-com.co.jp] of your address change Content-Length: 7142 samba@samba.org David Barroso <h4371719@alumnes.eup.udl.es> Jeremy Allison <jeremy@valinux.com> "Tulipant Gergely" <tulipant-gergely@dbrt.hu> Edwards Philip M Contr AFRL/SNRR <Philip.Edwards@wpafb.af.mil> Drenning Bruce <bdrenni@catholicrelief.org> Glenn

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 colleagues, the code below generated the two effects plots that I have attached. I hope they are not stripped. The original two models are as follows: green_shift_mod=glm(green_shift ~ educ+party_id+educ:party_id, family=binomial, data=x) carbon_tax_mod=glm(carbon_tax ~ educ+party_id+educ:party_id, family=binomial, data=x) Then, I try to plot the effects of party_id by education for

I am trying to use tree to partition a data set. The data set has 3924 observations. Partitioning seems to work for small subsets of the data, but when I use the entire data set, no partitioning occurs. The variables are: RESP respondent to a survey (0 = not a respondent, 1 = respondent) AGE_P Age (continuous) ORIGIN_I Hispanic Ethnicity (1 = Hispanic, 2 = non-Hispanic) RACRECI2 Race