similar to: interpret significance from the contr.poly() function

Displaying 20 results from an estimated 100 matches similar to: "interpret significance from the contr.poly() function"

2013 Feb 08
1
Contrasts for a data
Hi, I am using a data called Rail in the nlme package. The data contains two variables: Rail and Travel. >Rail Grouped Data: travel ~ 1 | Rail Rail travel 1 1 55 2 1 53 3 1 54 4 2 26 5 2 37 6 2 32 7 3 78 8 3 91 9 3 85 10 4 92 11 4 100 12 4 96 13 5 49 14 5 51 15 5 50 16 6 80
2000 Aug 13
2
Possible bug (PR#633)
2008 Jul 14
5
A question about using function plot
Hello, everyone! I have spent two hours to get what I wanted in a simple situation and have not been successful. The set up is extremely simple x = c(1,2,4,8) y = c(1,2,3,4) plot(x, y) What I need, however, is for the 4 points of 1,2,4,8 to be spaced evenly, not by their numerical scale. Also, there should not be any other values such as 5,6,7 marked on the axis. So I tried the following x =
2009 Nov 08
2
reference on contr.helmert and typo on its help page.
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').
2012 Oct 05
1
Setting the desired reference category with contr.sum
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: $contrasts [,1] [,2] [,3] [,4] [,5] 1 1 0 0 0 0 2 0 1 0
2010 Feb 23
0
Name for factor's levels with contr.sum
Hi R-useRs, after having read http://tolstoy.newcastle.edu.au/R/help/05/07/8498.html with the same topic but five years older. the solution for a contr.sum with names for factor levels for R version 2.10.1 will be to comment out the following line #colnames(cont) <- NULL in contr.sum i guess? by the way, with contrasts=FALSE colnames are set, so i don't know what the aim is to avoid
2013 Apr 27
1
Error in `contrasts<-`(`*tmp*`, value = contr.funs[1 + isOF[nn]]) :
i am getting the following error Error in `contrasts<-`(`*tmp*`, value = contr.funs[1 + isOF[nn]]) : contrasts can be applied only to factors with 2 or more levels can any on e suggest how to rectify [[alternative HTML version deleted]]
2002 Nov 25
1
Contr.poly for n > 100 (PR#2326)
Full_Name: David Clifford Version: Version 1.5.1 (2002-06-17) OS: Red Hat 7.3 Submission from: (NULL) (128.135.149.55) For n values above 100 there appears to be a bug in contr.poly(n). The contrast matrix should have rank n-1. Running the code below gives output (ie errors) at n=98, 100 and every value greater than 102. for(n in 2:150) { K <- contr.poly(n) rnk <-
2008 May 20
1
contr.treatments query
Hi Folks, I'm a bit puzzled by the following (example): N<-factor(sample(c(1,2,3),1000,replace=TRUE)) unique(N) # [1] 3 2 1 # Levels: 1 2 3 So far so good. Now: contrasts(N)<-contr.treatment(3, base=1, contrasts=FALSE) contrasts(N) # 1 2 # 1 1 0 # 2 0 1 # 3 0 0 whereas: contr.treatment(3, base=1, contrasts=FALSE) # 1 2 3 # 1 1 0 0 # 2 0 1 0 # 3 0 0 1 contr.treatment(3, base=1,
2005 Jul 13
1
Name for factor's levels with contr.sum
Good morning, I used in R contr.sum for the contrast in a lme model: > options(contrasts=c("contr.sum","contr.poly")) > Septo5.lme<-lme(Septo~Variete+DateSemi,Data4.Iso,random=~1|LieuDit) > intervals(Septo5.lme)$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
2010 Aug 29
2
glm prb (Error in `contrasts<-`(`*tmp*`, value = "contr.treatment") : )
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 =
2010 Sep 15
1
contr.sum, model summaries and `missing' information
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
2009 Jan 23
1
Interpreting model matrix columns when using contr.sum
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
2004 Aug 20
1
drop1 with contr.treatment
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
2012 Oct 27
1
contr.sum() and contrast names
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
2005 Jun 20
1
(no subject)
R friends, I am using R 2.1.0 in a Win XP . I have a problem working with lists, probably I do not understand how to use them. Lets suppose that a set of patients visit a clinic once a year for 4 years on each visit a test, say 'eib' is performed with results 0 or 1 The patients do not all visit the clinic the 4 times but they missed a lot of visits. The test is considered positive if it
2000 Apr 06
0
Please inform 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
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
2010 Jul 07
6
forcing a zero level in contr.sum
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]]
2009 Jul 30
2
lattice shingle plot axis annotation
Hello (R-)Experts I hope someone can help with this problem concerning axis annotation of a lattice shingle plot. I want a plot with three shingles to display some laboratory value over time. In the first panel over the first few days, then in the next panel some months, and in the last panel some years. In the following minimal example the axis annotation will be in days, but I'd like to
2011 Feb 02
2
unequally spaced factor levels orthogonal polynomial contrasts coefficients trend analysis
Hello [R]-help I am trying to find > a package where you can do ANOVA based trend analysis on grouped data > using orthogonal polynomial contrasts coefficients, for unequally > spaced factor levels. The closest hit I've had is from this web site: >(http://webcache.googleusercontent.com/search?q=cache:xN4K_KGuYGcJ:www.datavis.ca/sasmac/orpoly.html+Orthogonal+polynomial >l but I