similar to: Help with contrasts

Displaying 20 results from an estimated 40000 matches similar to: "Help with contrasts"

2011 May 18
1
Need expert help with model.matrix
Dear experts: Is it possible to create a new function based on stats:::model.matrix.default so that an alternative factor coding is used when the function is called instead of the default factor coding? Basically, I'd like to reproduce the results in 'mat' below, without having to explicitly specify my desired factor coding (identity matrices) in the 'contrasts.arg'. dd
2010 Apr 21
5
Bugs? when dealing with contrasts
R version 2.10.1 (2009-12-14) Copyright (C) 2009 The R Foundation for Statistical Computing ISBN 3-900051-07-0 R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type 'license()' or 'licence()' for distribution details. Natural language support but running in an English locale R is a collaborative project with
2005 Apr 13
2
multinom and contrasts
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
2001 Jun 15
1
contrasts in lm and lme
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") >
2001 Feb 08
2
Test for multiple contrasts?
Hello, I've fitted a parametric survival model by > survreg(Surv(Week, Cens) ~ C(Treatment, srmod.contr), > data = poll.surv.wo3) where srmod.contr is the following matrix of contrasts: prep auto poll self home [1,] 1 1 1.0000000 0.0 0 [2,] -1 0 0.0000000 0.0 0 [3,] 0 -1 0.0000000 0.0 0 [4,] 0 0 -0.3333333 1.0 0 [5,] 0 0
2004 Mar 03
1
Confusion about coxph and Helmert contrasts
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
2002 Dec 01
1
generating contrast names
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
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 =
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
2008 Aug 26
2
options("contrasts")
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
2012 May 11
1
set specific contrasts using lapply
I have the following data set > data A B X1 X2 Y 1 A1 B1 1.1 2.9 1.2 2 A1 B2 1.0 3.2 2.3 3 A2 B1 1.0 3.3 1.6 4 A2 B2 0.5 2.6 3.1 > sapply(data, class) A B X1 X2 Y "factor" "factor" "numeric" "numeric" "numeric" I'd like to set a specific type of contrasts to all the categorical factors
2006 Aug 17
1
Setting contrasts for polr() to get same result of SAS
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
2007 Oct 09
2
fit.contrast and interaction terms
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
2008 Oct 11
2
R vs SPSS contrasts
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
2000 Aug 27
1
under certain conditions, model.matrix appears to lack one column (PR#646)
Dear R Team, # Summary of the problem: setting contrasts as > contrasts(g) <- contr.treatment or > contrasts(g) <- matrix(c(1,-1,0),ncol=1) (i.e. without quotes around `contr.treatment' or `contr.sum', etc.) and fitting an lm model without an intercept results in a model matrix that lacks one column. (I do ask for forgiveness if this is not a bug but is due to my
2007 May 17
1
model.matrix bug? Nested factor yields singular design matrix.
Hi all, I believe this is a bug in the model.matrix function. I'd like a second opinion before filing a bug report. If I have a nested covariate B with multiple values for just one level of A, I can not get a non-singular design matrix out of model.matrix > df <- data.frame(A = factor(c("a", "a", "x", "x"), levels = c("x",
2006 Dec 14
2
Asymmetrical ANOVA / contrasts
Dear all, I have problems to code contrasts for performing an asymmetrical anova with aov(). I am using aov() because I want to get the Mean Squares for further analyses. I didn't find any solution to my problem in the help files of functions aov(), contrasts(), C(), etc. Let's say I have three locations, one with treatment P, and two with treatment C: >
2009 Sep 17
2
What does model.matrix() return?
Hi, I don't understand what the meaning of the following lines returned by model.matrix(). Can somebody help me understand it? What can they be used for? attr(,"assign") [1] 0 1 2 2 attr(,"contrasts") attr(,"contrasts")$A [1] "contr.treatment" attr(,"contrasts")$B [1] "contr.treatment" Regards, Peng > a=2 > b=3 > n=4
2002 Nov 07
4
Preferable contrasts?
Dear all, I'm working with Cox-regression, because data could be censored. But in this particular case not. Now I have a simple example: PRO and PRE are (0,1) coded. The response is not normal distributed. We are interested in a model which could describe interaction. But my results are depending strongly in the choose of the contrast option. It is clear that there is some dependence in
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,