search for: contr

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...

...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 many contributors. Type 'contributors()' for more information and 'citation()' on how to cite R or R packages in publications. Below are two cases where I don't seem to be getting contr.sum contrasts even though they were specified. Are these bugs? The first case is an interaction betwee...

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...

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 sh...

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 "contr.identity" and it shown at the bottom of the email for refer...

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 mat...

...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 contrast (category 1 vs category 4) with options(contrasts=c("contr.treatment", "contr.poly")), it makes the contrast but just for the basal category of x3, (coincident with the corresponding result of summary(reg3)), so that it is not what I am looking for, and i...

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 -0.39...

For ordered factor the natural contrast coding would be to parametrize by the succsessive differences between levels, which does not assume equal spacing of factor levels as does the polynomial contrasts (implicitly at least). This requires the contr.cum, which could be: contr.cum <- function (n, contrasts = TRUE) { if (is.n...

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 SPSS's 'constant' if I use contr.sum in R. Can someone please confirm that that is a correct match for the SP...

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').

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 1...

...2 a x 3 x x 4 x x So of course the full design matrix is singular, this is expected. > model.matrix(~ A * B, df) (Intercept) Aa Bb Aa:Bb 1 1 1 1 1 2 1 1 0 0 3 1 0 0 0 4 1 0 0 0 attr(,"assign") [1] 0 1 2 3 attr(,"contrasts") attr(,"contrasts")$A [1] "contr.treatment" attr(,"contrasts")$B [1] "contr.treatment" I'd like to drop the B main effect column, but get the unexpected result of a column of zeroes. > model.matrix(~ A * B - B, df) (Intercept) Aa Ax:Bb A...

This is my first post to this list so I suppose a quick intro is in order. I've been using SPLUS 2000 and R1.6.2 for just a couple of days, and love S already. I'm reading MASS and also John Fox's book - both have been very useful. My background in stat software was mainly SPSS (which I've never much liked - thanks heavens I've found S!), and Perl is my tool of choice for

...a. I want to repeat this process 999 times. However, i am getting an error when i use the "for i in" function. Is there any way to repeat this analysis 999 times. Here are the codes i used ; data4 <- matrix(c(sample(id),data1),203,3) a <- data4[,1] random.case=data4[a==0,] random.contr=data4[a==1,] random.case.locations<-list(x=random.case[,1],y=random.case[,2]) ppregion(xl=min(random.case[,1])-0.0001,xu=max(random.case[,1])+0.0001,yl=min(random.case[,2])-0.0001,yu=max(random.case[,2])+0.0001) random.l.case <- Kfn(random.case.locations,0.01) random.k.case <- ((random.l...

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, contrasts=TRUE) # 2 3 # 1 0 0 # 2 1 0 # 3 0 1 I can follow the last two f...

...t,dbp,dbp1) > sapply(dat1,data.class) Patient Visit Center Treatment dbp dbp1 "factor" "factor" "factor" "factor" "numeric" "numeric" After which the following code was run > library(nlme) # needed for the contrast contr.SAS > options(contrasts=c(factor="contr.SAS",ordered="contr.poly")) > res <- lm(dbp~Treatment+Visit+dbp1) > anova(res) > summary(res) and was repeated leaving out library(nlme) and replacing > options(contrasts=c(factor="contr.SAS",ordered=...

...2, 0.5, 1 y mi variable de respuesta es continua, todo dentro del set de datos llamado "datos". Mi asesor de tesis me recomienda que en vez de aplicar una anova de dos factores, que mejor aproveche que los factores A y B son continuos en vez de categóricos, y que mejor aplique un contraste polinomial para analizar los datos. Como se habrán dado cuenta, los niveles de los factores no son equidistantes, condición que según entiendo es necesaria para poder codificar los mismos con la función "contr.poly". Investigando por la red encontré que existe una forma de co...

Dear R users, When fitting a lme() object (from the nlme library), is it possible to test interactions *before* main effects? As I understand, R conventionally re-orders all terms such that highest-order interactions come last - but I??d like to know if it??s possible (and sensible) to change this ordering of terms. I??ve tried the terms() command (from aov) but I don??t know if something

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 0 0 3 0 0 1 0 0 4 0 0 0 1 0 5 0 0...