similar to: Error in `contrasts<-`(`*tmp*`, value = contr.funs[1 + isOF[nn]]) :

Hi, I have been running the same code without problem for the last few days, changing data sets etc with no issue. Today I changed the covariates for the model and am now getting this error message: Error in `contrasts<-`(`*tmp*`, value = contr.funs[1 + isOF[nn]]) : contrasts can be applied only to factors with 2 or more levels Everything in the code looks the same to me, but I'm a

I have a rather large data set (about 30 predictor variables) I need to preform a logistic regression on this data. My response variable is binary. My code looks like this: mylogit <- glm(Enrolled ~ A + B + C + ... + EE, data = data, family = binomial(link="logit")) with A,B,C, ... as my predictor variables. Some categorical, some continuous, some binary. I run the code and get

Is this a bug, or have I misunderstood the proper use of lm? Thanks, Whit code: x <- rnorm(50) y <- matrix(as.logical(round(runif(100),0)),ncol=2) NROW(x)==NROW(y) lm(x~y) > x <- rnorm(50) > y <- matrix(as.logical(round(runif(100),0)),ncol=2) > NROW(x)==NROW(y) [1] TRUE > lm(x~y) Error in "[[<-.data.frame"(`*tmp*`, nn, value = c(2, 1, 2, 1, 1, 1, 2, :

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 =

m1.logit <-lm(default ~ amt.fac + age.fac + duration + chk_acct + history + purpose + sav_acct + employment + install_rate + pstatus

Muy buenas, He usado muy poco R en mi vida, y ahora estoy intentando hacer una MANOVA, con dos variables independientes (Edad, que tiene 4 niveles y Genero que tiene 2) y 5 variables independientes. Le he puesto también que me haga el poshoc de las dos y la interacción. Decir también que me instalé el ULLRToolbox, que no se si tendrá algo que ver con el error que me da. Este es el código que uso

Hola: Probablemente está tomando 'Edad_Manova_18a54' y 'Genero2_1a2' como variables numéricas (integer). Deberías convertirlas en factor. Por ejemplo: Edad_Manova_18a54 <- factor(Edad_Manova_18a54) Genero2_1a2 <- factor(Genero2_1a2) Manova.fnc(datosPAS, variables=2:6, fac.inter=c('Edad_Manova_18a54','Genero2_1a2'),poshoc=c('todos')) Saludos,

Dear Abby, > On Aug 30, 2019, at 8:20 PM, Abby Spurdle <spurdle.a at gmail.com> wrote: > >> I think that it would be better to handle factors, character predictors, and logical predictors consistently. > > "logical predictors" can be regarded as categorical or continuous (i.e. 0 or 1). > And the model matrix should be the same, either way. I think that

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

Dear R-devel list members, I've discovered an inconsistency in how lm() and similar functions handle logical predictors as opposed to factor or character predictors. An "lm" object for a model that includes factor or character predictors includes the levels of a factor or unique values of a character predictor in the $xlevels component of the object, but not the FALSE/TRUE values

Hello everyone, I am trying to fit the following model All X. variables are continuous, while the conditions are categoricals. model <- lm(X2

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 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 

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 17.335324 25.088661 Variete2 -3.3794907 6.816101 17.011692 Variete3

Dear Bill, Thanks for pointing this difference out -- I was unaware of it. I think that the difference occurs in model.matrix.default(), which coerces character variables but not logical variables to factors. Later it treats both factors and logical variables as "factors" in that it applies contrasts to both, but unused factor levels are dropped while an unused logical level is not. I

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,

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]]

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

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