search for: xlevel

...The effect is evaluated at the mean age. > e Sex effect Sex F M 43.33083 44.48531 > > e$model.matrix (Intercept) Age SexM 1 1 130.5859 0 23 1 130.5859 1 To evaluate the effect at Age=120 I tried: e <-effect("Sex",mod,xlevels=list(Age=c(120))) but the effect was still evaluated at 130.5859. Is this an incorrect usage of xlevels? Thanks, David

I'm working on predict.survreg and am confused about xlevels. The model.frame method has the argument, but none of the standard methods (model.frame.lm, model.frame.glm) appear to make use of it. The documentation for model.matrix states: xlev: to be used as argument of model.frame if data has no "terms" attribute. But the terms attribute has...

...= runif(200,0.5,10) ) mx <- glm( D ~ ns(x,knots=1:2,Bo=c(0,5)) + f:I(x^2) , offset=log(Y) , family=poisson, data=dd) mi <- glm( D ~ ns(x,knots=1:2,Bo=c(0,5)) + f:I(x^2) + offset(log(Y)), family=poisson, data=dd) attr(mx$terms,"dataClasses") attr(mi$terms,"dataClasses") mi$xlevels mx$xlevels ...so far not quite there. Regards, Bendix Carstensen Senior Statistician Steno Diabetes Center Clinical Epidemiology Niels Steensens Vej 2-4 DK-2820 Gentofte, Denmark b at bxc.dk bendix.carstensen at regionh.dk http://BendixCarstensen.com ________________________________ Denne e...

Hi, Some try: > names(mi$xlevels) [1] "f" > all.vars(mi$formula) [1] "D" "x" "f" "Y" > names(mx$xlevels) [1] "f" > all.vars(mx$formula) [1] "D" "x" "f" When offset is indicated out of the formula, it does not work... Marc Le 0...

...#39;ve been wondering about this for a while. I am sorry to ask for your time, but can one of you help me understand this? This concerns duplicated labels, not levels, in the factor function. I think it is hard to understand that factor() fails, but levels() after does not > x <- 1:6 > xlevels <- 1:6 > xlabels <- c(1, NA, NA, 4, 4, 4) > y <- factor(x, levels = xlevels, labels = xlabels) Error in `levels<-`(`*tmp*`, value = if (nl == nL) as.character(labels) else paste0(labels, : factor level [3] is duplicated > y <- factor(x, levels = xlevels) > levels(y) &...

...lt;- cut(a, breaks=c(0,2,6,9,12), right=FALSE ) (x <- data.frame(a, a.cut)) # In 'x' the level "[0,2)" is "missing". AddMissingLevel <- function(xdf) { xfac <- factor( c("[0,2)", "[2,6)", "[6,9)", "[9,12)") ) xlevels <- levels(xfac) if(length(xlevels) != nlevels(factor(xdf$a.cut))) { v <- setdiff(xlevels, factor(xdf$a.cut)) u <- data.frame(a = 0, a.cut = v) x <- rbind(u, x) } return(x) } AddMissingLevel(x) Does a more general approach exist, e.g. using "exp...

..."), labels=c("Before Christ","After Christ")) y <- rnorm(6) m1 <- lm (y ~ xf ) plot(y ~ xf) abline (m1) ## Just a little problem the line does not "go through" the box ## plot in the right spot because contrasts(xf) is 0,1 but ## the plot uses xf in 1,2. xlevels <- levels(xf) newdf <- data.frame(xf=xlevels) ypred <- predict(m1, newdata=newdf) ##Watch now: the plot comes out "reversed", AC before BC plot(ypred ~ newdf$xf) ## Ah. Now I see: levels(newdf$xf) ## Why doesnt newdf$xf respect the ordering of the levels? -- Paul E. Joh...

...e 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 $xlevels component of the object, but not the FALSE/TRUE values for a logical predictor even though the latter is treated as a factor in the fit. For example: ------------ snip -------------- > m1 <- lm(Sepal.Length ~ Sepal.Width + Species, data=iris) > m1$xlevels $Species [1] "setosa&quot...

....vars(mg$terms) all.vars(mg$formula) ________________________________________ Fra: Marc Girondot <marc_grt at yahoo.fr> Sendt: 8. marts 2018 06:26 Til: Bendix Carstensen; r-help at r-project.org Emne: Re: [R] Names of variables needed in newdata for predict.glm Hi, Some try: > names(mi$xlevels) [1] "f" > all.vars(mi$formula) [1] "D" "x" "f" "Y" > names(mx$xlevels) [1] "f" > all.vars(mx$formula) [1] "D" "x" "f" When offset is indicated out of the formula, it does not work... Marc Le 0...

Dear list, I have a problem with freq from prettyR. Please have a look at my syntax with a litte example: library(prettyR) #Version 1 test.df<-data.frame(q1=sample(1:4,8,TRUE), gender=sample(c("f","m"),8,TRUE)) test.df freq(test.df) #No error message #Version 2 test.df<-data.frame(gender=sample(c("f","m"),8,TRUE), q1=sample(1:4,8,TRUE)) test.df

...ould argue about what's the better approach but not about what lm() does. BTW, I prefer treating a logical predictor as a factor because the predictor is essentially categorical. > > I note that the use factors and characters, doesn't necessarily > produce consistent output, for $xlevels. > (Because factors can have their levels re-ordered). Again, this misses the point: Both factors and character predictors produce elements in $xlevels; logical predictors do not, even though they are treated in the model as factors. That factors have levels that aren't necessarily ordered...

I was looking at how the model.frame method for lm works and comparing it to my own for coxph. The big difference is that I try to retain xlevels and predvars information for a new model frame, and lm does not. I use a call to model.frame in predict.coxph, which is why I went that route, but never noted the difference till now (preparing for my course in Nashville). Could someone shed light on the rationale for non-preservation? Terry...

...to mathematical function > names(aovmod) [1] "coefficients" "residuals" "effects" "rank" [5] "fitted.values" "assign" "qr" "df.residual" [9] "contrasts" "xlevels" "call" "terms" [13] "model" > names(lmmod) [1] "coefficients" "residuals" "effects" "rank" [5] "fitted.values" "assign" "qr"...

...d like to predict the response variable (Throughput) for unseen factor levels. When I try to predict I get the following error: > throughput.pred <- predict(throughput.fit,experiments,type="response") Error in model.frame.default(Terms, newdata, na.action = na.action, xlev = object$xlevels) : factor 'No_databases' has new level(s) 200, 400, 600, 800, 1000 Of course these are new factor levels, it is exactly what I am trying to achieve i.e. extrapolate the values of Throughput. Can anyone please advice? Below I include all details. Thanks in advance, Best regards, Giova...

...err <- attr(Terms,"specials")$Error - 1 tl <- attr(Terms, "term.labels")[-err] int <- attr(Terms, "intercept") facs <- attr(Terms, "factors")[-c(1,1+err), -err, drop=FALSE] vars <- rownames(facs) xl <- attr(object, "xlevels") if(!length(xl)) { # no factors in model return(as.list(coef(object))) } nxl <- setNames(rep.int(1, length(vars)), vars) tmp <- unlist(lapply(xl, length)) nxl[names(tmp)] <- tmp lterms <- apply(facs, 2L, function(x) prod(nxl[x > 0])) nl <-...

...sorry to ask for your > time, but can one of you help me understand this? > This concerns duplicated labels, not levels, in the factor function. > I think it is hard to understand that factor() fails, but levels() > after does not >> x <- 1:6 >> xlevels <- 1:6 >> xlabels <- c(1, NA, NA, 4, 4, 4) >> y <- factor(x, levels = xlevels, labels = xlabels) > Error in `levels<-`(`*tmp*`, value = if (nl == nL) > as.character(labels) else paste0(labels, : > factor level [3] is duplicated >> y &...

...~ x2 + x3 + x1, data = df, drop.unused.levels = TRUE) 11: eval(expr, envir, enclos) 10: eval(mf, parent.frame()) 9: lm(formula = y ~ x2 + x3 + x1, data = df, method = "model.frame") 8: eval(expr, envir, enclos) 7: eval(fcall, env, parent.frame()) 6: model.frame.lm(fob, xlev = object$xlevels) 5: model.frame(fob, xlev = object$xlevels) 4: stats:::add1.lm(object, scope = scope, scale = scale) 3: addterm.lm(fit, scope$add, scale = scale, trace = max(0, trace - 1), k = k, ...) 2: addterm(fit, scope$add, scale = scale, trace = max(0, trace - 1), k = k, ...) 1: stepAIC(mid, scope = lis...

...e package "lqs"? > names(slts) [1] "crit" "sing" "coefficients" "bestone" [5] "fitted.values" "residuals" "scale" "terms" [9] "call" "xlevels" "model" Also is there any plotting available for lqs? And trying plot on the lqs object doesn't work: > data(stackloss) > set.seed(123) > examplelqs <- lqs(stack.loss ~ ., data = stackloss) > plot (examplelqs) Error in as.double.default(x) : (list) objec...

...n predictor + offset : longer object length is not a multiple of shorter object length When predict used with model for which offset was defined in the formula, an error occurs predict( c1f.glm, newdata = nd ) Error in model.frame.default(Terms, newdata, na.action = na.action, xlev = object$xlevels) : variable lengths differ (found for 'offset(q25)') even if a column for offset is included in newdata, ndf <- cbind( nd, "offset(q25)" = rep( qlogis(0.25), length(cc) ) ) predict( c1f.glm, newdata = ndf ) Error in model.frame.default(Terms, newdata, na.action = na.acti...

...ression analysis in R. Using "names( blah.lm )" in R yields [1] "coefficients" "residuals" "effects" "rank" [5] "fitted.values" "assign" "qr" "df.residual" [9] "xlevels" "call" "terms" "model" In Splus, the same command yields [1] "coefficients" "residuals" "fitted.values" "effects" [5] "R" "rank" "assign&quo...