similar to: Mismatches in predict(newdata)

Hello! I'm still working on my problem, which also occurs with the predict.lm() function. - Providing newdata, which is a data.frame with all variables being "numeric", as str() shows, R tells me the following: ar1.xpred.test.pred <- predict(ar1.xpred.fitted, regdata.test, se.fit = FALSE) Fehler: variable 'lag(ret1)' was fitted with type "numeric" but type

Many thanks for your very useful comments and suggestions. Renaud 2006/5/30, Prof Brian Ripley <ripley at stats.ox.ac.uk>: > On Tue, 30 May 2006, Prof Brian Ripley wrote: > > > This is not really a bug. See > > > > http://developer.r-project.org/model-fitting-functions.txt > > > > for how this is handled in other packages. All model-fitting in R used =

I have a function that fits polynomial models for the orders in n: lmn <- function(d,n){ models=list() for(i in n){ models[[i]]=lm(y~poly(x,i),data=d) } return(models) } My data is: > d=data.frame(x=1:10,y=runif(10)) So first just do it for a cubic: > mmn = lmn(d,3) > predict(mmn[[3]]) 1 2 3 4 5 6 7 8

all.vars works fine, EXCEPT, it give a bit too much. I only want the regression variables, but in the following example I also get "k" the variable holding the chosen knots. Any machinery to find only "real" regression variables? cheers, Bendix library( splines ) y <- rnorm(100) x <- rnorm(100) k <- -1:1 ml <- lm( y ~ bs(x,knots=k) ) mg <- glm( y ~

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 07/03/2018 ? 06:20, Bendix Carstensen a ?crit?: > I would like

I would like to extract the names, modes [numeric/factor] and levels of variables needed in a data frame supplied as newdata= argument to predict.glm() Here is a small example illustrating my troubles; what I want from (both of) the glm objects is the vector c("x","f","Y") and an indication that f is a factor: library( splines ) dd <- data.frame( D =

Dear list, I have encountered a problem with predict.nls (Windows XP, R.2.5.0), but I am not sure if it is a bug... On the nls man page, an example is: DNase1 <- subset(DNase, Run == 1) fm2DNase1 <- nls(density ~ 1/(1 + exp((xmid - log(conc))/scal)), data = DNase1, start = list(xmid = 0, scal = 1)) alg = "plinear", trace =

Dear All, I have a question regarding predict.rpart. I use rpart to build classification and regression trees and I deal with data with relatively large number of input variables (predictors). For example, I build an rpart model like this rpartModel <- rpart(Y ~ X, method="class", minsplit =1, minbucket=nMinBucket,cp=nCp); and get predictors used in building the model like

Bug in predict.lm & poly The predict function doesn't work when used with poly and newdata. For example, I'd expect the following code to work, and plot a fitted cubic to the nearly straight line: x <- 1:10 y <- x + rnorm(10)/100 plot(x,y) fit <- lm(y ~ poly(x,3)) newx <- seq(1,10,len=100) lines(newx,predict(fit,newdata=data.frame(x=newx))) However, the plotted

I've tried to predict the values from a new data.frame using the nls.predict function and keep getting the error message: Error in if (sum(wrong) == 1) stop(gettextf("variable '%s' was fitted with class \"%s\" but class \"%s\" was supplied", : missing value where TRUE/FALSE needed I first thought that it was becuase there may have been something

Below is a SIMEX object that was generated with the "simex" function from the "simex" package applied to a logistic regression fit. From this mountain of information I would like to extract all of the values summarized in this line: .. ..$ variance.jackknife: num [1:5, 1:4] 1.684 1.144 0.85 0.624 0.519 ... Can someone suggest how to go about doing this? I can extract the

I am more than 'a little disappointed' that you expect a detailed explanation of the problems with your 'bug' report, especially as you did not provide any explanation yourself as to your reasoning (nor did you provide any credentials nor references). Note that 1) Your report did not make clear that this was only relevant to prediction intervals, which are not commonly used.

Dear R community, I have recently discovered the package oblique.tree and I must admit that it was a nice surprise for me, since I have actually made my own version of a kind of a classifier which uses the idea of oblique splits (splits by means of hyperplanes). So I am now interested in comparing these two classifiers. But what I do not seem to understand is why the function

Hi all I have run into a case where I don't understand why predict.lrm and predict.glm don't yield the same results. My data look like this: set.seed(1) library(Design); ilogit <- function(x) { 1/(1+exp(-x)) } ORDER <- factor(sample(c("mc-sc", "sc-mc"), 403, TRUE)) CONJ <- factor(sample(c("als", "bevor", "nachdem",

Dear All, I am attempting to make predictions based on a survreg() model with some censoring and a frailty term, as below: predict works fine on the original data, but not if I specify newdata. # a model with groups as fixed effect model1 <- survreg(Surv(y,cens)~ x1 + x2 + groups, dist = "gaussian") # and with groups as a random effect fr <- frailty(groups,

Hello! I'm puzzled by the following problem. It occurs while trying to predict responses in a test-dataset using a linear model fitted with regFit from the rMetrics "fRegression"-package. All goes well when I call "predict" using the training dataset. However, a call using the test-dataset retuns an error message - telling me that the latter dataset provides variables

Hello folks, it may be regarded as a user error to scale() your data prior to prcomp() instead of using its 'scale.' argument. However, it is a user thing that may happen and sounds a legitimate thing to do, but in that case predict() with 'newdata' can give wrong results: x <- scale(USArrests) sol <- prcomp(x) all.equal(predict(sol), predict(sol, newdata=x)) ## [1]

Dear community, I've a tree which included at first 23 variables. Then I've pruned this tree, and there are only 8 variables involved. I'd like to predict and only introduce in newdata the values of these 8 variables involved. However, as the tree was built with the 23, it asked me for 15 values, even if it doesn't need them. Is there a way to introduce only this 8 values?

Hi all - I'm stumped by the following mdl <- glm(resp ~ . , data = df, family=binomial, offset = ofst) WORKS yhat <- predict(mdl) WORKS yhat <- predict(mdl,newdata = df) FAILS Error in drop(X[, piv, drop = FALSE] %*% beta[piv]) : subscript out of bounds I've tried without offset, quoting binomial. The offset variable ofst IS in df. Previous postings indicate possible

Hello, Is this a bug or a feature? I am using R 2.7.1 on Apple OS X. > y <- matrix(1:3,nrow=3) # y is a single-column matrix > df <-data.frame(x=1:3,y=y) > sapply(df,data.class) x y "numeric" "numeric" > df$yy <- y > sapply(df,data.class) x y yy "numeric" "numeric" "matrix"