similar to: specifying model terms when using predict

Full_Name: Renaud Lancelot Version: Version 2.3.0 (2006-04-24) OS: MS Windows XP Pro SP2 Submission from: (NULL) (82.239.219.108) I think there is a bug in predict.lme, when a polynomial generated by poly() is used as an explanatory variable, and a new data.frame is used for predictions. I guess this is related to * not * using, for predictions, the coefs used in constructing the orthogonal

Say I fit a logistic model and want to calculate an odds ratio between 2 sets of predictors. It is easy to obtain the difference in the predicted logodds using the predict() function, and thus get a point-estimate OR. But I can't see how to obtain the confidence interval for such an OR. For example: model <- glm(chd ~age.cat + male + lowed, family=binomial(logit)) pred1 <-

Hi ! I am trying to get at the covariance of the predictions of a linear model. Suppose the we have: > x<-runif(1000) > y<-2 + 25x*x +rnorm(1000) > lm1 <-lm(y~x, data = data.frame(y = y, x=x)) > x.pred <-runif(10) > y.hat <- predict(lm1, newdata = data.frame(x=x.pred)) I was wondering how to get an estimate of the covariance of y.hat which would be a 10 x 10

Dear all; I have a dataframe with several columns. The columns are the elevation, volume and the area of the cells (which were placed inside a polygon). I have extracted them from DEM raster to calculate the volume under polygon and the elevation for a specific volume of the reservoir. > head(x6,2) Elevation Vol Area V_sum A_sum 1 2145 13990.38 85.83053 13990.38

Hello all, How do I actually use the output of predict.lm(..., type="terms") to predict new term values from new response values? I'm a chromatographer trying to use R (2.15.1) for one of the most common calculations in that business: - Given several chromatographic peak areas measured for control samples containing a molecule at known (increasing) concentrations, first

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

[Re-directing to R-devel, as I think this needs changes to the code.] Can I suggest a modification to stats:predict.princomp so that it will check for column (variable) names? In src/library/stats/R/princomp-add.R, insert the following after line 4: if (!is.null(cn <- names(object$center))) newdata <- newdata[, cn] Now Dana's example looks like: > predict(pca1, frz) Error in

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 am an R beginner and trying to run a market model using event study in R framework. First, I run a market model, that is lm(stock security~SP500 index, subset=Obs[197, 396]) ->result1 Then I get predict results for a new dataset using predict (result1, newdata=Obs[397,399]) ->pred1 Pred1 should have three numbers. Now I need to calculate abnormal return by the formula stock

Estimados compañeros hace unos días instale la versión última de R (3.4.1) en windows 10 y hoy me tiene loco el siguiente error cuando ejecuto predict para un árbol de clasificación-regresión. El problema da incluso con el ejemplo del comando predict library(tree) data(shuttle, package="MASS") shuttle.tr <- tree(use ~ ., shuttle, subset=1:253, mindev=1e-6,

Hi when I fit a glm by glm.fit(x,y,family = binomial()) and then try to use the object for prediction of newdata by: predict.glm(object, newdata) I get the error: Error in terms.default(object) : no terms component I know I can use glm() and a formula, but for my case I prefer glm.fit(x,y)... thanks for a hint christoph $platform [1] "i686-pc-linux-gnu" $arch [1]

Excuse the simple question... I'm not sure what I'm doing wrong with predict, but let me use this example: Suppose I do: dat<-matrix(c(0,0,10,20),2,byrow=T) lm1<-lm(dat[,2]~dat[,1]) Suppose I want to generate the linearly-interpolated y-values between the point (0,0) and (0,20) at every unit interval. I thought I just do: predict(lm1, data.frame(seq(0,10,1))) to get

Dear all, I'd like to plot the marginal effect of a variable in a multiplicative interaction regression, that is, the effect of a variable conditional on the values of another variable. As an illustration, given model lm1 lm1 <- lm(y ~ x*z) I'd like to get the effects of x on y conditional on the values of z, with the corresponding confidence intervals if possible. Does anyone know

Dear list, I'm a bit perplexed why the 95% confidence bands for the predicted probabilities for units where x=0 and x=1 overlap in the following instance. I've simulated binary data to which I've then fitted a simple logistic regression model, with one covariate, and the coefficient on x is statistically significant at the 0.05 level. I've then used two different methods to

Hi Everyone, I have been using the "tree" package for a while with no problems until now. When I run predict(tree, newdata), I get an error with the message "Corrupt tree" for about 50% of the trees that I generate with tree. For other trees, the predict function completes with no errors. I haven't identified a correlation between the corrupt trees and the working tree.

Dear, I am trying to get a prediction of my GAM on a response type. So that I eventually get plots with the correct values on my ylab. I have been able to get some of my GAM's working with the example shown below: * model1<-gam(nsdall ~ s(jdaylitr2), data=datansd) newd1 <- data.frame(jdaylitr2=(244:304)) pred1 <- predict.gam(model1,newd1,type="response")* The problem I am

Gracias Carlos miraré el fix y en su defecto intentaré mandar un mensaje al autor. De todas las maneras espero que alguien lo pruebe en windows 10 Saludos, Juan El 24/09/2017 a las 20:32, Carlos Ortega escribió: > Hola Juan Antonio, > > Acabo de instalar "tree" en Mac OSX y no aparece este error... > > > shuttle1 <- shuttle[254:256, ]  # 3 missing cases >

Hello r-help, I try to fit birds counts over years using glm. I have done (with Estate and year as factors): Model1 <- glm(Females~Estate+Year+offset = log(area)), family = quasipoisson(link = log), na.action = "na.exclude") After I have calculated the prediction using: Pred1 <- predict(Model1, type = "response", na.action = "na.exclude") My question

Hello, When calling several times the svm() function, I get different results. Do I miss something, or is there some random generation in the C library? In this second hypothesis, is it possible to fix an eventual seed? Thank you Pierre ### Example library('e1071') x = rnorm(100) # train set y = rnorm(100) c = runif(100)>0.5 x2 = rnorm(100)# test set y2 = rnorm(100) # learning a

Hello r-help, I'm trying to fit birds counts over years using glm. In fact I'm trying to reproduce an analysis already perform with genstat (attach document). I have done (with Estate and year as factors): Model1 <- glm(Females~Estate+Year+offset = log(area)), family = quasipoisson(link = log), na.action = "na.exclude") After I have calculated the prediction using: Pred1