similar to: predict function (PR#2958)

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

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

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

Hi I am having a particular problem with some glm models I am running. I have been adapting code from Bill Venables 'Programmers niche' in RNews Vol 2/2 to fit ca. 1000 glm models to a combination of species 0/1 data (as Y) and related physicochemical data (X), to automate the process of fitting this many models. I have successfully managed to fit all the models and have stored the

Estimado Oliver Nuñez Envío un ejemplo reproducible. Javier Marcuzzi # de donde tomo datos, y tiene el modelo (en el pdf) library(MCMCglmm) # librería con las funciónes que voy a usar library(nlme) datos0<-ChickWeight # creo algunos datos que agrego a los origonales Factor<-as.numeric(datos0$Chick) Factor[Factor > 0 & Factor <= 10] <- 'A' Factor[Factor > 10

Estimados Tengo un error que me desconcierta, es un código que simplifiqué de otro trabajo donde no hay problemas, sin embargo me da un error. Una diferencia es que en mi otro trabajo uso spline y ahora polinomio, este es de segundo grado y se encuentra tanto en efectos fijos como aleatorios, el modelo es correcto, corre con MCMCglmm pero no con nlme. grid <- expand.grid(Tiempo=4:6,

Hi R-community, I have the code as follows,i Fitted model as follows lbeer<-log(beer_monthly) t<-seq(1956,1995.2,length=length(beer_monthly)) #beer_monthly contains 400+ entries t2=t^2 beer_fit_parabola=lm(lbeer~t+t2) Below is not working for me. Please help me in preparing the new data set for the below prediction

Sometimes one might like to obtain pointwise bootstrap bias-corrected, accelerated (BCA) confidence intervals for a large number of statistics computed from a single dataset. For instance, one might like to get (so as to plot graphically) bootstrap confidence bands for the fitted values in a regression model. (Example: Chiu S et al., Early Acceleration of Head Circumference in Children with

Hi all, Very surprising (to me!) and mystifying result from predict.glm(): the predictions vary depending on whether or not I use ns() or splines::ns(). Reprex follows: library(splines) set.seed(12345) dat <- data.frame(claim = rbinom(1000, 1, 0.5)) mns <- c(3.4, 3.6) sds <- c(0.24, 0.35) dat$wind <- exp(rnorm(nrow(dat), mean = mns[dat$claim + 1], sd = sds[dat$claim + 1])) dat <-

Dear R-helpers, I try to perform glm's with negative binomial distributed data. So I use the MASS library and the commands: model_1 = glm.nb(response ~ y1 + y2 + ...+ yi, data = data.frame) and predict(model_1, newdata = data.frame) So far, I think everything should be ok. But when I want to perform a glm with a subset of the data, I run into an error message as soon as I want to predict

Full_Name: Murray H Smith Version: 1.7.1 OS: Windows2000 Submission from: (NULL) (202.36.29.1) This alleged bug is Windows specific and occurs when using Windows metafile plots. The problem does not occur in a Linux version. It does not occur in the pt rintout when a graphic is saved to a postscript file under Windows. The problem came to light when using plotmath to label a plot with

Full_Name: Murray Smith Version: 2.2.1 OS: Windows XP Submission from: (NULL) (202.36.29.1) order() will not allow a complex vector for its first argument. This contrasts with sort() which will happily sort a complex vector. While this may not be regarded as a bug by some, it is annoying for anyone who makes frequent use of complex numbers. The problem occured in the Windows version 2.2.1 of

Hi, I have a linear model ("mod1 <- lm(V3~V1+V2) and I would like to get the model's prediction at values of V1 and V2 not included in the original sample. samp <- read.table("data.dat",nrows=100) attach(samp) out.poly <- lm(V3 ~ V1 + V2) If I try to use out.poly to predict values for the function I run into problems. It seems that it isn't possible to use a new

hello, I'm trying to predict some values based on a linear regression model. I've created the model using one dataframe, and have the prediction values in a second data frame (call it newdata). There are 56 rows in the dataframe used to create the model and 15 in newdata. I ran predict(model1, newdata) and get the warning: 'newdata' had 15 rows but variable(s) found have 56 rows

Hey List, I did a multiple regression and my final model looks as follows: model9<-lm(calP ~ nsP + I(st^2) + distPr + I(distPr^2)) Now I tried to predict the values for calP from this model using the following function: xv<-seq(0,89,by=1) yv<-predict(model9,list(distPr=xv,st=xv,nsP=xv)) The predicted values are however strange. Now I do not know weather just the model does not fit

To my knowledge, this has not been reported previously, and doesn't seem to have been changed in R-devel or R-patched. If M is a matrix with coloumn names, and mod <- prcomp(M) # or princomp then predicting a single observation (row) with predict() gives the error Error in scale.default(newdata, object$center, object$scale) : length of 'center' must equal the number of

I am writing a function to assess the out of sample predictive capabilities of a time series regression model. However lm.predict isn't behaving as I expect it to. What I am trying to do is give it a set of explanatory variables and have it give me a single predicted value using the lm fitted model. > model = lm(y~x) > newdata=matrix(1,1,6) > pred =

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 =

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",

Hi, See below I reply your message for <https://stat.ethz.ch/pipermail/r-help/2008-April/160966.html>[R] predict.glm & newdata posted on Fri Apr 4 21:02:24 CEST 2008 You say it ##works fine but it does not: if you look at the length of yhat2, you will find 100 and not 200 as expected. In fact predict(reg1, data=x2) gives the same results as predict(reg1). So I am still looking for