similar to: predict.lm "variables found" question

Dear List, I am familier with binary models, however i am now trying to get predictions from a ordinal model and have a question. I have a data set made up of 12 categorical predictors, the response variable is classed as 1,2,3,4,5,6, this relates to threat level of the species ( on the IUCN rating). Previously i have combined levels 1 and 2 to form = non threatened and then combined 3-6 to

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 =

Hello, We have written a program (below) to model the effect of a covariate on observed values of a response variable (using only 80% of the rows in our dataframe) and then use that model to calculate predicted values for the remaining 20% of the rows. Then, we compare the observed vs. predicted values using a linear model and inspect that model's coefficients and its R2 value. We wish

Dear all, for a cross-validation I have to use predict.glmmPQL() , where the formula of the corresponding glmmPQL call is not given explicitly, but constructed using as.formula. However, this does not work as expected: x1<-rnorm(100); x2<-rbinom(100,3,0.5); y<-rpois(100,2) mydata<-data.frame(x1,x2,y) library(MASS) # works as expected model1<-glmmPQL(y~x1, ~1 | factor(x2),

Simple question. For a simple linear regression, I obtained the "standard error of predicted means", for both a confidence and prediction interval: x<-1:15 y<-x + rnorm(n=15) model<-lm(y~x) predict.lm(model,newdata=data.frame(x=c(10,20)),se.fit=T,interval="confidence")$se.fit 1 2 0.2708064 0.7254615

R-helpers, I am using the package "drc" to fit a 4 parameter logistic model. When I use the predict function to get prediction on a new dataset, I am not getting the requested confidence or prediction intervals. Any idea what is going on? Here is code to reproduce the problem: --- library(drc) # Fit model to existing dataset in package spinach.model <- drm(SLOPE~DOSE, data =

Hello Fellow R Users,I have spent the last week trying to find a work around to this problem and I can't seem to solve it. I simply want to plot my GEE model result with 95% confidence bands. I am using the geepack package to run a basic GEE model involving nestling weights, to a Gaussian distribution, with "exchangeable" error structure. I am examining how nestling weight varies

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

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

I would like to predict a matrix containing missing values according to a fitted linear model. The predicted values must have the same length as the number of observations in newdata, where missing predicted values (due to missing explanatory values) are replaced by NA. How can I achieve this? I tried the following example: > x <- matrix(rnorm(100), ncol=10) > beta <- rep(1, 10) >

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

Hi everybody, recently a member of the community pointed me to the useful predict.lm() comment. While I was toying with it, I stumbled across the following problem. I do the regression with data from five years. But I want to do a prediction with predict.lm for only one year. Thus my dataframe for predict.lm(mod, newdata=dataframe) is shorter than the orginial vector that I did the regression

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

Hi all, Suppose: y<-rnorm(100) x1<-rnorm(100) lm.yx<-lm(y~x1) To predict from a new data source, one can use: # works as expected dum<-data.frame(x1=rnorm(200)) predict(lm.yx, newdata=dum) Suppose lm.yx has been run and we have the lm object. And we have a dataframe that has columns that don't correspond by name to the original regressors. I very! naively assumed that doing

Hi, Please could someone explain how this element of predict.lm works? From the help file ` newdata An optional data frame in which to look for variables with which to predict. If omitted, the fitted values are used. ' Does this dataframe (newdata) need to have the same variable names as was used in the original data frame used to fit the model? Or will R just look across consecutive

Hi all - my first time here and am having an issue with the Predict function. I am using a tutorial as a guide, locate here: http://www.ats.ucla.edu/STAT/R/dae/mlogit.htm My code gives this error > newdata1$predicted <- predict(mlogit,newdata=newdata1,type="response") Error in `$<-.data.frame`(`*tmp*`, "predicted", value = c(0.332822934960197, : replacement has

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",

I'm trying to get predicted probabilities out of a regression model, but am having trouble with the "newdata" option in the predict() function. Suppose I have a model with two independent variables, like this: y=rbinom(100, 1, .3) x1=rbinom(100, 1, .5) x2=rnorm(100, 3, 2) mod=glm(y ~ x1 + x2, family=binomial) I can then get the predicted probabilities for the two values of

Dear all, the help file for predict.gbm states that "The predictions from gbm do not include the offset term. The user may add the value of the offset to the predicted value if desired." I am just not sure how exactly, especially for a Poisson model, where I believe the offset is multiplicative ? For example: library(MASS) fit1 <- glm(Claims ~ District + Group + Age +

Hi, I wanted to use the predict.lm() function to compare the empirical data with the predicted values. The problem is that I have NAs in my data. I wanted to cbind my data.frame with the empirical values with the vector I get from predict.lm. But they don't have the same length because predict.lm just skip NA-predictions. Is there a way to get a vector with predicted values of the same