similar to: glm predict on new data

I imitated predict.glm, my thing worked, now I need to revise. It would help me very much if someone would explain predict.glm line 28, which says object$na.action <- NULL # kill this for predict.lm calls I want to know 1) why does it set the object$na.action to NULL 2) what does the comment after mean? Maybe I need a pass by value lesson too, because I can't see how changing that

Hi, I split a data set into two partitions (80 and 42), use the first as the training set in glm and the second as testing set in glm predict. But when I call glm.predict, I get the warning message:  Warning message: 'newdata' had 42 rows but variable(s) found have 80 rows  ---------------------  s = sample(1:122)

Hello, I have tried reading the documentation and googling for the answer but reviewing the online matches I end up more confused than before. My problem is apparently simple. I fit a glm model (2^k experiment), and then I would like to predict the response variable (Throughput) for unseen factor levels. When I try to predict I get the following error: > throughput.pred <-

Hello together, I would like to predict my fitted values on a new dataset. The original dataset consists of the variable a and b (data.frame(a,b)). The dataset for prediction consists of the same variables, but variable b has a constant value (x) added towards it (data.frame (a,b+x). The prediction command returns the identical set of predicted values as for the original dataset yet I would have

Dear all, I am stumped at what should be a painfully easy task: predicting from an lm object. A toy example would be this: XX <- matrix(runif(8),ncol=2) yy <- runif(4) model <- lm(yy~XX) XX.pred <- data.frame(matrix(runif(6),ncol=2)) colnames(XX.pred) <- c("XX1","XX2") predict(model,newdata=XX.pred) I would have expected the last line to give me the

Maybe a useful addition to the predict functions would be to return the values of the predictor variables. It just (unless there are problems) requires an extra line. I have inserted an example below. "predict.glm" <- function (object, newdata = NULL, type = c("link", "response", "terms"), se.fit = FALSE,

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

How do I construct a figure showing predicted value plots for the dependent variable as a function of each explanatory variable (separately) using the results of a logistic regression? It would also be helpful to know how to show uncertainty in the prediction (95% CI or SE). Thanks- This email has been processed by SmoothZap - www.smoothwall.net

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

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 =

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

Dear R help, This seems to be a commonly asked question and I am able to run examples that have been proposed, but I can't seems to get this to work with my own data. Reproducible code is below. Thank you in advance for any help you can provide. The main problem is that I can not get the confidence lines to plot correctly. The secondary problem is that predict is not able to find my object

Hello there. I have fitted a rpart model. > rpartModel <- rpart(y~., data=data.frame(y=y,x=x),method="class", ....) and can use rpart$where to find out the terminal nodes that each observations belongs. Now, I have a set of new data and used predict.rpart which seems to give only the predicted value with no information similar to rpart$where. May I know how

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]

I am not sure whether it is a design decision or just an oversight. When I ask for the standard errors of the predictions with predict(budwm.lgt,se=TRUE) where budwm.lgt is a logistic fit of the budworm data in MASS, I got Error in match.arg(type) : ARG should be one of response, terms If one is to construct a CI for the fitted binomial probability, wouldn't it be more natural to do

I am fittling a spline to a variable in a regression model, I am then using the predict.glm funtion to make some predictions. When I use bs to fit the spline I don't have any problems using the predict.glm function however when I use ns I get the following error: Error in model.frame(formula, rownames, variables, varnames, extras, extranames, : variable lengths differ (found for

# Hello, # I have a data set that looks something like the following: site<-c(rep('a',5),rep('b',2),rep('c',4),rep('d',11)) year<-c(1980, 1981, 1982, 1993, 1995, 1980, 1983, 1981, 1993, 1995, 1999, c(1980:1990)) count<-c(60,35,36,12,8,112,98,20,13,15,15,65,43,49,51,34,33,33,33,40,11,0) data<-data.frame(site, year, count) # > site year count # 1

Dear all, I try to first link data in a data.frame() to a (polygon) read.shape() object and then to plot a polygon map showing the data in the data.frame. The first linking is called "link" in ArcView and "relate" in ArcMap. I use the code shown below, though without success. Help with this would be greatly appreciated. Thanks! Tord require(maptools) # Read shape file

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