similar to: what is set.fit in function predict.lm

Hi, don' t understand why the function fomula have this error, i enclose the parameter "a" with the function I() Thank Ruben x<-1:5 y<-c( 2 ,4 , 6 , 8 ,11) formu<-y~I(a*x) form<-formula(formu) dummy<-data.frame(x=x,y=y) fm<-lm(form,data=dummy) Error in unique(c("AsIs", oldClass(x))) : Object "a" not found

Hello, it wanted to know how I can extract of a dates frame the values peaks according to an interval that I establish. For example if dates are: 1 23 2 4 3 56 4 7 5 99 6 33 extract the date i wanted to divide into intervals of 2 an d to take alone the numbers 23, 56 and 99 of those 3 intervals. Thanks Ruben

Hi, probably just a quick question: can I somehow change the formula used with predict? E.g., the regression was run on "y ~ u + v + w" but for the prediction the term v should be removed from the formula contained in the regression object and only "y ~ u + w" be used. I could use model.matrix etc. to do the predictions but it would be very helpful to know a simpler way.

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, it want to know if there is a library that transform a file in excell (*. xls) to text file(*.txt, *,prn, *.csv). Thanks

Hello, my question is if I have two axes in a graphics as I can put him a different color al axis and to the number of the scale. ? Thanks. Ruben

Hi, i 'dont understand how to take a general formula, view this: x<-1:5 y<-c(0,1,1.7,2,2.1.4) dummy<-data.frame(x=x,y=y) formula<-"y~A*log(x)/log(2)" formu<-as.formula(formula) fm<-lm(formu,data=dummy) Error in eval(expr, envir, enclos) : Object "A" not found but A is the parameter of fitting, why is this?Thanks Ruben

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, sorry but i don't understand how to make a function with as.function() formula<-"2+3*x" formu<-as.symbol(formula) > formu 2+3*x formul<-as.function(alist(x=,formu)) curve(formul,1,5,col="blue") Error in xy.coords(x, y, xlabel, ylabel, log) : x and y lengths differ > typeof(formul) [1] "closure" and not plot the curve function, Why?

I'm making some functions to illustrate regressions and I have been staring at termplot and predict.lm and residuals.lm to see how this is done. I've wondered who wrote predict.lm originally, because I think it is very clever. I got interested because termplot doesn't work with interactive models: > m1 <- lm(y ~ x1*x2) > termplot(m1) Error in `[.data.frame`(mf, , i) :

Following an example on p 111 in 'Nonlinear Regression with R' by Ritz & Streibig, I have been fitting nls models using square brackets with the grouping variable inside. In their book is this example, in which 'state' is a factor indicating whether a treatment has been used or not: > Puromycin.m1 <- nls(rate ~ Vm[state] * + conc/(K[state] + conc), data = Puromycin,

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

The option se.fit in predict.nls is currently ignored. Is there any other function available to calculate the error in the predictions? Thanks, Manuel ______________________________________________ LLama Gratis a cualquier PC del Mundo. Llamadas a fijos y m??viles desde 1 c??ntimo por minuto.

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]

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

Dear R-people ... I'm a new user. I can't get predict.lm() to produce predictions for new independent data. There are some messages in archived help about this problem, but I still don't see my error after reviewing those. I understand that the new independent data must have the same name(s) as used when the model was made. In the example below, predict.lm produces the

Just a short reminder / question -- We've had one posting >> Date: Sun, 17 Aug 1997 19:51:20 -0700 >> From: Kung-Sik Chan <kchan@stat.uiowa.edu> >> To: r-help@stat.math.ethz.ch >> Subject: R-beta: bug report with a predict.default that would "work" with (some) lm objects, and I think it was said that predict.lm "is being" written (Peter

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

Hi All, I created a two variable lm() model slm<-lm(y[1:3000,8]~y[1:3000,12]+y[1:3000,15]) I made two predictions predict(slm,newdata=y[201:3200,]) predict(slm,newdata=y[601:3600,]) there is no error message for either of these. the results are identical, and identical to slm$fitted as well. if this is not the right way to apply the model coefficients to a new set of inputs, what is

For a given linear regression, I wish to find the 2-tailed t-dist probability that Y-hat <= newly observed values. I generate prediction intervals in predict() for plotting, but when I calculate my t-dist probabilities, they don't agree. I have researched the issues with variance of individual predictions and been advised to use the variance formula below (in the code). I presume my