similar to: predict.lm

Michael Haenlein wrote Dear all, I would like to use predict.lm to obtain a set of predicted values based on a regression model I estimated. When I apply predict.lm to two vectors that have the same values, the predicted values will be identical. I know that my regression model is not perfect and I would like to take account of the error inherent in the model

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

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

I've got a small problem. I have some observational data (environmental samples: abiotic explanatory variable and biological response) to which I've fitted both a multiple linear regression model and also a gam (mgcv) using smooths for each term. The gam clearly fits far better than the lm model based on AIC (difference in AIC ~ 8), in addition the adjusted R squared for the gam is

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.

Hello there, I have a linear regression model for which I estimated heteroskedasticity-robust (Huber-White) standard errors using the coeftest function in the lmtest-package. Now I would like to inspect the predicted values of the dependent variable for particular groups and include a confidence interval for this prediction. My question: is it possible to estimate confidence intervals for the

Since I read the list in digest form (and was out ill yesterday) I'm late to the discussion. There are 3 steps for predicting survival, using a Cox model: 1. Fit the data fit <- coxph(Surv(time, status) ~ age + ph.ecog, data=lung) The biggest question to answer here is what covariates you wish to base the prediction on. There is the usual tradeoff between too few (leave out something

I'm fitting a Poisson gam model, say model<-gam(a65tm~as.factor(day.week )+as.factor(week)+offset(log(pop65))+s(time,k=10,bs="cr",fx=FALSE,by=NA,m=1),sp=c( 0.001),data=dati1,family=poisson) Currently I've difficulties in obtaining right predictions by using gam.predict function with MGCV package in R version 2.2.1 (see below my syntax).

Dear useRs, I am using the gamm function in the mgcv package to model a smooth relationship between a covariate and my dependent variable, while allowing for quantification of the subjectwise variability in the smooths. What I would like to do is to make subjectwise predictions for plotting purposes which account for the random smooth components of the fit. An example. (sessionInfo() is at

Hi I have a problem with lm and predict I have us [1] 2789.53 3128.43 3255.03 3536.68 3933.18 4220.25 4462.83 4739.48 [9] 5103.75 5484.35 5803.08 5995.93 6337.75 6657.40 7072.23 7397.65 [17] 7816.83 8304.33 8746.98 9268.43 9816.98 10127.95 10469.60 10960.75 [25] 11685.93 12433.93 13194.70 13843.83 us.p [1] 227.62 229.92 232.13 234.25 236.31 238.42 240.59 242.75 244.97

Dear all, I'm fitting a linear model with numerous lag terms of the response variable [i.e. y(t-1), y(t-2),y(t-3)...,] and other explanatory variables [x(1), x(2), x(3),....]- which go into my design matrix X. I'm fitting the linear model: lm(Y ~ X, ...). I would like to use the predict.lm function however the future predictions of Y are dependent upon previous predictions of Y [i.e.

Folks, I'm doing fine with using orthogonal polynomials in a regression context: # We will deal with noisy data from the d.g.p. y = sin(x) + e x <- seq(0, 3.141592654, length.out=20) y <- sin(x) + 0.1*rnorm(10) d <- lm(y ~ poly(x, 4)) plot(x, y, type="l"); lines(x, d$fitted.values, col="blue") # Fits great! all.equal(as.numeric(d$coefficients[1] + m

Hello dear r-help group I am turning for you for help with FAQ number 7.31: "Why doesn't R think these numbers are equal?" http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f *My story* is this: I wish to run many lm predictions and need to have them run fast. Using predict.lm is relatively slow, so I tried having it run faster by

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

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 all, I've come across an online posting http://www.biostat.wustl.edu/archives/html/s-news/2001-10/msg00119.html that described how to get confidence intervals for predicted values from predict.glm. These instructions were meant for S-Plus. Yet, it generally seems to work with R too, but I am encountering some problems. I am explaining my procedure in the following and would be most

Hello everybody, In order to predict income for different time points, I fitted a linear model with polynomial effects using BMA (bicreg(...)). It works fine, the results are consistent with what we are looking for. Now, we would like to predict income for a future time point t_next and of course draw the prediction interval around the estimated value for this point t_next. I've found the

Hello R Gurus, I am perplexed by the different results I obtained when I ran code like this: set.seed(100) test1<-randomForest(BinaryY~., data=Xvars, trees=51, mtry=5, seed=200) predict(test1, newdata=cbind(NewBinaryY, NewXs), type="response") and this code: set.seed(100) test2<-randomForest(BinaryY~., data=Xvars, trees=51, mtry=5, seed=200, xtest=NewXs, ytest=NewBinarY) The

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

Dear all, I'm struggling with predicting "expected time until death" for a coxph and survreg model. I have two datasets. Dataset 1 includes a certain number of people for which I know a vector of covariates (age, gender, etc.) and their event times (i.e., I know whether they have died and when if death occurred prior to the end of the observation period). Dataset 2 includes another