similar to: Error in step()

I have two questions about the built-in function step. Ultimately I want to apply a lm fitting and subsequent step procedure to thousands of data sets groups by a factor defined as a unique ID. Q1. The code below creates a data.frame comprising three marginally noisy surfaces. The code below fails when I use step in a function but summary seems to show the model fits are legitimate. Any ideas on

Full_Name: Jerome Asselin Version: 1.3.1 OS: MacOS 9.2 Submission from: (NULL) (142.103.173.46) The step() function attempts to calculate the deviance of fitted models even if does not really need it. As a consequence, the step() function gives an error when it is used with coxph(). (There is currently no method to calculate the deviance of coxph() fits.) The code below gives an example of how

Hi.., is there a function in R to fit bivariate splines ? I came across 'polymars' (POLSPLINE) and 'mars' (mda) packages. Are these the one to use or are there other specific commands? Thanks. Harsh

dears R-users, I'm interested in model selection problem, and i have faced some problems that i would like to ask for help. well, this is a very small example with 4 variable (just one var. is the response - z) with 100 individuals i would like to do a stepwise search, for the "best" model, and a use BIC criteria. I know when I have a lot of variables, let's say 120, I know,

Could anyone tell me what am I doing wrong? > pro<-function(indep,dep){ + d<-data.frame(indep) + form<-formula(lm(dep~.,data=d)) + forward<-step(lm(dep~X1,data=d),scope=form,trace=0,direction='f') + return(forward) + } > pro(m,q) Error in inherits(x, "data.frame") : Object "d" not found Where q is a vector with the dependent variable's

Hi, (R version 2.15.0) I am running a pgm with 1 response (earlier standardized Y) and 44 independent vars (Xi) from the same data =a2: When I run the 'lm' function on single Xi at a time, the beta coefficient for let's say X1 is = -0.08 (se=0.03256) But when I run the same Y with 44 Xi-s with the 'step' function (because I left direction parameter empty, I assume a backward

Dear R-list, I'm not sure what I've found about a function in DAAG package is a bug. When I was using cv.lm(DAAG) , I found there might be something wrong with it. The problem is that we can't use it to deal with a linear model with more than one predictor variable. But the usage documentation hasn't informed us about this. The code illustrates my discovery: > library(DAAG)

I posted this question way down at teh end of another thread realted to an error in step, but that was stupid since it really is another matter altogether. I should have posted it separately, as I have now done. The code below creates a data.frame comprising three marginally noisy surfaces. The code below (including a fix courtesy of David Winsemius that avoids a step function error through use

Hi.. is there a limit on the number of explanatory variables in nls ? i have a dataframe with the columns names x1,x2..,x300 when i run nls it gives the error: " x181 not found" thought it does run when i have x1,x2,...,x170 variables. Thanks Harsh --------------------------------- [[alternative HTML version deleted]]

Hello, I met this problem with the function step( ) when I was trying to do forward variable selection. Below is the code I used and the error message. I don't why I am getting this error message. Could someone help me out. I noticed that I got this error message whenever it chose a model with an interaction term X_p :X_q and if I put X_q before X_p in the upper scope formula, like

On Wed, 5 Oct 2005 Hallgeir.Grinde at elkem.no wrote: > And some more informastion I forgot. > R does not crash if I write out the formula: > > set.seed(123) > x1 <- runif(1000) > x2 <- runif(1000) > x3 <- runif(1000) > x4 <- runif(1000) > x5 <- runif(1000) > x6 <- runif(1000) > x7 <- runif(1000) > x8 <- runif(1000) > y <-

Dette er en melding med flere deler i MIME-format. --=_alternative 004C4E4A00257091_= Content-Type: text/plain; charset="US-ASCII" Yes. so (x1*x2*x3*x4*x5*x6*x7*x8)^2 = (x1+x2+x3+x4+x5+x6+x7+x8)^8 ? and there is a difference in (x1*x2*x3*x4*x5*x6*x7*x8)^2 and (x1*x2*x3*x4*x5*x6*x7*x8) althoug the resulting formulas are the same, or? This fikses my problem, but R still crashes for the

On Wed, 5 Oct 2005 Hallgeir.Grinde at elkem.no wrote: > Yes. > so (x1*x2*x3*x4*x5*x6*x7*x8)^2 = (x1+x2+x3+x4+x5+x6+x7+x8)^8 ? Yes in the sense that the simplified formula given by terms() is the same. > and there is a difference in > (x1*x2*x3*x4*x5*x6*x7*x8)^2 > and > (x1*x2*x3*x4*x5*x6*x7*x8) > althoug the resulting formulas are the same, or? The first is reduced to the

Dette er en melding med flere deler i MIME-format. --=_alternative 004613C000257091_= Content-Type: text/plain; charset="US-ASCII" And some more informastion I forgot. R does not crash if I write out the formula: set.seed(123) x1 <- runif(1000) x2 <- runif(1000) x3 <- runif(1000) x4 <- runif(1000) x5 <- runif(1000) x6 <- runif(1000) x7 <- runif(1000) x8 <-

I am trying to run separate regressions for different groups of observations using the lapply function. It works fine when I write the formula inside the lm() function. But I would like to pass formulae into lm(), so I can do multiple models more easily. I got an error message when I tried to do that. Here is my sample code: #generating data x1 <- rnorm(100,1) x2 <- rnorm(100,1) y <-

Dear Berwin, Simply stacking the problems and treating the resulting observations as independent will give you the correct coefficients, but incorrect coefficient variances and artificially zero covariances. The approach that I suggested originally -- testing a linear hypothesis using the coefficient estimates and covariances from the multivariate linear model -- seems simple enough. For

SImplify your call to lm using the "." argument instead of manipulating formulas. > strt <- lm(y1 ~ ., data = dat) and you do not need to explicitly specify the "1+" on the rhs for lm, so > frm2<-as.formula(paste(trg," ~ ", paste(xvars,collapse = "+"))) works fine, too. Anyway, doing this gives (but see end of output)" bst <-

Hi All Maybe I dont understand it, but I would have expected that the design matrix has as many rows as there were observations available to fit the model. Below a small artificial dataset created, then one model fitted and the design matrix outputted, having 27 rows. Then I delete 6 obs, and fit the model on these 21 obs, but the design matrix that comes out has 26 rows? Thanks for your

Hello, After I use the lm() function to perform a multiple linear regression, and then use the step function to eliminate variables that predict the weakest, I need to export the final equation to a spreadsheet or a text file. Below is some sample code. In the end I want to export the coefficients to a spreadsheet. Will you please direct me to the appropriate syntax? Thanks for your time, --Eric

Hi R-helpers, I have a character object called dd that has 32 elements each of which is a model formula contained within quotation marks. Here's what it looks like: > dd [1] "lm(y ~ 1,data=Cement)" "lm(y ~ X,data=Cement)" "lm(y ~ X1,data=Cement)" [4] "lm(y ~ X2,data=Cement)" "lm(y ~