similar to: Obtaining the adjusted r-square given the regression

Hello Alexandra, R2 is only defined for regressions with intercept. See a decent econometrics textbook for its derivation. HTH, Bernhard -----Urspr??ngliche Nachricht----- Von: Alexandra R. M. de Almeida [mailto:alexandrarma at yahoo.com.br] Gesendet: Mittwoch, 11. Januar 2006 03:48 An: r-help at stat.math.ethz.ch Betreff: [R] Obtaining the adjusted r-square given the regression coefficients

Hi people, I want to obtain the adjusted r-square given a set of coefficients (without the intercept), and I don't know if there is a function that does it. Exist???????????????? I know that if you make a linear regression, you enter the dataset and have in "summary" the adjusted r-square. But this is calculated using the coefficients that R obtained,and I want other coefficients

Dear Rhelpers, Is there a faster way than below to set a vector based on values from another vector? I'd like to call a pre-existing function for this, but one which can also handle an arbitrarily large number of categories. Any ideas? Cat=c('a','a','a','b','b','b','a','a','b') # Categorical variable

Dear R-Help, I wish to simulate a process so that it has certain properties in the frequency domain. What I attempted was to generate a random time-series signal, use spec-pgram(), apply a function in the frequency domain, and then inverse transform back to the time-domain. This idea does not seem as straight forward in practice as I anticipated. e.g. x<-ts(rnorm(1000, 0,1), frequency=256)

I want to compute B=A^{1/2} such that B*B=A. For example a=matrix(c(1,.2,.2,.2,1,.2,.2,.2,1),ncol=3) so > a [,1] [,2] [,3] [1,] 1.0 0.2 0.2 [2,] 0.2 1.0 0.2 [3,] 0.2 0.2 1.0 > a%*%a [,1] [,2] [,3] [1,] 1.08 0.44 0.44 [2,] 0.44 1.08 0.44 [3,] 0.44 0.44 1.08 > b=a%*%a i have tried to use singular value decomposion > c=svd(b) > c$u%*%diag(sqrt(c$d))

Dear List-mates, I'm trying to read a tk window menu from a file using {svWidgets} and 'menus.txt' but am receiving Warnings without seeing the desired consequences of the call. library(svWidgets) tkWinAdd("KSesnMain",title="kLab Session Manager for R", pos="+0+0") MenuRead(file="menus.txt") Warning messages: 1: Unrecognized menu type for

Dear list, How is the tolerance for a model parameter in an lm() call extracted? I did not see a solution in the documentation for lm(), or predict(), nor in the archives using 'tolerance' as the search string. I also checked into the nlme package, though nothing popped out at me. Sincerely, KeithC.

Dear R-Help, I am working on a paper in an R course for large file support in R using scan(), relational databases, and XML. I have never used SQL or heirarchical document formats such as XML (except where it occurs without user interaction), and knowledge in RDBs and XML is lacking in my program. I have tried finding a working example for the novices-novice on the topic, read many postings, the

Dear list mates, Are {utils} dialog box functions, and winMenuAdd... functions used to change (e.g. Console) menus, platform dependent? I'm writing a script loaded with .First that provides first time users in a lab course with the ability to select, load, and change between what I called a 'session' (more specific save that focuses on the session object I defined rather than the

Dear List, I was trying to verify that I could use extend.series in the wavelets package and kept getting an error when trying to use method="zero". I'm not seeing where my syntax has gone awry. According to the documentation, [see ?extend.series] " method: A character string indicating which extension method to use. Possible values are

R-Help Please can you tell me if there is a function within R that will return the day of any given date? Using 19 March 2006 as an example: given 20060319 (or any other date format) will return the day Sunday. Many thanks Fiona.

Dear Prof Ripley, First of all, unless you are an english professor, then I do not think you have any business policing language. I'm still very much a student, both in R, and regarding signal analysis. My competence on the subject as compared too your own level of expertise, or my spelling for that matter, may be a contension for you, but it would have been better had you kept that opinion

Dear list, What on earth is spec.pgram() normalized too? If you would like to skip my proof as to why it's not normed too the mean squared or sum squared amplitude of the discrete function a[], feel free too skip the rest of the message. If it is, but you know why it's not exact in spec.pgram() when it should be, skip the rest of this message. The issue I refer herein refers only too a

Dear List-mates, I have been trying to set up a 4x8 trellis plot (that is, 4 columns, 8 rows), and would like to have the axis limits set based on the data range of rows (for ylim) and columns (for xlim). I've been using the call: foo<-xyplot(y~x|Epoch+Subject, type=c("l","r"), par.strip.text=list(cex=0.5), ...) and updating to see different effects of scale

I have some data fit to a smooth.spline object as follows: (x=vector of data for the predictor variable, y=vector of data for the response variable) fit <- smooth.spline(x,y) Now, given a spline fit point y_new, I want to be able to find out what value of x_new yielded this fit value. How to do so? (This problem is the inverse of the predict.smooth.spline function, which takes x_new as input

I do not see an R-squared value after preforming the glm regression. Is there a separate command for this? Thanks -- View this message in context: http://r.789695.n4.nabble.com/Obtaining-R-squared-value-in-Logistic-Regression-tp4648954.html Sent from the R help mailing list archive at Nabble.com.

Hello, I have the following model: lm.7 <- lm(Y ~ F + C1 + C2 , data = EM4) F is a 4-level factor, the rest are covariates centered at their mean (Y is a two-column matrix). I have tried to find functions to give the model-adjusted means (adjusted at the covariates'means) and their standard deviations for each. (That is, what I believe is called in SAS "least square or LS-means,

Thanks for sharing the questions and responses! Is it possible to appreciate how much the coefficients matter in one or the other model? Say, using Biau's example, using coxph, as.factor(grade2 == "high")TRUE gives hazard ratio 1.27 (rounded). As clinician I can grasp this HR as 27% relative increase. I can relate with other published results. With survreg the Weibull model gives a

Dear R help list, I am modeling some survival data with coxph and survreg (dist='weibull') using package survival. I have 2 problems: 1) I do not understand how to interpret the regression coefficients in the survreg output and it is not clear, for me, from ?survreg.objects how to. Here is an example of the codes that points out my problem: - data is stc1 - the factor is dichotomous

I have run several glm models and would like to obtain the adjusted R squared for the model. What is the appropriate command for obtaining an adjusted R squared?