similar to: explanation of lm's coefficients

I'm using Linux version 2.4.20-8smp (gcc version 3.2.2, RedHat 9). I installed R-1.7.1-1.src.rpm following INSTALL. However, no X11 device is found when I run R. > X11() Error in X11() : X11 is not available In the configuration, I found "checking for X... no" How do I configure so that X11 device capability is installed? -- ________________________________________ Justin

I would like to call both p-values and R-squared values from lm's in a function. I can get the p-values from coef(summary(name.lm))[r,c], however, I cannot figure out how to call the R-squared values without manually calling the summary and inserting them in the script - which negates the value of automating the process through a function. Thanks, Mike [[alternative HTML version deleted]]

Simple example: # Classification Tree with rpart library(rpart) # grow tree fit <- rpart(Kyphosis ~ Age + Number + Start, method="class", data=kyphosis) Now I would like to know how can I measure the "importance" of each of my three explanatory variables (Age, Number, Start) in the model? If this was a regression model, I could have looked at p values from the

Dear friends, After running the lm() model, we can get summary resluts like the following: Coefficients: Estimate Std. Error t value Pr(>|t|) x1 0.11562 0.10994 1.052 0.2957 x2 -0.13879 0.09674 -1.435 0.1548 x3 0.01051 0.09862 0.107 0.9153 x4 0.14183 0.08471 1.674 0.0975 . x5 0.18995 0.10482 1.812 0.0732 . x6 0.24832 0.10059 2.469 0.0154 * x7

Hello, Does anyone know of a way I can speed this up? Basically I'm attempting to get the data item on the same row as the report date for each report date available. In reality, I have over 11k of columns, not just A, B, C, D and I have to do that over 100 times. My solution is slow, but it works. The loop is slow because of merge. # create sample data z.dates =

Hola a todos, Ajusté un modelo lineal mixto en el cual tengo una interacción significativa entre dos facotres, cada uno con 3 niveles (osea un diseño 3x3). Esta es la tabla de medias             P1                P 2             P3 M1      220.66       311.85        260.80 M2      348.57        89.33         191.71 M3      103.57      381.37         511.62 Con el fin de interpretar

Hello all, Does anyone know how to constrain/force specific coefficients when running lm()? I need to run recresid() {strucchange package} on the residuals of forecast.lm, but forecast.lm's coefficients must be determined by parameter.estimation.lm I could estimate forecast.lm without lm() and use some other kind of optimisation, but recresid() requires an object with class lm.

Dear R-devel list, I realized that removing a predictor in lm through the "-"'s operator in formula() does not affect the complete cases that are considered. A minimal example is: summary(lm(Wind ~ ., data = airquality)) # 42 observations deleted due to missingness summary(lm(Wind ~ . - Ozone, data = airquality)) # still 42 observations deleted due to missingness, even if only 7

>From http://bugs.r-project.org/cgi-bin/R: If you are not sure whether you have observed a bug or not, it is a good idea to ask on the mailing list R-Help by sending an e-mail to r-help at stat.math.ethz.ch rather than submitting a bug report. I'm wondering whether to submit a bug report on this: ============================================================== >

The merge of the format changes and new psychoacoustics is complete. There's still tuning work left to do (most notably there's a somewhat higher-than-acceptable level of background low-midrange/upper bass noise; this is due mostly to incomplete psychoacoustic modelling data in the 100Hz-500Hz range) however, at this point, pending bugfixes, the Vorbis 1.0 stream format is frozen.

lm's "predict" function offers an "interval" parameter to choose between 'confidence' and 'prediction' bands. In the package "robust" and for "lmRob", there is also a "predict" but it lacks such a parameter, and the documented "type" parameter has only "response" offerred. Is there some way of obtaining

Hi: I think I found an error in plot.lm with the option which=5, of course I can be wrong , as usually happen, but I had work on it for a while and show it to some other people that work with R, and so far I don't see what I can be interpreting wrong. I also worked over the plot.lm's code and change some lines to get what I call "the right plot", if any body is

Dear R-developers, In the 'lm' documentation, the '-' operator is only specified to be used with -1 (to remove the intercept from the model). However, the documentation also refers to the 'formula' help file, which indicates that it is possible to subtract any term. Indeed, the following works with no problems (the period '.' stands for 'all terms except the

Hi, I am a very occasional user of R, and will be grateful for some help in constructing a regression across groups. Here is an example: library(MASS) attach(cats) Sex[120:144]<-factor(TG) #Renaming some males to transgender, to create 3 groups, male, female and transgender out<-lm(Bwt~Sex/Hwt) #Gives me 3 separate linear regressions for groups M, F and TG What I now want to do

R folks, As an R novice, I struggle with the mystery of subsetting. Textbook and online examples of this seem quite straightforward yet I cannot get my mind around it. For practice, I'm using the code in MASS Ch. 6, "whiteside data" to analyze a different data set with similar variables and structure. Here is my data frame: ###subset one of three cases for the variable

Hi, [System info: R 1.6.1 compiled from source (no args to configure) on Redhat 7.1 Linux i/386] I have some spatial statistics programs that recently (since R 1.6.1) have generated SIGSEGV. These R programs use the splancs library as well as some C code of my own for Voronoi analysis. The program has a big loop to fit various model parameters against a spatial distribution of biological

Hello, I have been running linear regressions on large data sets. As 'lm' saves a great deal of extraneous (for me) data including the residuals, fitted.values, model frame, etc., I generally set these to NULL within the object before saving off the model to a file. In the below example, however, I have found that depending on whether or not I run 'lm' within another function or

I havent been able to find a solution to my original post: http://www.ruby-forum.com/topic/63110 I will really appreciate some insight/advice Thank you. -- Posted via http://www.ruby-forum.com/.

Hello, I recently stumbled upon the fact that different 32-bit R Windows versions produce slightly different results when calling 'lm'. More precisely, this seems to be caused by a change in the transition from R 3.2.x to R 3.3.0. A minimal example, for reproducing this is "lm(y~x, data.frame(x=c(1,2,3), y=c(1,2,3)))". In R 3.3.0 the result is Call: lm(formula = y ~ x, data =

Hi all, I had samba-1.9.18p10 working fine on my SGI IRIX 6.4 box. I am trying to install samba-2.0.3 Installation went through without a glitch and testparm went through fine as well, with the old smb.conf So I start samba as a daemon: nmbd -D smbd -D But when I do "smbclient -L klebit" an erros occurs: Added interface ip=128.252.252.152 bcast=128.252.252.255 nmask=255.255.255.0