similar to: Bias-corrected percentile confidence intervals

Is there any way to put an argument into an object name. For example, say I have 5 objects, model1, model2, model3, model4 and model5. I would like to make a vector of the r.squares from each model by code such as this: rsq <- summary(model1)$r.squared for(i in 2:5){ rsq <- c(rsq, summary(model%i%)$r.squared) } So I assign the first value to rsq then cycle through models 2 through

I?ve been using Frank Harrell?s rms package to do bootstrap model validation. Is it the case that the optimum penalization may still give a model which is substantially overfitted? I calculated corrected R^2, optimism in R^2, and corrected slope for various penalties for a simple example: x1 <- rnorm(45) x2 <- rnorm(45) x3 <- rnorm(45) y <- x1 + 2*x2 + rnorm(45,0,3) ols0 <- ols(y

R-Listers: I am doing a quasi-maximum likelihood estimation and I get a "subscript out of bound" error message, Typically I would think this means that a subscript used in the function is literally out of bounds however I don't think this is the case. All I change in the code is a constant, that is hard-wired in (not data dependent and not parameter dependent), furthermore,

Dear UseR, I get an error when I run "R CMD check" on my .tar.gz file package, and I don't understand why since I don't obtain any error with "R CMD check" on the package directory. Do you have any idea ? $ sudo ./R-2.11.0/bin/R CMD check eqtl_1.1.tar.gz and $ sudo ./R-2.11.0/bin/R CMD --check-subdirs=no eqtl_1.1.tar.gz return an Error * checking for working

Hello, I've a problem with a self written routine taking a lot of memory (>1.2Gb). Maybe you can suggest some enhancements, I'm pretty sure that my implementation is not optimal ... I'm creating many linear models and store coefficients, anova p-values ... all I need in different lists which are then finally returned in a list (list of lists). The input is a matrix with 84 rows

Hello list, I have a linear regression: mylm = lm(y~x-1) I've been reading old mail postings as well as the plotmath demo and I came up with a way to print an equation resulting from a linear regression: model = substitute(list("y"==slope%*%"x", R^2==rsq), list(slope=round(mylm$coefficients[[1]],2),rsq=round(summary(mylm)$adj.r.squared, 2))) I have four models and I

Several ways: 1. Read ?randomForest, especially the `Value' section. 2. Look at str(myforest.rf). 3. Look at print.randomForest. If the forest has 100 trees, then the mse and rsq are vectors with 100 elements each, the i-th element being the mse (or rsq) of the forest consisting of the first i trees. So the last element is the mse (or rsq) of the whole forest. HTH, Andy > From: David

Signed-off-by: Ilia Mirkin <imirkin at alum.mit.edu> --- Untested beyond compiling a few shaders to see if they look like they might work. nvdisasm agrees with envydis's decoding of these things. Will definitely get ahold of a G200 to run tests on before pushing this. .../drivers/nouveau/codegen/nv50_ir_emit_nv50.cpp | 94 ++++++++++++++++++---

Hi there, I am stuck trying to solve what should be a fairly easy problem. I have a data frame that essentially consists of (ID, time as seqMonth, variable, value) and i want to find the regression coefficient of value vs time for each combination of ID and Variable. I have tried several approaches and none of them seems to work as i expected. For example, i have tried:

Hi, I'm a new user of R and I'm trying to make a linear model from this kind of dataset x [1] 16.87 19.93 25.85 20.94 17.06 19.49 19.93 25.45 27.74 20.15 25.81 21.06 17.17 20.03 25.50 27.79 20.44 16.88 19.93 25.79 z<-x-10 y [1] 0.80 1.27 2.22 1.32 0.90 1.18 1.84 2.41 2.97 1.25 2.07 1.41 1.14 1.66 2.59 3.51 1.53 0.81 1.26 2.30 plot(x,y) I want to be able to force the line of

Does this give correct results for special floats (0, infs)? We tried to improve (for single floats) x86 rcp in llvmpipe with newton-raphson, but unfortunately not being able to give correct results for these two cases (without even more additional code) meant it got all disabled in the end (you can still see that code in the driver) since the problems are at least as bad as those due to bad

Hi, I have a dataframe basically like this: > head(asturias.gen2011[,c(1,4,9:14)]) municipio total upyd psoe pp iu fac tipo 440 Allande 2031 1.44 31.10 39.75 4.01 21.62 1000-10000 443 Aller 12582 1.37 33.30 37.09 15.53 10.35 10000-50000 567 Amieva 805 1.48 32.69 37.36 6.15 20.16 <1000 849

Hello, Using this data: http://n4.nabble.com/file/n1676330/US_Final_Values.txt US_Final_Values.txt and the following code i got the image at the end of this message: US.final.values<-read.table("c:/tmp/US_Final_Values.txt",header=T,sep=" ") US.nls.1<-nls(US.final.values$ECe~a*US.final.values$WTD^b+c,data=US.final.values,start=list(a=2.75,b=-0.95,c=0.731),trace=TRUE)

Dear all, I want to improve my adj - R sq. I 've chequed some established models and they introduce two times the same variable, one transformed, and the other not. It also improves my adj - R sq. But, isn't this bad for the collinearity? Do I interpret coefficients as usual? Estimate Std. Error t value Pr(>|t|) (Intercept) 1.73140 7.22477 0.240

Hi, I am fitting a weighted least square regression and trying to compute SSE,SST and SSReg but I am not getting SST = SSReg + SSE and I dont know what I am coding wrong. Can you help please? xnam <-colnames(X) # colnames Design Matrix fmla1 <- as.formula(paste("Y ~",paste(xnam, collapse=

I'm trying to get the coefficient of partial determination for each of three independent variables. I've tried mvr in package pls.pcr. I'm a little confused by the output. I'm curious how I can order the LV's according to their names rather than their relative contribution to the regression. For instance, using the crabs data from MASS I made a regression of FL~RW+noise

I am combining many different random forest objects run on the same data set using the combine ( ) function. After combining the forests I am not sure whether the variable importance, local importance, and rsq predictors are recalculated for the new random forest object or are calculated individually for each tree ensemble? Is it possible to calculate these predictors for the new random forest

Dear Random Forests gurus, I have a question about R^2 provided by randomForest (for regression). I don't succeed in finding this information. In the help file for randomForest under "Value" it says: rsq: (regression only) - "pseudo R-squared'': 1 - mse / Var(y). Could someone please explain in somewhat more detail how exactly R^2 is calculated? Is "mse"

Hi, I ran a linear regression on a data set and got an Rsq of about .27. The plot looks as though an exponential curve would be a better fit. What codes do I use to do this? Thank you, Ryan Munroe [[alternative HTML version deleted]]

>I am trying to use the Hmisc function transace to transform predictors > > test<-cbind(flowstress,pressres,alloy) > xtrans<-transace(x,binary=pressres',monotonic='flowstress', categorical='alloy') > > >and I am getting the following message?? >Error in ace(x[, -i], x[, i], monotone = im, categorical = ic) : > unused argument(s) (monotone ...)