similar to: Error in f(x, ...) : subscript out of bounds

Hi all, myint=function(mu,sigma){ integrate(function(x) dnorm(x,mu,sigma)/(1+exp(-x)),-Inf,Inf)$value } mymu=seq(-3,3,length(1000)) mysigma=seq(0,1,length(500))[-1] k=1 v=c() for (j in 1:length(mymu)) { for (i in 1:length(mysigma)) { v[k]=myint(mymu[j],mysigma[i]) k=k+1 } } Basically, I want to investigate for what values of mu and sigma, the integral is divergent. Is there another way

Hi, I was very much hoping someone could help me with the following. I am trying to convert some SAS NLMIXED code to NLME in R (v.2.1), but I get an error message. Does anyone have any suggestions? I think my error is with the random effect "u" which seems to be parametrized differently in the SAS code. In case it's helpful, what I am essentially trying to do is estimate parameters

Hi folks, I'm trying to estimate bias-corrected percentile (BCP) confidence intervals on a vector from a simple for loop used for resampling. I am attempting to follow steps in Manly, B. 1998. Randomization, bootstrap and monte carlo methods in biology. 2nd edition., p. 48. PDF of the approach/steps should be available here: https://wyocoopunit.box.com/s/9vm4vgmbx5h7um809bvg6u7wr392v6i9 If

Dear All I am trying to estimate a mixture of regression and get the following error using the mixreg package: Error in y - yhat : non-conformable arrays The instruction I used were: x <- as.matrix(LRHUN) y <- as.matrix(LRINTER) TS <- list(list(beta=c(3.0,1.0),sigsq=1,lambda=0.4), list(beta=c(0.0,1.0),sigsq=1,lambda=0.6)) prova <- mixreg(x,y, ncomp=2, theta.start=TS)

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

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

I want to run some R script using the inline package (which allows to create and run inline C++ code in my humble understanding). So, after loading the required packages and copy and paste the example that runs C code (in the Reference Manual as a PDF), I have a compilation error. Any body has ever tried this inline package? -- View this message in context:

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