similar to: optimizing lm with multiple very similar RHSs

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 <-

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 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

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 <-

> From: Peter Dalgaard > > Hallgeir.Grinde at elkem.no writes: > > > 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)

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 have a sparse contingency table (most cells are 0): > xtabs(~.,data[,idx:(idx+4)]) , , x3 = 1, x4 = 1, x5 = 1 x2 x1 1 2 3 1 0 0 31 2 0 0 112 3 0 0 94 , , x3 = 2, x4 = 1, x5 = 1 x2 x1 1 2 3 1 0 0 0 2 0 0 0 3 0 0 0 , , x3 = 3, x4 = 1, x5 = 1 x2 x1 1 2 3 1 0 0 0 2 0 0 0 3 0 0 0 , , x3 = 1, x4

Hi, Using S=1000 and simdata <- replicate(S, generate(3000)) #If you want both "m1" and "m0" #here the missing values are 0 res1<-sapply(seq_len(ncol(simdata.psm1)),function(i) {x1<-merge(simdata.psm0[,i],simdata.psm1[,i],all=TRUE); x1[is.na(x1)]<-0; x1}) res1[,997:1000] #????? [,1]???????? [,2]???????? [,3]???????? [,4]??????? #x1??? Numeric,3000 Numeric,3000

Hi, I wonder if anyone can help me with this issue. I am using R2BayesX. It seems that the model can maximally contain 20 interactions. When the number of interaction terms exceed 20, the code stops working. Here is a piece of toy code. rm(list=ls()) library(BayesX) library(R2BayesX) #data generating model f2<-function(x1,x2,x3,x4) { y<-2*sin(pi*x1)*1.5+exp(2*x2)/3+2 * sin(4 * pi * (x3

I'm using specify.model for the sem package. I can't figure out how to represent the residual errors for the observed variables for a CFA model. (Once I get this working I need to add some further constraints.) Here is what I've tried: model.sa <- specify.model() F1 -> X1,l11, NA F1 -> X2,l21, NA F1 -> X3,l31, NA F1 -> X4,l41, NA F1 -> X5, NA, 0.20

Hi, I am new to R for solving optimization problems, I have set of communication channels with limited capacity with two types of costs, fixed and variable cost. Each channel has expected gain for a single communication. I want to determine optimal number of communications for each channel maximizing ROI)return on investment) with overall budget as constraint.60000 is the budget allocated.

Dear R-users I have a probelm running stepAIC in R1.7.1 I wrote a program which used stepAIC as a part of it, and it worked fine while I was using the previous version of R1.7.0. However, I found the program did not work any more. Now, R produces a message which tells "Error in as.data.frame.default(data) : can't coerce function into a data.frame" every time I run the part of

I have an application that would generate a cross-tabulation in array format in R. In particular, my application would give me a result similar to that of : array(5,c(2,2,2,2,2)) The above could be seen as a cross-tabulation of 5 variables with 2 levels each (could be 0 and 1). In this case, the data were such that each cell has exactly 5 observations. I Now, I want the output to look like 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

Dear R-List, I tried to fit a partial credit model using the "pcmdat" from eRm-package comparing the results of mirt, eRm, ltm and winsteps. The results where quite different, though. I cannot figure out what went wrong and I do not know which result I can rely on. This is what I did in R library(mirt) #load(file="u3.RData")

I got my posting bounced and sorry if I accidentally post twice. I have been looking at 'lars' pkg and got puzzled by the behavior of function 'lars'. I want to do weighted lasso regression and can't get a match from lars output with lm output. Here is an example: y = rnorm(10) x = matrix(runif(50),nrow=10) X = data.frame(y,x) z = runif(10) X = data.frame(y,x,z) X$z = X$z /

Dear R community members and R experts I am stuck at a point and I tried with my colleagues and did not get it out. Sorry, I need your help. Here my data (just created to show the example): # generating a dataset just to show how my dataset look like, here I have x variables # x1 .........to X1000 plus ind and y ind <- c(1:100) y <- rnorm(100, 10,2) set.seed(201) P <-

Hello I have data like this x1 x2 x3 x4 x5 I want to create a matrix similar to a correlation matrix, but with the difference between the two values, like this x1 x2 x3 x4 x5 x1 x2-x1 x3-x1 x4-x1 x5-x1 x2 x3-x2 x4-x2 x5-x2 x3 x4-x3 x5-x3 x4 x5-x4 x5 Then I

#### I checked through every 3 factor * 3 loading case. #### While, 4 factor * 3 loading failed. #### the data is 6 factor * 3 loading require(sem); cor18<-read.moments(); 1 .68 1 .60 .58 1 .01 .10 .07 1 .12 .04 .06 .29 1 .06 .06 .01 .35 .24 1 .09 .13 .10 .05 .03 .07 1 .04 .08 .16 .10 .12 .06 .25 1 .06 .09 .02 .02 .09 .16 .29 .36 1 .23 .26 .19 .05 .04 .04 .08 .09 .09 1 .11 .13 .12 .03 .05 .03

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 <-