similar to: WordPerfect--any version working?

I want to test Wordperfect Suite 2000 in Wine. The Corel version of Wine should be available somewhere. But i could not find it.... Any ideas where i can get it? The link corel gave seems to be dead. "Q. Where can I get the source for Corel's version of Wine? A. The source for Corel's version of Wine is available from opensource.corel.com. This site, devoted to the Open Source

Hi List, Being new to R, I am trying to apply boot.stepAIC() for Model selection by bootstrapping the stepAIC() procedure. I had gone through the discussion in various thread on the variable selection methods. Understood the pros and cons of various method, also going through the regression modelling strategies in rms. I want to read Final model or Formula or list of variables automatically

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

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

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

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 <-

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

> 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-users I would like to apply classification and regression tree(CART) to the following data. I have some question on using 'tree' package. The data contains one response variable Y and five explanatory variables. The explanatory variable "x2" is categorical and not ordinal. But, the result obtained after running following R code has indicated that x2 is regard as

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

Yo copio y pego este código y me sale correctamente. Se me ocurre que pueda deberse a la versión de R ¿cuál usas? El 10/09/2020 a las 17:51, Samura . escribió: > Gracias por las respuestas. > > Probé lo de hacer la función y no me salía. Pensaba que hacía algo mal. > Ahora con el código de Marcelino tampoco me sale. > > col1 <- c('x1', 'x2', 'x11',

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'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

Hola, me gustar?a hacer algo como en el siguiente ejemplo A un df a?adirle una columna que es la transformaci?n de otra, en plan a todo lo que sea x1, x2, x3 lo llamo prueba 1 todo lo que sea x4,x5,x6 lo llamo prueba 2 el resto de x las dejo como est?n. Ser?a algo as? col1 <- c('x1', 'x2', 'x11', 'x1','x33', 'x1','x4', 'x5',

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 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 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 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