similar to: Counting the number of integers at one swoop

Dear R users I'd like to generate two sets of random numbers with a fixed correlation coefficient, say .4, using R. Any suggestion will be greatly appreciated. Regards, Kathryn Lord -- View this message in context: http://r.789695.n4.nabble.com/generate-two-sets-of-random-numbers-that-are-correlated-tp3735695p3735695.html Sent from the R help mailing list archive at Nabble.com.

Dear R users, I'd like to compute X like below. X_{i,t} = 0.1*t + 2*X_{i,t-1} + W_{i,t} where W_{i,t} are from Uniform(0,2) and X_{i,0} = 1+5*W_{i,0} Of course, I can do this with "for" statement, but I don't think it's good idea because "i" and "t" are too big. So, my question is that Is there any better idea to avoid "for" statement

Dear R users... I'd like to change this character vector, "zz", zz <- c("12","56","89") to the following numeric matrix. [,1] [,2] [1,] 1 2 [2,] 5 6 [3,] 8 9 Actually, "zz" vector has a long length. Any comments will be greatly appreciated. Kathryn Lord -- View this message in context:

Dear R users, I'd like to make this data rem.y = c(-1,0,2,4,5) from y = c(-1,-1,0,2,2,2,2,4,4,5,5,5,5,5). That is, I need to remove repeated values. Here is my code, but I don't think it is efficient. How could I improve this? #------------------------------------------------------------------------ y = c(-1,-1,0,2,2,2,2,4,4,5,5,5,5,5) n=length(y) for (i in 1:n) #

Dear R users, I am looking for more efficient way to compute the followings -------------------------------------------------------------------------- a <- matrix(c(1,1,1,1,2,2,2,2),4,2) b <- matrix(c(1,2,3,4),4,1) Eventually, I want to get this matrix, `c`. c <- matrix(c(1/1,1/2,1/3,1/4,2/1,2/2,2/3,2/4),4,2) --------------------------------------------------------------------------

Hi, R users, Here is an example. k <- c(1,2,3,4,5) i <- c(0,1,3,2,1) if k=1, then j=0 from i if k=2, then j=0, 1 from i if k=3, then j=0, 1, 2, 3 from i if k=4, then j=0, 1, 2 from i if k=5, then j=0, 1 from i so i'd like to create a list like below. > list k j 1 1 0 2 2 0 3 2 1 4 3 0 5 3 1 6 3 2 7 3 3 8 4 0 9 4 1 10 4 2 11 5 0 12 5 1 I tried expand.grid, but I

Dear R users... I made this by help of one of R users. _________________________________________________________________ X=matrix(seq(1,4), 2 , 2) B=matrix(c(0.6,1.0,2.5,1.5) , 2 , 2) func <- function(i,y0,j) { y0*exp(X[i,]%*%B[,j]) } list1 <- expand.grid( i=c(1,2) , y0=c(1,2) , j=c(1,2) ) results <- do.call( func , list1 )

Dear R users I am trying to use OPTIMX(OPTIM) for nonlinear optimization. There is no error in my code but the results are so weird (see below). When I ran via OPTIM, the results are that Initial values are that theta0 = 0.6 1.6 0.6 1.6 0.7. (In fact true vales are 0.5,1.0,0.8,1.2, 0.6.) -------------------------------------------------------------------------------------------- >

Dear R users When I use OPTIM with BFGS, I've got a significant result without an error message. However, when I use OPTIMX with BFGS( or spg), I've got the following an error message. ---------------------------------------------------------------------------------------------------- > optimx(par=theta0, fn=obj.fy, gr=gr.fy, method="BFGS", >

Dear R users... I need to split this matrix(or dataframe), for example, z <- matrix(c(13,1,1,1,1,12,0,0,0,0,8,1,0,1,1,8,0,1,0,0, 10,1,1,1,1,3,0,1,0,0,3,1,0,1,1,6,1,1,1,1),8,5,byrow = T) > z [,1] [,2] [,3] [,4] [,5] [1,] 13 1 1 1 1 [2,] 12 0 0 0 0 [3,] 8 1 0 1 1 [4,] 9 0 1 0 0 [5,] 10 1 1 1 1

Dear R users, I have the list as follows; #------------------------------------------------------ > z [[1]] [[1]][[1]] matrix(A) [[1]][[2]] matrix(B) [[1]][[3]] matrix(C) [[2]] [[2]][[1]] matrix(D) [[2]][[2]] matrix(E) [[2]][[3]] matrix(F) #--------------------------------------------- I'd like to compute matrix(A)+matrix(B)+matrix(C)+matrix(D)+matrix(E)+matrix(F) In

Dear R users, is there way to ignore an error and go back to 1st line? I mean, #------------------------------- while (or repeat) ------------ { 1 2 . . . 6 } #----------------------------- For example, if I have an error in the 6th line, then I'd like to go back to the 1st line. I've already tried "try", but it didn't work. Any suggestion will be greatly

Hi, R users, Here is an example. k <- c(1,2,3,4,5) i <- c(0,1,3,2,1) if k=1, then i=0 if k=2, then i=0, 1 if k=3, then i=0, 1, 2, 3 if k=4, then i=0, 1, 2 if k=5, then i=0, 1 so i'd like to create a list like below. > list k i 1 1 0 2 2 0 3 2 1 4 3 0 5 3 1 6 3 2 7 3 3 8 4 0 9 4 1 10 4 2 11 5 0 12 5 1 I tried expand.grid, but I can't. Any suggestion will be

Dear R users... I made the R-code for this double summation computation http://www.nabble.com/file/p19213599/doublesum.jpg ------------------------------------------------- Here is my code.. sum(sapply(1:m, function(k){sum(sapply(1:m, function(j){x[k]*x[j]*dnorm((mu[j]+mu[k])/sqrt(sig[k]+sig[j]))/sqrt(sig[k]+sig[j])}))})) ------------------------------------------------- In fact, this is

Dear R users, I have two n*1 integer vectors, y1 and y2, where n is very very large. I'd like to compute elbp = 4^(y1) * 5^(y2) * sum_{i=0}^{max(y1, y2)} [{ (y1-i)! * (i)! * (y2-i)! }^(-1)]; that is, I need to compute "elbp" for each (y1, y2) pair. So I made R code like below, but I don't think it's efficient Would you plz tell me how to avoid this "for"

Dear R users, I try to compute this summation, http://www.nabble.com/file/p25054272/dd.jpg where f(y|x) = Negative Binomial(y, mu=exp(x' beta), size=1/alp) http://www.nabble.com/file/p25054272/aa.jpg http://www.nabble.com/file/p25054272/cc.jpg In fact, I tried to use "do.call" function to compute each u(y,x) before the summation, but I got an error, "Error in X[i, ]

Ok, so I'm slowly figuring out what a factor is, and was able to follow the related thread about finding a mode by using constructs like my_mode = as.numeric(names(table(x))[which.max(table(x))]) Now, suppose I want to keep looking for other modes? For example, Rgames> sample(seq(1,10),50,replace=TRUE)->bag Rgames> bag [1] 2 8 8 10 7 3 2 9 8 3 8 9 6 6 10 10 7 1

"Not that it really matters, but" Can someone explain how the row numbers get assigned in the following sequence? It looks like something funky happens when rbind() coerces 'bar' into a dataframe. In either sequence of rbind below, once you get past the first two rows, the row numbers count normally. Rgames> (foo<-data.frame(x=5,y=4,r=3)) x y r 1 5 4 3 Rgames>

I feel bad even asking, but: Rgames> data(OrchardSprays) Rgames> model<-lm(decrease~.,data=OrchardSprays) Rgames> model Call: lm(formula = decrease ~ ., data = OrchardSprays) Coefficients: (Intercept) rowpos colpos treatmentB treatmentC 22.705 -2.784 -1.234 3.000 20.625 treatmentD treatmentE treatmentF treatmentG treatmentH

Hi all, I'm working with a bunch of large graphs, and stumbled across something useful. Probably many of you know this, but I didn't and so others might benefit. Using pch="." speeds up plotting considerably over using symbols. > x <- runif(1000000) > y <- runif(1000000) > system.time(plot(x, y, pch=".")) user system elapsed 1.042 0.030 1.077