similar to: do.call or something instead of for

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 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'd like to count the number of integers in a vector y. Here is an example. y <- c(0,1,1,3,3,3,5,5,6) In fact, I know how to count the number of specific number in y. sum(y==0) -> 1 sum(y==1) -> 2 sum(y==2) -> 0 sum(y==3) -> 3 sum(y==4) -> 0 sum(y==5) -> 2 sum(y==6) -> 1 However, in one computation I want to get this vector [1,2,0,3,0,2,1]. Thank

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) --------------------------------------------------------------------------

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 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 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.) -------------------------------------------------------------------------------------------- >

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 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, ]

Dear R users, I am trying to figure out the control parameter in "optim," especially, "fnscale" and "parscale." In the R docu., ------------------------------------------------------ fnscale An overall scaling to be applied to the value of fn and gr during optimization. If negative, turns the problem into a maximization problem. Optimization is performed on

Dear R users, I have 32 observations in data x. After sorting this, I want to compute means and variances of 3 groups divided by "nr". Actually, the number of groups is flexible. Any suggestion will be greatly appreciated. Kathryn Lord --------------------------------------------------------------------------- x=rnorm(32) y=sort(x) nr=matrix(c(12,11,10,10,10,11),2,3) > nr

...as promised. This describes the mdct used in my d.m.l patch. I think it is the same as the Lee fast-dct. I typed it in a kind of pseudo-TeX, 'cause the ascii art would kill me. Hope you can read TeX source; if not, ask someone who can to make a .ps/.gif/.whatever of the TeX output, and put it on a webpage or something. I'm to lazy to do it (and besides, I don't have access to TeX,

Dear all this is a question of model specification in lme which I'd for which I'd greatly appreciate some guidance. Suppose I have data in long format gene treatment rep Y 1 1 1 4.32 1 1 2 4.67 1 1 3 5.09 . . . . . . . . . . . . 1 4 1 3.67 1 4 2 4.64 1 4 3 4.87 .