similar to: Random Sample

Hi all. I'm looking for robust ways of building lagged variables in a dataset with multiple individuals. Consider a dataset with variables like the following: ## set.seed(123) d <- data.frame(id = rep(1:2, each=3), time=rep(1:3, 2), value=rnorm(6)) ## >d id time value 1 1 1 -0.56047565 2 1 2 -0.23017749 3 1 3 1.55870831 4 2 1 0.07050839 5 2 2 0.12928774 6

I am writing a function where the arguments are names of objects or variable names in a data frame. To convert the strings to the objects I am using eval(parse(text=name)): f.graph.two.vbs<-function(dataname,v1){ val<-paste(dataname,v1,sep="$") val<-eval(parse(text=val)) val } However running this returns an error:

Once again I need to tap into the collective knowledge here. Let's say I have the following columns and data below Y X1 X2 X3 X4 I would like to generate additional new columns and column names (ie the data would be squared - and I'd like the column names to reflect this) like: Y X1 X2 X3 X4 X1^2 X2^2 X3^2 X4^2 I believe I can compute the values correctly with the code below, but I

Hi folks, I want to sort a matrix row-by-row and create a new matrix that contains the corresponding colnames of the original matrix. E.g. > set.seed(123) > a <- matrix(rnorm(20), ncol=4); colnames(a) <- c("A","B","C","D") > a A B C D [1,] -0.56047565 1.7150650 1.2240818 1.7869131 [2,]

Hello useRs, I'm using the command "solnp" in package "Rsolnp" to solve a general nonlinear programming problem. But I got an error that " the leading minor of order 15 is not positive definite ". Can anybody tell what may cause this error? Does it have something to do with the starting values? Thanks a lot!!! Xiaoxi [[alternative HTML version

Hello R users, Can anyone recommend me any package that can be used to solve linear programming subject to nonlinear equality/inequality constraints. Rdonlp2 is now unavailable due to licensing issue. Thanks, Xiaoxi [[alternative HTML version deleted]]

Sorry, another newbie question :-( I loaded a data set with 10 rows and 30 columns. The first column is characters for names of car manufacturers: Jeep Nissan Toyota1 Toyota2 Etc. How can I replace "Toyota2" with "Scion"? Thanks again [[alternative HTML version deleted]]

Hello R users, I have two quick questions while using "lpSolve" package for linear programming. (1) the result contains both characters and numbers, e.g., Success: the objective function is 40.5, but I only need the number, can I only store the number? (2) How to set boundaries for variables? e.g., all variable are positive. Thanks a lot! Xiaoxi

Hello I have a vector: set.seed(123) > n<-3 > x<-rnorm(n); x [1] -0.56047565 -0.23017749 1.55870831 I like to create a matrix with elements containing variances and covariances of x. That is var(x[1]) cov(x[1],x[2]) cov(x[1],x[3]) cov(x[2],x[1]) var(x[2]) cov(x[2],x[3]) cov(x[3],x[1]) cov(x[3],x[2]) var(x[3]) And I like to do it with "apply". Thanks. On 10/4/2024 6:35

Hello all, Can anybody tell me how to implement the heterogeneous bootstrap algorithm by the FEAR package. I only know there is a command (boot.sw98) for homogeneous bootstrap. Thanks a lot. Xiaoxi _________________________________________________________________ Hotmail is redefining busy with tools for the New Busy. Get more from your inbox.

OK. Thanks to all. Suppose I have two vectors, x and y. Is there a way to do the covariance matrix with ?apply?. The matrix I need really contains the deviation products divided by the degrees of freedom (n-1). That is, the elements (1,1), (1,2),...,(1,n) (2,1), (2,2),...., (2,n) .... (n,1),(n,2),...,(n,n). > Hello, > > This doesn't make sense, if you have only one vector you

Hello, If you have a numeric matrix or data.frame, try something like cov(mtcars) Hope this helps, Rui Barradas ?s 10:15 de 04/10/2024, Steven Yen escreveu: > On 10/4/2024 5:13 PM, Steven Yen wrote: > >> Pardon me!!! >> >> What makes you think this is a homework question? You are not >> obligated to respond if the question is not intelligent enough for you.

Hello, This doesn't make sense, if you have only one vector you can estimate its variance with var(x) but there is no covariance, the joint variance of two rv's. "co" or joint with what if you have only x? Note that the variance of x[1] or any other vector element is zero, it's only one value therefore it does not vary. A similar reasonong can be applied to cov(x[1],

It's still hard to figure out what you want. If you have two vectors you can compute their (2x2) covariance matrix using cov(cbind(x,y)). If you want to compute all pairwise squared differences between elements of x and y you could use outer(x, y, "-")^2. Can you explain a little bit more about (1) the context for your question and (2) why you want/need to use apply() ? On