similar to: creating lagged variables

Hello R users, Here is my question about generating random sample. How to set the random seed to recreate the same random numbers? For example, 10 random numbers is generated from N(0,1), then "runif(10)" is used.What if I want to get the same 10 random numbers when I run runif(10) again? Is it possible?I think .Random.seed should be used here. Thanks. Xiaoxi [[alternative

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,]

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

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]]

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:

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, 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],

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

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

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.