similar to: Fwd: request: most repeated sequnce

Dear R community Hope every one be in best of his/her health. I have a situation in which there are s-sectors. Each sector is further divided into r-rows and c-columns. All it makes an array having dimension (r,c,s). e.g. x=c(1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,2,2,2,3,3,0,0,0,0,0,0,0,0,0,0,1,1,1,2,2,3,3,3,4,4,4,0,0,0,0,0,0,0,1,1,1,2,2,2,3,3,3,4,4,4,

Dear R community I have stored the results of arrays in a list consist of J-components (say 200 components). Each component containing same no of columns but may be different no of rows. e.g [[1]] [,1] [,2] [,3] [,4] [,5] [1,] 4 0 0 0 0 [2,] 4 3 4 0 0 [3,] 4 3 4 0 0 [4,] 4 3 0 0 0 [[2]] [,1] [,2] [,3] [,4] [,5]

Dear R community I have a problem regarding which of the column in a matrix contains all of zero elements. e.g. x=c(3,3,3,3,0,0,0,0,5,5,5,5,8,8,8,8); x=matrix(x, nrow=4) the output is > x [,1] [,2] [,3] [,4] [1,] 3 0 5 8 [2,] 3 0 5 8 [3,] 3 0 5 8 [4,] 3 0 5 8 In this case the required column is second so the result should be "2".

Respected R helpers/ users I am one of the new R user. I have a problem regarding to know which of the integer in each column of the following matrix is in majority. I want to know that integer e.g. in the first column 1 is in majority. Similarly in the third column 4 is in majority. So what is the suitable way to get the desired integer for each column. I am looking for some kind reply. Thanks

--- John Kane <jrkrideau at yahoo.ca> wrote: > Date: Fri, 20 Apr 2007 10:47:45 -0400 (EDT) > From: John Kane <jrkrideau at yahoo.ca> > Subject: Re: [R] Character coerced to factor and I > cannot get it back > To: jim holtman <jholtman at gmail.com> > > Thanks Jim, > > I can live with it as a factor or I will do as you > suggest. What is bothering

Dear R users I tried a lot to solve the following problem but could not. I have two arrays having same order i.e 1 by 150. j=10; ss=150; r=array(0 , c( j , ss )); rr=array(0 , c( j , ss )); r1=array(0 , c( j-1 , ss )); r2=array(0 , c( j-1 , ss )); r3=array(0 , c( 2 , ss )) for(i in 1:j-1){ r1[ i , ] <- r[ j+1, ]-r[ j, ]; r2[ i , ] <- rr[ j+1, ]-rr[ j, ]

Dear friends I have an array consist of r-rows and c-columns e.g. x=c(1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4,0,0,0,0,0,0,0,0); x1=array(x, dim=c(4,6)) output is > x1 [,1] [,2] [,3] [,4] [,5] [,6] [1,] 1 2 3 4 0 0 [2,] 1 2 3 4 0 0 [3,] 1 2 3 4 0 0 [4,] 1 2 3 4 0 0 How can i ignore columns having zero sums? Help in this regard

Dear R community I have a problem to access particular list. I have a code given below where there is recursive process. It is not possible to run it because there are few other functions involved inside like sv, LN, RN etc. k=0; n=0; variable=c(); vr<-list() func <- function(data,testdata) { . . if(......){ n<<-n+1; vr[[n]] <<- variable; print(vr)

Dear friends There is a list of arrays comprising different no of rows and columns even sometimes NULL, such as [[2]] given below. How can we ignore [[2]] or others like this in the complete list. Any help in this regard is needed. Thanks [[1]] [,1] [,2] [1,] 3 1 [2,] 3 1 [3,] 3 1 [[2]] NULL [[3]] [,1] [,2] [,3] [,4] [,5] [,6] [,7] [1,] 3 1

ok, sorry, my mistake was the C printf. Thank you for your good answer Regards ----- Message d'origine ---- De : JS Ubei <jsubei at yahoo.fr> ? : jim holtman <jholtman at gmail.com> Cc : r-help at r-project.org Envoy? le : Jeudi, 17 Juillet 2008, 15h25mn 07s Objet : [R] Re : float and double precision with C code thank you for your quick answer, I'm far of the digits

hi All, ok fine got it. The design of R is such that it will work only if data fits in the main memory. This issue has been taken up many times but it seems it would be very difficult to change the core code hmmm. ok fine. hence if i want to work then i will have to either partition the data or work with columns. thanks to you all - Sayonara With Smile & With Warm Regards :-) G a u

hi All, ok fine got it. The design of R is such that it will work only if data fits in the main memory. This issue has been taken up many times but it seems it would be very difficult to change the core code hmmm. ok fine. hence if i want to work then i will have to either partition the data or work with columns. thanks to you all - Sayonara With Smile & With Warm Regards :-) G a u

Dear R users I have a problem regarding an addition of an extra "row" to a matrix. e.g. i have a matrix a <- matrix(1:6,2,3) > a [,1] [,2] [,3] [1,] 1 3 5 [2,] 2 4 6 I want to add a matrix having just one row. e.g. b <- matrix(7:9,1,3) > b [,1] [,2] [,3] [1,] 7 8 9 Now i want to get result like this [,1] [,2] [,3] [1,] 1 3 5

Does something like this get you close: x <- list() keys <- LETTERS[1:6] # create for (i in keys){ x[[i]] <- data.frame(a=1:5, b=1:5, c=1:5) } # output output <- file('tempxx.txt', 'w') for (i in keys){ write.table(i, row.names=FALSE, col.names=FALSE, file=output, quote=FALSE) write.table(x[[i]], file=output, quote=FALSE) } close(output) On Wed, Jul 2,

Dear R users I have a very basic question. I tried but could not find the required result. using dat <- pima f <- table(dat[,9]) > f 0 1 500 268 i want to find that class say "0" having maximum frequency i.e 500. I used >which.max(f) which provide 0 1 How can i get only the "0". Thanks and best regards Muhammad Azam Ph.D. Student Department of

Dear R community Hope all of you are fine. I have a question regarding the an error message. Actually, I am trying to generate classification trees using "tree" package. It works well but for some datasets e.g., wine, yeast, boston housing etc. it gives an error message. Error in tree(V14 ~ ., data = training.data, method = c("recursive.partitioning"), : maximum

Dear R community I have a question regarding the value of cost complexity parameter "k" used in "tree" package for pruning purpose. Any help in finding the optimum value of "k" is requested. Please give some suggestion in this regard. In the example below i used k=0 but i don't know why? But if i use k=NULL, then it will not plot the resultant tree.

Dear R community I am trying to assign alphabets to integer values 1, 2, 3 etc. in y given below. Can any body suggest some simple way to do the same job? ds=iris; dl=nrow(ds) c1=ds[,1]; c2=ds[,2]; c3=ds[,3]; c4=ds[,4]; c5=ds[,5]; iris=cbind(c1,c2,c3,c4,c5) y=iris[,5] y1=which(y==1); y[y1] <- c("a"); y2=which(y==2); y[y2] <- c("b"); y3=which(y==3); y[y3] <-

Hi Jim, What you told me works well. I have tried using 'eval(parse(text="1:20 = x"))' as well as 'try(parse(text="1:20 = x"))' before but not using eval anfd try functions together. I just added try function following your suggestion, e.g. 'try(eval(parse(text="1:20 = x")))'and then it worked well. Thank you very much for the nice

thank you for your quick answer, I'm far of the digits capacity and my values are not the result of a computation. I'm developping a R package to acces a specific data source. And I need precision a few better. How can I do ? When I try this In R console, this is correct and what I need : > my_value <- 29.958334 > my_value == 29.958334 [1] TRUE But I need to do the first