similar to: loop problem

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 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 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

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 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

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 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)

---------- Forwarded message ---------- From: jim holtman <jholtman at gmail.com> Date: Sun, Sep 7, 2008 at 11:42 AM Subject: Re: [R] request: most repeated sequnce To: Muhammad Azam <mazam72 at yahoo.com> This should do it for you: > 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 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

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

This message also reports wrong estimates produced by lmRob.fit.compute() for nested factors when using the correct contrast matrix. And in these respects, I have found that S-Plus behaves the same way as R. Using the three available contrast types (sum, treatment, helmert) with lm() or lm.fit(), but just contr.sum with rlm() and lmRob(), and small examples, I generated contrast matrices for

I hope you'll all forgive me for displaying my severe lack of knowledge on this topic, and I can't really provide much in the way of reproducible code. A colleague of mine has asked if I know how to import an S object into R. The object is stored in a file named 'pre3.f' When I open the file as a text document I get what's printed below. Being one of those young bucks with

Dear R users, I am trying to have an xyplot of a data set which has the following variables: case (n=10,20,30) parameter (parm=a,b) group (grp=g1,g2) y (y values) x (x=2,4,8) My plot should be parameter by case such that I have 2 rows (each row= each parameter) and 3 columns (each column=each case). My R-code is as follows but I am not able to get what I want to: tp1.sim <- xyplot(y~ x | case

I'm trying to collapse the following table along the sub-group factor. In this case, collapsing means taking the average ages within a subgroup and creating a new table. I seem to be running into trouble trying to create this new data frame. I can use the ave() function to find averages within a subgroup but how do I maintain the Group/Gender factors after collapsing? (see bottom table) Can

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

Hi, R-users, If I have a data frame like this: >x<-data.frame(g=c("g1","g2","g1","g1","g2"),v=c(1,7,3,2,8)) g v 1 g1 1 2 g2 7 3 g1 3 4 g1 2 5 g2 8 It contains two groups, g1 and g2. Now for each group I want the max v: > aggregate(x$v,list(g=x$g),max) g x 1 g1 3 2 g2 8 Beautiful. But what if I want to keep the row index of (g1

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 all, I''m trying to write a function, that will take as an argument, some aligned genome sequences, and using a sliding window, do pairwise comparisons of sequence similarity. Coding the sliding window I think I can manage but what I''m trying to get to grips with is getting it so as every pairwise comparison is made, no matter how many genomes are added, from 3 to N. So if

Hello, I have the following data.frame structure(list(Study = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L,