similar to: request: most repeated component of a list

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 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 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 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 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)

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] <-

Dear friends Hope you all are fine. Suppose we have a list of arrays. a1=c(4,4,4,4,0,4,4,4,0,3,3,0,0,0,0,0); a1=array(a1,dim=c(4,4)); a2=c(4,4,4,4,0,4,4,4,0,3,3,0,0,0,0,0); a2=array(a2,dim=c(4,4)); a3=c(4,4,4,4,0,3,3,4,0,4,4,0,0,0,0,0); a3=array(a3,dim=c(4,4)); a4=c(4,4,4,4,4,0,3,3,3,3,0,4,4,4,0,0,0,0,0,0); a4=array(a4,dim=c(5,4)); a5=c(4,4,4,4,4,0,4,4,4,4,0,3,3,3,0,0,1,1,0,0);

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 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

---------- 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 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.

Hello, I'm using a very large data set (n > 100,000 for 7 columns), for which I'm pretty happy dealing with pairwise-deleted correlations to populate my correlation table. E.g., a <- cor(cbind(col1, col2, col3),use="pairwise.complete.obs") ...however, I am interested in the number of cases used to compute each cell of the correlation table. I am unable to find such a

Dear R members I have a problem regarding storing the lists. Let L=number of distinct values of any predictor (say L=5) P=number of predictors (say P=20) g1 <- c() for(i in 1:P){ if(L > 1){ for(j in 1:(L-1)){ g <- .... g1[j] <- g } } g2[]=sort.list(g1) } Now the question is: What should we use inside brackets of g2[....], whether

Dear R-friends I have three matrices e.g. var <- matrix(c(4,4,4,4,0,4,4,4,0,3,3,0),nrow=4); val <- matrix(c(0.6,0.6,0.6,0.6,0,1.6,1.6,1.6,0,4.9,4.9,0),nrow=4); nod <- matrix(c(-1,-1,1,1),ncol=1) > var [,1] [,2] [,3] [1,] 4 0 0 [2,] 4 4 3 [3,] 4 4 3 [4,] 4 4 0 > val [,1] [,2] [,3] [1,] 0.6 0.0 0.0 [2,] 0.6 1.6 4.9 [3,] 0.6 1.6

hi all. perhaps someone can help me with subsetting here............ i'm trying to use the nested() function in the bipartite package. my problem is that i have a list of 10,000 matrices and i want the output to be a vector of the nested() function on each of the 10,000 matrices (so i can use these numbers in a statistical test). here is what i'm working with currently: #to store

Dear Colleagues, Occasionally I deal with computer-generated (i.e., websurvey) data files that haven't quite worked correctly. When I try to read the data into R, I get something like this: Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : line 26 did not have 648 elements ...is there any way to get R to tell me how many elements line 26 *did* have? That

Dear List, I neeed a grouped list with two sort of categorical data. I have a data .frame like this. year cat. b c 1 2006 a1 125 212 2 2006 a2 256 212 3 2005 a1 14 12 4 2004 a3 565 123 5 2004 a2 156 789 6 2005 a1 1 456 7 2003 a2 786 123 8 2003 a1 421 569 9 2002 a2 425 245 I need a list with the sum of b and c for every year and every cat (a1, a2 or a3) in this year. I had used the tapply