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---------- 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, + 0,0,0,0,0,0,1,2,2,2,2,2,0,3,3,0,4,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,...

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 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 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 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".

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 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)

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

I want to take the first row of each unique ID value from a data frame. For instance > ddTable <- data.frame(Id=c(1,1,2,2),name=c("Paul","Joe","Bob","Larry")) I want a dataset that is Id Name 1 Paul 2 Bob > unique(ddTable) Will give me all 4 rows, and > unique(ddTable$Id) Will give me c(1,2), but not accompanied by the name column.

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

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 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 Petr Thanks for the response. Hope it will now help me to proceed. best regards M.Azam ________________________________ From: Petr PIKAL <petr.pikal@precheza.cz> Cc: R Help <r-help@r-project.org>; r-help-bounces@r-project.org Sent: Monday, October 27, 2008 8:43:27 AM Subject: Odp: [R] Request: Most repeated sequence considering combinations at each row Hi not sure if this is

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 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 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