similar to: request: which integer in each column is in majority

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 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 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 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 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 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 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 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 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 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 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 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

I applied bootpred for multinomial logistic reg. (with nnet package). I used same as theta.fit and theta.predict of R for my data. but give me error. Can I do this with response vriable;7 levels predictor variables:5 (1 classifier, 4 continuous)?   Thanks alot Azam   [[alternative HTML version deleted]]

Hi   I want to do leave-one-out cross-validation for multinomial logistic regression in R. I did multinomial logistic reg. by package nnet in R. How I do validation? by which function? response variable has 7 levels   please help me   Thanks alot Azam [[alternative HTML version deleted]]

>> Hi, I am trying to draw a random sample from an household survey with sample weight. Is there any function in R or Splus which allows this. Regards, ******************************************************* Mehtabul Azam Department of Economics Southern Methodist University Dallas TX 75275-0496 Tel: (214) 214 938 3906 Email: mazam at smu.edu <mailto:mazam at smu.edu>

Dear All   Pleas help me to increase the memory in R.   In order increase memory, I read the FAQ and follow the instruction as below     Close R, then right-click on your R program icon (the icon on your desktop or in your programs directory). Select ``Properties'', and then select the ``Shortcut'' tab. Look for the ``Target'' field and after the closing quotes around the

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);