similar to: request: maximum depth reached problem

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 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 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

Hi, When I call the *switch* function first time, it works. but when I call it at the second time, it does nothing. The version I use is R version 2.9.0 Under development (unstable) (2009-02-21 r47969) here is the output: > organism="human" > species <- switch(organism, human <- "Hs", fly <- "Dm", mouse <-

Hi, Few days ago I have asked about spatial join on the minimum distance between 2 sets of points with coordinates and attributes in 2 different data frames. Simon Knapp sent code to do it when calculating distance on a sphere using lat, long coordinates and I've change his code to use Euclidian distances since my data had UTM coordinates. Typically one data frame has around 30 000 points

I have a problem with running the tree package (dec.8, 2001) for R. The problem is, it will only give me 5/6 terminal node and then stop, while using Splus's tree on the same data with the same specification give me hundreds of nodes. Here's a little more background info: R-1.4.1 Solaris 5.7 rpart (most recent version) tree (..) Splus 6.0 Solaris 5.7 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] <-

---------- 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, +

Hello to everyone, I am trying to compute a mean and show it to the screen and then save to later be used as a boxplot.I have used the following code: dat<-read.table(file="C:\\Documents and Settings\\Owner\\My Documents\\Yeast\\Yeast.txt",header=T,row.names=1) file.show(file="C:\\Documents and Settings\\Owner\\My Documents\\Yeast\\Yeast.txt") x<-dat[2,23:46]

Hello everyone, I am trying to subset a matrix with the following code: dat<-read.table(file="C:\\Documents and Settings\\Owner\\My Documents\\Yeast\\Yeast.txt",header=T,row.names=1) file.show(file="C:\\Documents and Settings\\Owner\\My Documents\\Yeast\\Yeast.txt") Z.matrix<-as.matrix(dat) Y<-array(dat,dim(,[46:63]) but I keep getting the error message: error in dim

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, ]

This must be a very simple question, but I can't find any information on it elsewhere, sorry. When extracting information from a list using column headers, how do I get R to interpret something as a variable rather than a string? For example: xx$"YAL002" works, but this doesn't: gene <- "YAL002" xx$gene neither do xx$parse(gene) xx$eval(gene)

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)

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]]

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