similar to: Using str() in a function.

Dear R-helper, How could I count only some variable was exist after running sample (random) function. For example, > testx <- factor(c("Game","Paper","Internet","Time","Money")) > for(i in 1:2) { + x <- sample(testx,replace=TRUE) + print(x) + } [1] Money Money Time Internet Time Levels: Game Internet

Hello Please see the example below > class(testX) [1] "matrix" > class(testX[1,]) [1] "numeric" Why not matrix? What am I missing here? Is there a way to keep the same class? The reason for the question is that I want to implement a k-step ahead prediction for my own routines and R wrecks does not seem to like [1,] as shown below. >

I've found a problem when using categorical variables in lmp() from package lmPerm According to help(lmp): "This function will behave identically to lm() if the following parameters are set: perm="", seq=TRUE, center=FALSE.") But not in the case of including categorical variables: require(lmPerm) set.seed(42) testx1 <- rnorm(100,10,5) testx2 <-

Hi, I have a question about using lm on matrix, have to admit it is very trivial but I just couldn't find the answer after searched the mailing list and other online tutorial. It would be great if you could help. I have a matrix "trainx" of 492(rows) by 220(columns) that is my x, and trainy is 492 by 1. Also, I have the newdata testx which is 240 (rows) by 220 (columns). Here is

All, For my understanding, I wanted to see if I can get glmnet predictions using both the predict function and also by multiplying coefficients by the variable matrix. This is not worked out. Could anyone suggest where I am going wrong? I understand that I may not have the mean/intercept correct, but the scaling is also off, which suggests a bigger mistake. Thanks for your help. Juliet Hannah

Hi all I am using the caret package and having difficulty in obtaining the results using regression, I used the glmnet to model and trying to get the coefficients and the model parameters I am trying to use the extractPrediction to obtain a confusion matrix and it seems to be giving me errors. x<-read.csv("x.csv", header=TRUE); y<-read.csv("y.csv", header=TRUE);

ok, i'm NO expert but I can learn a LOT of things. I have used wine very sparingly in the past, enough to know that IE, outlook and a few other "Win Only" apps will run. So when wine 1.9 was released i figure i HAD to have it. but BOY its a lot different. Installation was smooth, MDV rpms work as expected, but thats where it ends. I use the same IE5.5 install source I have used

Dear list, I have a problem on persp() x <- u1data #first coloum in attached data y <- u2data #second coloum in attached data f <- function(x,y){qgev(pnorm(rhoF*qnorm(pnorm((qnorm(y)-rho2*qnorm(x)/sqrt(1-rho2^2)))) +sqrt(1-rhoF^2)*qnorm(0.95)),-0.3935119, 0.4227890, 0.2701648)} z <- outer(x,y,f) persp(x,y,z) The R will display: "Error in persp.default(x, y,

Hi all, I have some troubles in writing XXInstFormat.td for my little backend. I don't know how should I start writing up this file. Is there any general rule? Thanks. Regards, chenwj -- Wei-Ren Chen (陳韋任) Computer Systems Lab, Institute of Information Science, Academia Sinica, Taiwan (R.O.C.) Tel:886-2-2788-3799 #1667 Homepage: http://people.cs.nctu.edu.tw/~chenwj

Dear all, I am confused with the behaviour of survfit with newdata option. I am using the latest version R-2-15-0. In the simple example below I am building a coxph model on 90 patients and trying to predict 10 patients. Unfortunately the survival curve at the end is for 90 patients. Could somebody please from the survival package confirm that this behaviour is as expected or not - because I

Dear folks: Let?s suppose I want a function to print return the name of the object passed to it. > myname <- function(object) {out<-deparse(substitute(object)); out} This works fine on a single object: > O1 <-c(1:4) > myname(O1) [1] "O1" However it does not work if you use lapply to pass it the same object from a list: > O2 <-c(1:4) > object.list <-

thanks for your attention. temporarily with the following script I'm looking at myself. #!/bin/bash INTERVAL=180 LOG=true function log { if $LOG; then echo $1 fi } while true; do testx=`ps -auxw | grep "smbd" | wc -l` testy=`netstat -ap | grep samba | grep sock | wc -l` if test "$testx" -gt "50" -a "$testy" -gt "50" ;then echo "samba

Readers, How to solve this unicode input error please? > postscript("~/tmp/test.eps",width=100/25.4,height=100/25.4,horizontal=FALSE,onefile=TRUE,paper="special") > testx<-seq(1:5) > testy<-seq(1:5) > plot(testy~testx) > mtext('text (O?)\n more text',side=3,line=1) Warning messages: 1: In mtext("text (O?)\n more text", side = 3, line =

Hi all, I am trying to create a function that will open connections to all files of one type within the working directory. I've got the function to open the connections, but I am having a bugger of a time trying to get these connections named as objects in the workspace. I am at the point where I can do it outside of the function, but not inside, using assign. I'm sure I'm

I suggest u grep how this file is used. Since it is included in xxInstrInfo.td and this is its only use, we can conclude that this file does nothng more than defining some helper tablegen classes and we thus can totally move these definitions into xxInstrInfo.td and then remove XXInstFormat.td Regards. 在 2013-2-6 下午4:52，"陳韋任 (Wei-Ren Chen)" <chenwj at iis.sinica.edu.tw>写道： >

Dear mailing list I have a data that have 20,000 rows and 20 columns. Io wonted to extract the 10th row only. Example the 10th, 20th, 30th 40th…..20000 th. can you please help me how do I do that.Than kyou. Example is below. Inpute: AG GG GG AG CC CC CC CC CT CC CT CT GG GG GG GG CC CC CC CC GG GG GG GG CC CC CC CC GG CG CG GG GG GG GG GG *CC CC CC CC* AA AG AG AA AA AA AA AA GG AG AG GG GG AG AG

Dear All I have a data.frame like that: structure(list(name = c("Gga_rs10722041", "Gga_rs10722249", "Gga_rs10722565", "Gga_rs10723082", "Gga_rs10723993", "Gga_rs10724555", "Gga_rs10726238", "Gga_rs10726461", "Gga_rs10726774", "Gga_rs10726967", "Gga_rs10727581", "Gga_rs10728004",

Hello! I am writing some numbers and character vectors to an ascii file and would like to get rid of [1] ... as shown bellow (a dummy example) R> runif(20) [1] 0.653574 0.164053 0.036031 0.127208 0.134274 0.103252 0.506480 0.547759 [9] 0.912421 0.584382 0.987208 0.996846 0.666760 0.053637 0.327590 0.370737 [17] 0.505706 0.412316 0.887421 0.812151 I have managed to work up to remove quotes

Hi, I am trying to split multiple columns. One column works just fine, but I want to do it for multiple columns??? Example > a ID V2 V3 V4 V5 V6 V7 V8 V9 V10 1 PBBA0644 -- GG AA -- AA -- AA GG GG 2 PBBA1010 -- GG AA -- AA -- AA GG GG 3 0127ATPR -- GG AA -- AA -- AA GG GG 4 0128EHAB -- GG AA -- AG -- AA AG GG 5 PBBA0829 -- GG AA -- AA -- AA GG AG

Dear list, I am trying to evaluate how many elements in a vector equal a certain value. The vectors are the columns of a data.frame, read in using read.table(): > dim(data) [1] 2600 742 > data[1:5,1:5] SNP001 SNP002 SNP003 SNP004 SNP005 1 GG AA TT TT GG 2 GG AA TC TT GG 3 GG AC CC TT GG 4 AG AA TT TT GG 5