[R] SVM

Hello, I have 12 sample each sample has got 1000 observation, i.e I have a matrix X with 1000 rows and 12 columns! m <- svm(t(X)) p <- predict (m) Can anyone tell me how to use svmtrain() in R! Many Yhanks, Samuel [[alternative HTML version deleted]] ___

Re: [R] SVM

wrote: From: Steve Lianoglou Subject: Re: [R] SVM To: "Samuel Okoye" Cc: r-h...@stat.math.ethz.ch Date: Thursday, September 17, 2009, 8:36 AM Hi, On Sep 17, 2009, at 7:39 AM, Samuel Okoye wrote: > Hello, > > I have 12 sample each sample has got 1000 observation, i.e I have

Re: [R] SVM

TRUE samples in group A and FALSE samples in group B? Best wishes, Samuel PS: I will not give up to understand statistics! --- On Thu, 9/17/09, Steve Lianoglou wrote: From: Steve Lianoglou Subject: Re: [R] SVM To: "Samuel Okoye" Cc: r-h...@stat.math.ethz.ch Date: Thursday, September 1

[R] simulation of censoring data

Dear all, I would like to understand the effect of censoring in a linear model therefore I am doing the following    n <- 20  cen <- sample(x=c(0,1), size=n-1, replace=TRUE, prob=c(0.8,0.2))  cen <- c(cen,0)  x <- rnorm(n,0,1)  tt <- -1+x+log(-log(1-runif(n)))+0.5767  y <- c()  for(i in 1:n){    

[R] variation in one variable

/09, Samuel Okoye wrote: From: Samuel Okoye Subject: SVM To: r-h...@stat.math.ethz.ch Date: Thursday, September 17, 2009, 4:39 AM Hello, I have 12 sample each sample has got 1000 observation, i.e I have a matrix X with 1000 rows and 12 columns! m <- svm(t(X)) p <- predict (m) Can anyon

[R] lasso cross-validation

Dear all, I am using cv.glmnet in r and I have the following question: The default is 10-fold cross-validation, but it is not clear to me how many times are repeated? Is it 50 repeats? I am sore if me question will be very easy for some people! Many thanks in advance, Samuel [[altern

[R] glm

Hi, I have the following data data1 <- data.frame(count = c(0,1,1,2,4,5,13,16,14), weeks = 1:9,     treat=c(rep("1mg",3),rep("5mg",3),rep("10mg",3))) and I am using library(splines) to fit glm.m <- glm(count~bs(weeks)+as.factor(treat),family=poisson,data=data1) and I am inte

Re: [R] glm

Tue, 22/6/10, Joris Meys wrote: From: Joris Meys Subject: Re: [R] glm To: "Samuel Okoye" Cc: r-help@r-project.org Date: Tuesday, 22 June, 2010, 9:50 On Tue, Jun 22, 2010 at 1:00 AM, Samuel Okoye wrote: > Hi, > > I have the following data > > data1 <- data.frame

[R] UniCox in R

Hello, I am interested in Figure 2 in http://www-stat.stanford.edu/~tibs/ftp/cus.pdf Can anyone tell please how to create this plot? Many thanks Samuel [[alternative HTML version deleted]] __ R-help@r-project.org mailing list

Re: [R] UniCox in R

holtman wrote: From: jim holtman Subject: Re: [R] UniCox in R To: "Samuel Okoye" Cc: r-help@r-project.org Date: Wednesday, 8 December, 2010, 5:25 Do a little reading on how to use the graphics commands.  I would look at plot lines segments a combination of those will let you easily

[R] AUC

Hello, Is there is any R function computes the AUC for paired data? Many thanks, Samuel [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting

[R] Problem with library(SSPA)

Hello, I have the fellowing problem and I am thankful for any advice! Regards, Samuel >   source("http://bioconductor.org/biocLite.R";) BioC_mirror = http://www.bioconductor.org Change using chooseBioCmirror(). > biocLite("SSP

Re: [R] Problem with library(SSPA)

I can't find his email and I have asked the same question to bioconduc...@stat.math.ethz.ch Regards, Samuel --- On Wed, 9/6/10, Uwe Ligges wrote: From: Uwe Ligges Subject: Re: [R] Problem with library(SSPA) To: "Samuel Okoye" Cc: r-help@r-project.org Date: Wednesday, 9 J

Re: [R] Problem with library(SSPA)

Dear Maarten, I have problem to use your package and I would be very thankful if you could help me to solve this. Regards, Samuel --- On Wed, 9/6/10, Uwe Ligges wrote: From: Uwe Ligges Subject: Re: [R] Problem with library(SSPA) To: "Samuel Okoye" Cc: r-help@r-project.org Date:

[R] Power calculation

Hello, Is there any R function which does power calculation for unbalanced groups (n1 neq n2)? Since power.t.test has n Number of observations (per group). Many thanks, Samuel [[alternative HTML version deleted]] __ R-help@r-project

[R] glm

Dear all, I am using glm with quasibinomial. What does the following error message mean: Error in eval(expr, envir, enclos) : y values must be 0 <= y <= 1 Does it mean that the predictor variable should only have zero and one or it is also possible to have continuous values between zero and one

[R] SAS data

Hello, I am trying to read the SAS file MyData.sa7bdat in R! This file is saved under D:\data! I therefore wrote > path <-"D:/SasData" > sashome <- "C/Progra, Files/SAS Institute/9_1/SAS" > sascmd <- file.path(sashome, "sas.exe") > MyData <- read.ssd(path, "MyData", sascmd=sascm

[R] ANCOVA

Hello,   Could you please help me in the following question: I have 16 persons 6 take 0.5 mg, 6 take 0.75 mg and 4 take placebo! Can I use the ANCOVA and t-test in this case? Is it possible in R?   Thank you in advance, Samuel [[alternative HTML version deleted]] ___

[R] date

Hello, I have got the following problem: > setwd("C:/temp") > library(xlsReadWrite) > MyData <- read.xls(file="Mappe1.xls", colNames = TRUE,dateTimeAs = > "isodatetime") > attach(MyData) > MyData name value times 1 A1 2 2006-05-12 2 A2 3 2006-05-16 3 A3

[R] plot or boxplot!

Hello, if we suppose that times <- c("2006-05-14", "2006-06-12", "2006-06-12", "2006-05-14", "2006-05-14", "2006-06-12") value <- c(2,3,1,4,3,1) then with plot(times, value) we have two boxplots in one graph for 2006-05-14 and 2006-06-12 respectively! Is it possibl

[R] sort time

Hello, I have got the following problem: > times <- c("02.07.2007", "03.07.2007","03.09.2007", "04.07.2007","05.07.2007") > mode(times) [1] "numeric" > tim <- as.character(times) > mode(tim) [1] "character" > sort(times) [1] "02.07.2007" "03.07.2007" "03.09.2007" "04.07.2007" "05.07.2007" I

[R] problem with times

Hello, I have got the following problem: > times <- c("02.07.2007", "03.07.2007","03.09.2007", "04.07.2007","05.07.2007") > mode(times) [1] "numeric" > tim <- as.character(times) > mode(tim) [1] "character" > sort(times) [1] "02.07.2007" "03.07.2007" "03.09.2007" "04.07.2007" "05.07.2007" Is it p

[R] y_hat

Hello, suppose one has the following values x1 <- rnorm(10,5,1) x2 <- rgamma(10,5,1) y <- rnorm(10,4,1) mydat <- data.frame(y,x1,x2) then one can use glm like mod <- glm(y~x1+x2, data=mydat, family=gaussian) But how could I estimate y_hat? Thanks alot! Sam --