similar to: ROC plot for KNN

Hello. I found the behavior of knn( http://stat.ethz.ch/R-manual/R-devel/library/class/html/knn.html) function looking very strange. Consider the toy example. > library(class) > train <- matrix(nrow=5000,ncol=2,data=rnorm(10000,0,1)) > test <- matrix(nrow=10,ncol=2,data=rnorm(20,0,1)) > cl <- rep(c(0,1),2500) > knn(train,test,cl,1) [1] 1 1 0 0 1 0 1 1 0 1 Levels: 0 1 It

Hi, I was trying to get the optimal 'k' for the knn. To do this I was using the following function : knn.cvk <- function(datmat, cl, k = 2:9) { datmatT <- (datmat) cv.err <- cl.pred <- c() for (i in k) { newpre <- as.vector(knn.cv(datmatT, cl, k = i)) cl.pred <- cbind(cl.pred, newpre) cv.err <- c(cv.err, sum(cl != newpre)) }

Hello, I am feeling that this question can have a very simple answer, but I can't find it. I need to use the function knn.dist from knnflex library. Whatever I try, I get the error: Error in as.vector.dist(x, "character") : unused argument(s) ("character") First example: > a<-NULL > a<-rbind(a,c(5.2,-8.1)) > a<-rbind(a,c(8.8,-16.1)) >

ear all, This is my first time using this listserv and I am seeking help from the expert. OK, here is my question, I am trying to use impute.knn function in impute library and when I tested the sample code, I got the error as followingt: Here is the sample code: library(impute) data(khanmiss) khan.expr <- khanmiss[-1, -(1:2)] ## ## First example ## if(exists(".Random.seed"))

hi, r users i have a problem with KNN. i have 2 datasets, X0 and X1. >dim(X0) >1471*13 dim(X1) >5221*13 and for every instances in the dataset X1, i want to find the nearest neighbour(1nn) in the dataset X0. and i dont have the true classifications of dataset X1. but the function knn() need true classifications(cl) to do prediction. i just curious if there are some other function

The function knn.cv in the class package doesn't have error checking to ensure that the length of the classlabel argument is equal to the number of rows in the test set. If the classlabel is short, the result is often a segfault. > library(class) > dat <- matrix(rnorm(1000), nrow=10) > cl <- c(rep(1,5), rep(2,5)) > cl2 <- c(rep(1,5), rep(2,4)) > knn.cv(dat, cl) [1] 2

I am trying to use knn to do a nearest neighbor classification. I tried using my dataset and got an error message so I used a simple example to try and understand what I was doing wrong and got the same message. Here is what I typed into R: try [,1] [,2] [,3] [,4] r "A" "A" "T" "G" r "A" "A" "T" "G" f

Hello, I want to do regression or missing value imputation by knn. I searched r-help mailing list. This question was asked in 2005. ksmooth and loess were recommended. But my case is different. I have many predictors (p>20) and I really want try knn with a given k. ksmooth and loess use band width to define neighborhood size. This contrasts to knn's variable band width via fixing a

Hi Everybody, i again have the problem in using tune.knn(), its giving an error saying missing values are not allowed.... again here is the script for BreastCancer Data, library(e1071) library(mda) trdata<-data.frame(train,row.names=NULL) attach(trdata) xtr <- subset(trdata, select = -Class) ytr <- Class bestpara <-tune.knn(xtr,ytr, k = 1:25, tunecontrol = tune.control(sampling

hi, i am getting an error when calling the impute.knn function (see the screenshot below). what is the problem here and how can it be solved? screenshot: ################## *** caught segfault *** address 0x513c7b84, cause 'memory not mapped' Traceback: 1: .Fortran("knnimp", x, ximp = x, p, n, imiss = imiss, irmiss, as.integer(k), double(p), double(n), integer(p),

Hi, I'm involved in a bioinformatics project at my university, and we're doing a comparison paper between some methods of classification of nc-RNA. I've been encharged of ploting the ROC curves' graphs. But I'm new on working with R and I'm having some difficulty with the prediction-class. I don't get where the values of ROCR.simple$predictions, for example, came from

How can I do a simple k nearest neighbor regression in R? My training data have 1 predictor and 1 outcome, both are numeric. I also need to use FPE and SC to find the optimal model. I know there is knn() in class package, but it's for knn classification. I also find a kknn package. What function should I use? Thanks in advance! Menghui

Could anybody help me out please? > cl<-as.factor(traindata[,13]) > knn(traindata[1:295,2], newdata[1:32,2], cl,k=2, prob=TRUE) Error in knn(traindata[1:295, 2], newdata[1:32, 2], cl, k = m, prob = TRUE) : Dims of test and train differ Both traindata and newdata have 13 elements. Only one of the first 12 elemnets is needed to predict the 13 element. What's the problem of

Hello all, am a newby in R, am trying the knn function, and am doing just a stupid test : > knn(c(1,2,3,4,5,6), c(3), k=4 ,prob=TRUE,factor(c(1:6))) the result is unstable !! i have each time different result : [1] 5 attr(,"prob") [1] 0.1666667 Levels: 1 2 3 4 5 6 [1] 4 attr(,"prob") [1] 0.1666667 Levels: 1 2 3 4 5 6 [1] 1 attr(,"prob") [1] 0.1666667 Levels:

The name conflicts in Epi and ROC packages (2 'ROC' functions are the problem) cause the following code to work once, but not twice: library(MASS); data(cats); x = cats[,2] y = ifelse(cats[,1]=='F',0,1) library(Epi); ROC(x,y,grid=0)$AUC library(ROC); AUC(rocdemo.sca(y, x, dxrule.sca)) What is the standard way of resolving name conflicts? Ask maintainers to resolve

Maybe you should show your colleague how to access help pages in R? Right in ?knn, it says: prob: If this is true, the proportion of the votes for the winning class are returned as attribute 'prob'. so 1.0 mean all three NNs are of the `winning'; i.e., predicted, class, and 0.66667 means 2 out of the 3 NNs are of the winning class, etc. Andy > From: Ko-Kang

Hello I am trying to make a graph of 10 different lines built each from 4 different segments and to add a darker line that will represent the average of all graphs - all in the same plot.Actually each line is a ROC plot The code I'm using for plotting one line is as follows: logit.roc.plot <- function(r, title="ROC curve") { old.par <- par(no.readonly = TRUE);

I might be missing something but I thought that AUC was a measure for comparing ROC curves, so there is nothing else needed to "compare" them. The larger AUC is the higher correlation of 2 variables compared. No other measures or calculations are needed. Jarek Tuszynski -----Original Message----- From: r-help-bounces at stat.math.ethz.ch [mailto:r-help-bounces at stat.math.ethz.ch] On

Dear all, Some functions like 'ROC(Epi)' can be used to perform ROC analyssi, but it needs us to specify the fitting model in the argument. Now i have got the predicted p-values (0,1) for the 0/1 response variable using some other approach, see the following example dataset: id mark predict.pvalue 1 1 0.927 2 0 0.928 3 1 0.928 ..................

Hi, I was lately debugging parts of my 'colAUC' function in caTools package, and in a process looked into other packages for calculating Areas Under ROC Curves (AUC). To my surprise I found at least 6 other functions: * wilcox.test * AUC from ROC package, * performance from ROCR package, * auROC from limma package, * ROC from Epi package, * roc.area from verification