similar to: How to compare areas under ROC curves calculated with ROCR package

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 I am trying to compare the performances of several methods using the AUC0.1 and not the whole AUC. (meaning I wanted to compare to AUC's whose x axis only goes to 0.1 not 1) I came to know about the Mcneil Hanley test from Bernardo Rangel Tura and I referred to the original paper for the calculation of "r" which is an argument of the function cROC. I can only find the

Please help. I found the code in the archive. The author of this script says: "The first function (seROC) calculate the standard error of ROC curve, the second function (cROC) compare ROC curves." Can some one explain to me what are the na, nn and r parameters which are used as the input to the following two functions? Thanks much in advance. > From: Bernardo Rangel Tura >

Dear R-helper, I would like to compare the AUC of two logistic regression models (same population). Is it possible with R ? Thank you Roman Rouzier [[alternative HTML version deleted]]

Hi, I am quite new to R and I have two questions regarding ROCR. 1. I have tried to understand how to extract area-under-curve value by looking at the ROCR document and googling. Still I am not sure if I am doing the right thing. Here is my code, is "auc1" the auc value? " pred1 <- prediction(resp1,label1) perf1 <- performance(pred1,"tpr","fpr") plot(

Guys, I used Random Forest with a couple of data sets I had to predict for binary response. In all the cases, the AUC of the training set is coming to be 1. Is this always the case with random forests? Can someone please clarify this? I have given a simple example, first using logistic regression and then using random forests to explain the problem. AUC of the random forest is coming out to be

Here is a snippet to show what i'm trying to do. library(randomForest) library(ROCR) library(caret) data(iris) iris <- iris[(iris$Species != "setosa"),] fit <- randomForest(factor(Species) ~ ., data=iris, ntree=50) train.predict <- predict(fit,iris,type="prob")[,2]

Hi, I recently bought a Nitram Elite 2005 UPS and i installed nut to monitor it. Unfortunatly, i often have timeout errors (20 timeout per days): Broadcast Message from aurelien at berkelium (no tty) at 18:05 CET... Communications with UPS Nitram at localhost lost Broadcast Message from aurelien at berkelium (no tty) at 18:05 CET... Communications with UPS Nitram(AT)localhost

Hi, I recently bought a Nitram Elite 2005 UPS and i installed nut to monitor it. Unfortunatly, i often have timeout errors (20 timeout per days): Broadcast Message from aurelien at berkelium (no tty) at 18:05 CET... Communications with UPS Nitram at localhost lost Broadcast Message from aurelien at berkelium (no tty) at 18:05 CET... Communications with UPS Nitram(AT)localhost

Dear sirs, I have tried to find a command to fit(plot) a regression line but I haven't found any command on the html.help. My question is if I am using a regression for instance reg<-lm(y~x) and I would like to make a regression line, to see how well my data is fit. Can you please help me with this?? Yours Sincerely Soheila Zad

Hi everybody, I recently bought a UPS Cyber Power DX600E ad I'm trying to manage it through Nut under FreeBSD. Unfortunately, usbhid-ups returns an error when I do: [root at berkelium /usr/local/etc/nut]# /usr/local/libexec/nut/usbhid-ups -u root -DD -a onduleur Network UPS Tools - Generic HID driver 0.34 (2.4.1) USB communication driver 0.31 debug level is '2'

Pleased to announce that the next version (2.3.0) of our package ?concurve? is out on CRAN (https://cran.r-project.org/package=concurve). In addition to plotting confidence (consonance) curves, it can now plot two functions next to one another to see the amount of overlap. It can also plot likelihood and deviance functions along with consonance densities and distributions. And it can utilize

Pleased to announce that the next version (2.3.0) of our package ?concurve? is out on CRAN (https://cran.r-project.org/package=concurve). In addition to plotting confidence (consonance) curves, it can now plot two functions next to one another to see the amount of overlap. It can also plot likelihood and deviance functions along with consonance densities and distributions. And it can utilize

I am a novice user of R and am stumbling on how to order a dataset produced during my session. I have a 1863 row X 14 column dataset that I want to put out to a file. I want the output sorted by the first column and then by the second column both in ascending order. The first column is character and the second is numeric (I hope). I used an "as.numeric" function to assign that

The Wine maintenance release 5.0.1 is now available. What's new in this release (see below for details): - Fix compilation with gcc 10 - Add some timezones - Various bug fixes The source is available from the following locations: https://dl.winehq.org/wine/source/5.0/wine-5.0.1.tar.xz http://mirrors.ibiblio.org/wine/source/5.0/wine-5.0.1.tar.xz Binary packages for various

Dear all, I am looking at habituation of dogs trotting on a treadmill. I record the ground reaction force and I analyze it with several discrete variables (maximum, minimum,...) For each variable, I get between 40 and 50 data per sample. I record data at time 1 min, 2 min, and 4 min a day, and I have 4 days of measurement (one day a week). That means I have 12 samples : Day1_Min1, Day1_Min2,

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Dear R-gurus, I have an interpretation problem regarding lme models. I am currently working on dog locomotion, particularly on some variation factors. I try to figure out which limb out of 2 generated more dispersed data. I record a value called Peak, around 20 times for each limb with a record. I repeat the records during a single day, and on several days. I tried to build two models, one

If you used all 28 animals to find the model out of a group of candidate models, I would have my reservations about this 'validation'. Any confidence intervals you get from the final model are bound to be overly optimistic because you haven't accounted for the degrees of freedom chewed up during the model fitting/finding process. Matt Austin Statistician Amgen One Amgen Center

The Wine development release 5.3 is now available. What's new in this release (see below for details): - More work towards Ucrtbase runtime support. - Full support for Unicode normalization. - Improvements in Shell Folders handling. - Various bug fixes. The source is available from the following locations: https://dl.winehq.org/wine/source/5.x/wine-5.3.tar.xz