similar to: Cross-validation question

The function crossval (in library bootstrap) works well for the first degree polynomial model, but in the case of the second degree model I got an error message (see below). I would be very greatfull if somebody could give some advices for the following: > library(bootstrap) > > x<-c(22,23.4,24.9,28.5,29.8,31.6,34.2,36.4,37.7,39) >

I am trying to understand what is the difference between linear and polynomial kernel: linear: u'*v polynomial: (gamma*u'*v + coef0)^degree It would seem that polynomial kernel with gamma = 1; coef0 = 0 and degree = 1 should be identical to linear kernel, however it gives me significantly different results for very simple data set, with linear kernel

Hey all, first time poster here. I'm new to R and working on my first real programming and forecasting asignment. I'm using unemployment data from 1948-2012. I successfully completed part a and the linear fit for part b, but i am really struggling fitting a polynomial with a power greater than 2 to my forecast. I'll upload my R code at the bottom. Any help is very much appreciated!

Hi, everyone: I ask for help about translating a stata program into R. The program perform cross validation as it stated. #1. Randomly divide the data set into 10 sets of equal size, ensuring equal numbers of events in each set #2. Fit the model leaving out the 1st set #3. Apply the fitted model in (2) to the 1st set to obtain the predicted probability of a prostate cancer diagnosis. #4. Repeat

Dear R users: Recently, I tried to write a program to calculate cross-validated predicted value. My sources are as follows. However, the R reported an error. Could you please check the sources? Thanks. set.seed(100) x<-rnorm(100) y<-sample(rep(0:1,50),replace=T) dat<-data.frame(x,y) library(rms) fito<-lrm(y~x) preo<-predict(fito) pre<-matrix(NA,nrow=100,ncol=200) for (i in

> On Jul 12, 2017, at 6:58 PM, Ng, Kelvin Sai-cheong <kscng at connect.hku.hk> wrote: > > Dear all, > > I am using the pls package of R to perform partial least square on a set of > multivariate data. Instead of fitting a linear model, I want to fit my > data with a quadratic function with interaction terms. But I am not sure > how. I will use an example to

I am accustomed to the LIBSVM package, which provides cross-validation on training with the -v option % svm-train -v 5 ... This does 5 fold cross validation while building the model and avoids over-fitting. But I don't see how to accomplish that in the e1071 package. (I learned that svm(... cross=5 ...) only _tests_ using cross-validation -- it doesn't affect the training.) Can

?? If I haven't misunderstood, they are completely different! 1) NIR must be a matrix, or poly(NIR,...) will fail. 2) Due to the previously identified bug in poly, degree must be explicitly given as poly(NIR, degree =2,raw = TRUE). Now consider the following example: > df <-matrix(runif(60),ncol=3) > y <- runif(20) > mdl1 <-lm(y~df*I(df^2)) > mdl2

Hi All, I have been trying to get this LOO-Cross Validation method to work on R for the past 3 weeks but have had no luck. I am hoping someone would kindly help me. Essentially I want to code the LOO-Cross Validation for the 'Local Constant' and 'Local Linear' constant estimators. I want to find optimal h, bandwidth. Thank you very much! -M -- View this message in context:

Hi, I ran two svm models in R e1071 package: the first without cross-validation and the second with 10-fold cross-validation. I used the following syntax: #Model 1: Without cross-validation: > svm.model <- svm(Response ~ ., data=data.df, type="C-classification", > kernel="linear", cost=1) > predict <- fitted(svm.model) > cm <- table(predict,

Hi, I ran two svm models in R e1071 package: the first without cross-validation and the second with 10-fold cross-validation. I used the following syntax: #Model 1: Without cross-validation: > svm.model <- svm(Response ~ ., data=data.df, type="C-classification", > kernel="linear", cost=1) > predict <- fitted(svm.model) > cm <- table(predict,

Dear list, I am getting weired with fitting data with a 1/x-polynomial. Suggest I have the following data: x <- c(1,2,3,4,5,6,7) y <- c(100,20,4,2,1,.3,.1) I may fit this with a linear model fit1 = lm(y ~ I(x)) Getting plot out of this model I applied library(polynom) pol1 = polynomial(fit1$coefficients) f1 = as.function(pol1) plot(x,y) lines(x, f1(x), col = 2) Clearly, this model

Hi I am trying to validate my regression results using the leave one out crosss validation method. Is any script available in R to use this method for a linear regression equation? Both R or SPLUS would do. any clues on how to write the script will also help. Thanks a lot, A [[alternative HTML version deleted]]

Below. -- Bert Bert Gunter On Thu, Jul 13, 2017 at 3:07 AM, Luigi Biagini <luigi.biagini at gmail.com> wrote: > I have two ideas about it. > > 1- > i) Entering variables in quadratic form is done with the command I > (variable ^ 2) - > plsr (octane ~ NIR + I (nir ^ 2), ncomp = 10, data = gasTrain, validation = > "LOO" > You could also use a new variable

Hi, all When I am using mle.cv(wle), I find a interesting problem: I can't do leave-one-out cross-validation with mle.cv(wle). I will illustrate the problem as following: > xx=matrix(rnorm(20*3),ncol=3) > bb=c(1,2,0) > yy=xx%*%bb+rnorm(20,0,0.001)+0 > summary(mle.cv(yy~xx,split=nrow(xx)-1,monte.carlo=2*nrow(xx),verbose=T), num.max=1)[[1]] mle.cv: dimension of the split subsample

Dear R users, I know cross-validation does not work in rpart with user defined split functions. As Terry Therneau suggested, one can use the xpred.rpart function and then summarize the matrix of the predicted values into a single "goodness" value. I need only a confirmation: set for example xval=10, if I correctly understood a single column of the matrix obatined by xpred.rpart gives

<< repost because previous attempt was not plain text, sorry! >> Hi Folks, I have a pretty simple problem: after building a multivariate linear model, I need to report my 95% confidence interval for predictions based on future observations. I know that one option is to use predict(interval="prediction") but I'm curious about less parametric ways to get an estimate. I

Hi Bert, Ok, to your initial point, the key nuance is that if 'x' is a vector, you can leave the 'degree' argument unnamed, however, if 'x' is a matrix, you cannot. That aspect of the behavior does not seem to change if poly() is called stand alone or, as suggested in ?poly, within a formula to be parsed. Working on tracing through the code using debug(), the error is

> On Jul 13, 2017, at 10:43 AM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > > poly(NIR, degree = 2) will work if NIR is a matrix, not a data.frame. > The degree argument apparently *must* be explicitly named if NIR is > not a numeric vector. AFAICS, this is unclear or unstated in ?poly. I still get the same error with: library(pld) data(gasoline) gasTrain <-

Hi Everyone, I want to generate some random numbers according to some empirical distribution. Therefore I am looking for the inverse of an empirical cumulative distribution function. I haven't found any in R. Can anyone give a pointer? Thanks, Geoffrey _______________________________________________________=0A= =0A= =0A= The information in this email or in any file attached