similar to: running crossvalidation many times MSE for Lasso regression

No error message shown Please include the error message so that it is not necessary to rerun your code. This might enable someone to see the problem without running the code (e.g. downloading packages, etc.) -- Bert On Sun, Oct 22, 2023 at 1:36?PM varin sacha via R-help <r-help at r-project.org> wrote: > > Dear R-experts, > > Here below my R code with an error message. Can

?s 20:12 de 23/10/2023, varin sacha via R-help escreveu: > Dear R-experts, > > I really thank you all a lot for your responses. So, here is the error (and warning) messages at the end of my R code. > > Many thanks for your help. > > > Error in UseMethod("predict") : > ? no applicable method for 'predict' applied to an object of class

Dear R-experts, I really thank you all a lot for your responses. So, here is the error (and warning) messages at the end of my R code. Many thanks for your help. Error in UseMethod("predict") : ? no applicable method for 'predict' applied to an object of class "c('matrix', 'array', 'double', 'numeric')" > mean(unlist(lst)) [1] NA

For what it's worth it looks like spm2 is specifically for *spatial* predictive modeling; presumably its version of CV is doing something spatially aware. I agree that glmnet is old and reliable. One might want to use a tidymodels wrapper to create pipelines where you can more easily switch among predictive algorithms (see the `parsnip` package), but otherwise sticking to glmnet

I forgot, you should also set.seed() before calling boot() to make the results reproducible. Rui Barradas On 5/22/2018 10:00 AM, Rui Barradas wrote: > Hello, > > If you want to bootstrap a statistic, I suggest you use base package boot. > You would need the data in a data.frame, see how you could do it. > > > library(boot) > > bootMedianSE <- function(data,

Hello, Right! I copied from the OP's question without thinking about it. Corrected would be bootMedianSE <- function(data, indices){ d <- data[indices, ] fit <- rq(crp ~ bmi + glucose, tau = 0.5, data = d) ypred <- predict(fit) y <- d$crp median((y - ypred)^2) } Sorry, rui Barradas On 5/22/2018 11:32 AM, Daniel Nordlund wrote: > On 5/22/2018

On 5/22/2018 2:32 AM, Rui Barradas wrote: > bootMedianSE <- function(data, indices){ > ???? d <- data[indices, ] > ???? fit <- rq(crp ~ bmi + glucose, tau = 0.5, data = d) > ???? ypred <- predict(fit) > ???? y <- d$crp > ???? median(y - ypred)^2 > } since the OP is looking for the "median squared error", shouldn't the final line of the

I realize this may be more of a math question. I have the following optim: optim(c(0.0,1.0),logis.op,x=d1_all$SOA,y=as.numeric(md1[,i])) which uses the following function: logis.op <- function(p,x,y) { ypred <- 1.0 / (1.0 + exp((p[1] - x) / p[2])); res <- sum((y-ypred)^2) return(res) } I would like to add weights to the optim. Do I have to alter the logis.op function by

Hello, If you want to bootstrap a statistic, I suggest you use base package boot. You would need the data in a data.frame, see how you could do it. library(boot) bootMedianSE <- function(data, indices){ d <- data[indices, ] fit <- rq(crp ~ bmi + glucose, tau = 0.5, data = d) ypred <- predict(fit) y <- d$crp median(y - ypred)^2 } dat <-

Dear R-experts, I am trying to bootstrap (and average) the median squared error evaluation metric for a robust regression. I can't get it. What is going wrong ? Here is the reproducible example. ############################# install.packages( "quantreg" ) library(quantreg) crp <-c(12,14,13,24,25,34,45,56,25,34,47,44,35,24,53,44,55,46,36,67) bmi

DeaR list, I'm trying to represent some information via 3D plots. My data and session info are at the end of this message. So far, I have tried scatterplot3d (scatterplot3d), persp3d (rgl), persp (graphics) and scatter3d (Rmcdr) but any of them gave me what I'd like to have as final result (please see [1] for a similar 3D plot changing PF by ypred, pdn by h4 and pup by h11). In general

Hi, forget about the below details. It is not related to the fact that the function is returned from a function. Sorry about that. I've been troubleshooting soo much I've been shoting over the target. Here is a much smaller reproducible example: x <- 1:10 y <- 1:10 + rnorm(length(x)) sp <- smooth.spline(x=x, y=y) ypred <- predict(sp$fit, x) # [1] 2.325181 2.756166 ...

Dear R-help, I have fitted a glm logistic function to dichotomous forced choices responses varying according to time interval between two stimulus. x values are time separation in miliseconds, and the y values are proportion responses for one of the stimulus. Now I am trying to extrapolate x values for the y value (proportion) at .25, .5, and .75. I have tried several predict parameters, and they

Hello, Everybody: This may not be a "bug", but for me it is an unexpected outcome. A factor variable's levels do not retain their ordering after the levels function is used. I supply an example in which a factor with values "BC" "AD" (in that order) is unintentionally re-alphabetized by the levels function. To me, this is very bad behavior. Would you agree? #

Hello, I am using logistic discriminant analysis to check whether a known classification Yobs can be predicted by few continuous variables X. What I do is to predict class probabilities with multinom() in nnet(), obtaining a predicted classification Ypred and then compute the percentage P(obs) of objects classified the same in Yobs and Ypred. My problem now is to figure out whether P(obs) is

Hi all, I'm relatively new to using R, and have been trying to fit an L1 regularization path using coxpath from the glmpath library. I'm interested in using a cross validation framework, where I crossvalidate on a training set to select the lambda that achieves the lowest error, then use that value of lambda on the entire training set, before applying to a test set. This seems to entail

Full_Name: Noel O'Boyle Version: 2.1.0 OS: Debian GNU/Linux Sarge Submission from: (NULL) (131.111.8.96) (1) Description of error The 10-fold CV option for the svm function in e1071 appears to give incorrect results for the rmse. The example code in (3) uses the example regression data in the svm documentation. The rmse for internal prediction is 0.24. It is expected the 10-fold CV rmse

1. This is _not_ a bug in R itself. Please don't use R's bug reporting system for contributed packages. 2. This is _not_ a bug in svm() in `e1071'. I believe you forgot to take sqrt. 3. You really should use the `tot.MSE' component rather than the mean of the `MSE' component, but this is only a very small difference. So, instead of spread[i] <- mean(mysvm$MSE), you

Dear R-experts, Doing cross-validation for 2 robust regressions (HBR and fast Tau). I can't get the 2 errors rates (RMSE and MAPE). The problem is to predict the response on the testing data. I get 2 error messages. Here below the reproducible (fictional example) R code. #install.packages("MLmetrics") # install.packages( "robustbase" ) # install.packages(

Folks; I am having a problem with the cv.glm and would appreciate someone shedding some light here. It seems obvious but I cannot get it. I did read the manual, but I could not get more insight. This is a database containing 3363 records and I am trying a cross-validation to understand the process. When using the cv.glm, code below, I get mean of perr1 of 0.2336 and SD of 0.000139. When using a