similar to: Unexpected behavior in factor level ordering

Last Friday, I noticed that it is difficult to work with regression models in which there are factors. It is easier to do the old fashioned thing of coding up "dummy" variables with 0-1 values. The predict function's newdata argument is not suited to insertion of hypothetical values for the factor, whereas it has no trouble with numeric variables. For example, if one uses a

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

To extwnd on Martin 's explanation : In factor(), levels are the unique input values and labels the unique output values. So the function levels() actually displays the labels. Cheers Joris On 15 Jun 2017 17:15, "Martin Maechler" <maechler at stat.math.ethz.ch> wrote: >>>>> Paul Johnson <pauljohn32 at gmail.com> >>>>> on Wed, 14 Jun

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 ...

Suppose plotx <- "someName" modx <- "otherName" plotxRange <- c(10,20) modxVals <- c(1,2,3) It often happens I want to create a dataframe or object with plotx or modx as the variable names. But can't understand syntax to do that. I can get this done in 2 steps, creating the data frame and then assigning names, as in newdf <- data.frame( c(1, 2, 3, 4),

Dear R-helpers, I am stuck on an error in R: When I run my code (below), I get this error back: Error in names(x) <- value : 'names' attribute must be the same length as the vector Then when I use traceback(), R gives me back this in return: `colnames<-`(`*tmp*`, value = c(""Item", "Color" ,"Number", "Size")) I'm not exactly

How does one remove a column from a data frame when the name of the column to remove is stored in a variable? For Example: colname <- "LOT" newdf <- subset(olddf,select = - colname) The above statement will give an error, but thats what I'm trying to accomplish. If I had used: newdf <- subset(olddf,select = - LOT) then it would have worked, but as I said the column

I'm pasting below a working R file featuring a function I'd like to polish up. I'm teaching regression this semester and every time I come to something that is very difficult to explain in class, I try to simplify it by writing an R function (eventually into my package "rockchalk"). Students have a difficult time with predict and newdata objects, so right now I'm

Hi all, I still have problems with the predict function by setting up the values on which I want to predict ie: original df: p1 (193 obs) variates y x1 x2 rm(list=ls()) x1<-rnorm(193) x2<-runif(193,-5,5) y<-rnorm(193)+x1+x2 p1<-as.data.frame(cbind(y,x1,x2)) p1 y x1 x2 1 -0.6056448 -0.1113607 -0.5859728 2 -4.2841793 -1.0432688 -3.3116807 ...... 192

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

I have a data frame with 155,000 rows. One of the columns represents the user id (of which about 10,000 are unique). I am able to isolate 1000 of these user ids (stored in a list) that I want to eliminate from the data set, but I don't know of an efficient way to do this. Certainly this would be slow: newdf<-df for(i in listofbadusers) { newdf<-subset(tmp,uid!=i) } is there a better

Dear R-experts, Here below my R code with an error message. Can somebody help me to fix this error?? Really appreciate your help. Best, ############################################################ #?MSE CROSSVALIDATION Lasso regression? library(glmnet) ?

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

Dear R devel I've been wondering about this for a while. I am sorry to ask for your time, but can one of you help me understand this? This concerns duplicated labels, not levels, in the factor function. I think it is hard to understand that factor() fails, but levels() after does not > x <- 1:6 > xlevels <- 1:6 > xlabels <- c(1, NA, NA, 4, 4, 4) > y <- factor(x,