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) ?

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,

Hi, Some try: > names(mi$xlevels) [1] "f" > all.vars(mi$formula) [1] "D" "x" "f" "Y" > names(mx$xlevels) [1] "f" > all.vars(mx$formula) [1] "D" "x" "f" When offset is indicated out of the formula, it does not work... Marc Le 07/03/2018 ? 06:20, Bendix Carstensen a ?crit?: > I would like