similar to: Strange results : bootrstrp CIs

Dear Duncan, Dear Ivan, I really thank you a lot for your response. So, if I correctly understand your answers the problem is coming from this line: coef(lm(Score~ Time + factor(Country)),data=data[idx,]) This line should be: coef(lm(Score~ Time + factor(Country),data=data[idx,])) If yes, now I get an error message (code here below)! So, it still does not work. Error in t.star[r, ] <-

It took me a little while to figure this out, but: the problem is that if your resampling leaves out any countries (which is very likely), your model applied to the bootstrapped data will have fewer coefficients than your original model. I tried this: cc <- unique(e$Country) func <- function(data, idx) { coef(lm(Score~ Time + factor(Country, levels =cc),data=data[idx,])) } but lm()

? Sat, 13 Jan 2024 20:33:47 +0000 (UTC) varin sacha via R-help <r-help at r-project.org> ?????: > coef(lm(Score~ Time + factor(Country)),data=data[idx,]) Wrong place for the data=... argument. You meant to give it to lm(...), but in the end it went to coef(...). Without the data=... argument, the formula passed to lm() picks up the global variables inherited by the func() closure.

Sorry, didn't cc this to the list. -------- Forwarded Message -------- Subject: Re: [R] Strange results : bootrstrp CIs Date: Sat, 13 Jan 2024 17:37:19 -0500 From: Duncan Murdoch <murdoch.duncan at gmail.com> To: varin sacha <varinsacha at yahoo.fr> You can debug things like this by setting options(error = recover). That will drop into the debugger when the error occurs.

Well, this would seem to work: e <- data.frame(Score = Score , Country = factor(Country) , Time = Time) ncountry <- nlevels(e$Country) func= function(dat,idx) { if(length(unique(dat[idx,'Country'])) < ncountry) NA else coef(lm(Score~ Time + Country,data = dat[idx,])) } B <- boot(e, func, R=1000) boot.ci(B, index=2, type="perc")

I am trying to install the package boss but i am getting error below: Please advice install.packages("boss") --- Please select a CRAN mirror for use in this session --- CRAN mirror 1: 0-Cloud 2: Argentina (La Plata) 3: Argentina (Mendoza) 4: Australia (Canberra) 5: Australia (Melbourne) 6: Austria 7: Belgium 8:

Dear list I am trying to install the systemfit package under unix, install.packages(systemfit) the installation failed. I am attaching the error and version information below, (if dependencies=TRUE, much more error) any help appreciated best, yong ===================================================================== > install.packages("systemfit") Warning in

Dear R, I'm trying to plot the first principle component of an analysis vs the first variable but am having trouble. I have no trouble doing the initial plot but have difficulty thereafter. First I want to highlight some points of the following data set list(running) [[1]] X100m X200m X400m X800m X1500m X5K X10K Marathon Argentina 10.39 20.81 46.84 1.81

Anyone able to get me a comp/highly discounted ticket to this? $150 just to visit the exhibition halls sounds crazy? Dean > -----Original Message----- > From: Jeff Pulver [mailto:jeff@pulver.com] > Sent: Tuesday, 23 August 2005 11:47 AM > To: mailinglist1 > Subject: Register Today for Fall 2005 VON: "The Destination for IP > Communications" > > Hi There, >

R-experts, I have fitted many different lines. The fast-tau estimator (yellow line) seems strange to me?because this yellow line is not at all in agreement with the other lines (reverse slope, I mean the yellow line has a positive slope and the other ones have negative slope). Is there something wrong in my R code ? Is it because the Y variable is 1 vector and should be a matrix ? Here is the

Dear R-experts, Here below my R code. The warning message is not a problem to me but there is an error message more problematic. I understand the error message but I don't know if it is possible to correct the error and if yes, how to correct it. Many thanks. n <- 60 b <- runif(n, 0, 5) a <- runif(n, 0, 5) z <- rnorm(n*0.95,2,3) + rnorm(n*0.05,2,9) y_model <- 0.1 * b - 0.5 *

At the cost of breaking the thread I'm going to change your subject and replace 'Principle' by 'Principal'. I just can't stand it any longer... OK, here is how I would solve your other problems. First put > wh <- c("USA", "New Zealand", "Dominican Republic", "Western Samoa", "Cook Islands") > ind

Hi Rui, Many thanks. The R code works BUT the results I get are quite weird I guess ! MIC = 0.2650 Normal 95% CI = (0.9614, 1.0398) The MIC is not inside the confidence intervals ! Is there something wrong in the R code ? Here is the reproducible example : ########## C=c(2,4,5,6,3,4,5,7,8,7,6,5,6,7,7,8,5,4,3,2) D=c(3,5,4,6,7,2,3,1,2,4,5,4,6,4,5,4,3,2,8,9) library(minerva) mine(C,D)$MIC

On Sat, 13 Jan 2024 17:59:16 -0500 Duncan Murdoch <murdoch.duncan at gmail.com> wrote: <SNIP> > My guess is that one of the bootstrap samples had a different > selection of countries, so factor(Country) had different levels, and > that would really mess things up. > > You'll need to decide how to handle that: If you are trying to > estimate the coefficient for

On 31/03/2018 11:57 AM, varin sacha via R-help wrote: > Dear R-experts, > > Here below my reproducible R code. I want to add many straight lines to a plot using "abline" > The last fit (fast Tau-estimator, color yellow) will not appear on the plot. What is going wrong ? > Many thanks for your reply. > It's not quite reproducible: you forgot the line to create

Dear R-experts, Here below the reproducible example. I am trying to get the average of the 100 results coming from the "lst" function. I have tried lst$mean and mean(lst). It does not work. Any help would be highly appreciated. #################### ?## R script for getting MedAe and MedAeSQ from HBR model on Testing data install.packages("robustbase") install.packages(

On 13/01/2024 8:58 p.m., Rolf Turner wrote: > On Sat, 13 Jan 2024 17:59:16 -0500 > Duncan Murdoch <murdoch.duncan at gmail.com> wrote: > > <SNIP> > >> My guess is that one of the bootstrap samples had a different >> selection of countries, so factor(Country) had different levels, and >> that would really mess things up. >> >> You'll

Is there any good free / accurate online resources with detailed country numbering plans? Failing that let's get something running ourselves. I was also thinking maybe people present could contribute some information on this list for now. The countries I am after are below. To start this off I will provide the information for Australia +61 and New Zealand +64. NZ Cellular: area code 21

Dear R-experts, Here below is my code, I would like to add a vertical line on my plot, showing the median and I would like to place some values on this graph as well, i.e. 4.3 and -8.4. How can I do ? Many thanks for your reply. A=c(1,2.3,4,3.5,4.3,2.5,6.3,-0.1,-1.5,3.7,-2.3,-3.5,5.4,3.2, -10.5,-8.4,-9.4) d <- density(A) plot(d) median(A)

You need to pay attention to the documentation more closely. If you don't know what something means, that is usually a signal that you need to study more... in this case about the difference between an input variable and a design (model) matrix. This is a concept from the standard linear algebra formulation for regression equations. (Note that I have never used RobPer, nor do I regularly