similar to: R-help Digest, Vol 250, Issue 13

Yep, that did it. I didn't know that you could have pipelines within pipelines. Thanks, again, for all your help. -Kevin On 5/16/23 11:44, Rui Barradas wrote: > ?s 15:29 de 16/05/2023, Kevin Zembower via R-help escreveu: >> Hello, >> >> I's still working with my tsibble of weight data for the last 20 years. >> In addition to drawing an overall trend line,

Rui, thanks so much for your help. Your explanation and example were clear and concise. Thanks for taking the time and effort to help me. -Kevin On 5/12/23 16:06, Rui Barradas wrote: > ?s 14:24 de 12/05/2023, Kevin Zembower via R-help escreveu: >> Hello, I'm trying to create a line graph with a legend, but have no >> success controlling the legend. Since nothing I've

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.

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

Hello, While exploring if rgl is along the lines of what I need, I checked out demo("rgl") and didn't find quite what I'm looking for, and am therefore seeking additional help/suggestions. The application is geared towards displaying a 3D rendering of a contaminant plume in the subsurface with the following highlights: Once the plume was rendered as a 3D object, a

Hello, I's still working with my tsibble of weight data for the last 20 years. In addition to drawing an overall trend line, using lm, for the whole data set, I'd like to draw short lines that would recompute lm and draw it, say, just for the years from 2010:2015. Here's a short example that I think illustrates what I'm trying to do. The commented out sections show what

You need: myCor <- function(data, index){ mine(data[index, ])$MIC[1, 2] } results=boot(data = cbind(C,D), statistic = myCor, R = 2000) boot.ci(results,type="all") Look at the differences between: mine(C, D) and mine(cbind(C, D)) The first returns a value, the second returns a symmetric matrix. Just like cor() David L. Carlson Department of Anthropology Texas A&M

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()

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")

Dear Sacha, use glm() in this case. I'd rather code the covariable as TRUE / FALSE or as a factor. Best regards, ir. Thierry Onkelinx Statisticus / Statistician Vlaamse Overheid / Government of Flanders INSTITUUT VOOR NATUUR- EN BOSONDERZOEK / RESEARCH INSTITUTE FOR NATURE AND FOREST Team Biometrie & Kwaliteitszorg / Team Biometrics & Quality Assurance thierry.onkelinx at inbo.be

Varin In the teaching you to fish department, you can find those dependencies Duncan appeared to pull out of thin air by looking at the CRAN contributed packages web pages, which can easily be found with Google. Start at the matie page and follow the dependency links and look at dependencies. In particular the error message mentioning the proxy package is a clue that dependencies might be a

Hello, After the plot just do abline(v = median(A)) As for how to plot points, see, well, ?points(). Hope this helps, Rui Barradas Em 22-10-2017 16:33, varin sacha via R-help escreveu: > 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

? 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.

Hi, a is of length 60. b is of length 60. z is of length 57. What do you expect to have happen when you create y_model ? What happens to those other 3 observations? Sarah On Tue, Oct 27, 2020 at 3:07 PM varin sacha via R-help <r-help at r-project.org> wrote: > > Dear R-experts, > > Here below my R code. The warning message is not a problem to me but there is an error message

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

Hello, I'm trying to create a line graph with a legend, but have no success controlling the legend. Since nothing I've tried seems to work, I must be doing something systematically wrong. Can anyone point this out to me? Here's my data: > weights # A tibble: 1,246 ? 3 Date J K <date> <dbl> <dbl> 1 2000-02-13 133 188 2

On 5/8/2018 12:26 PM, varin sacha via R-help wrote: > > 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

On Mon, 4 Apr 2016, varin sacha via R-help wrote: > Hi Deepak, > > In econometrics there is another test very often used : the white test. > The white test is based on the comparison of the estimated variances of > residuals when the model is estimated by OLS under the assumption of > homoscedasticity and when the model is estimated by OLS under the > assumption of

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 <-

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