similar to: Add a vertical line and some values on a plot

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

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

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

Dear R-Experts, Here below is my R code (reproducible example) to calculate the confidence intervals around the spearman coefficient. ########## 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) cor(C,D,method= "spearman") library(boot) myCor=function(data,index){ cor(data[index, ])[1,2] } results=boot(data=cbind(C,D),statistic=myCor, R=2000)

Hello, First of all, when I tried to use function mic I got an error. mic(cbind(C, D)) Error in mic(cbind(C, D)) : could not find function "mic" So I've changed your function myCor and all went well, with a warning relative to BCa intervals. myCor <- function(data, index){ mine(data[index, ])$MIC } results=boot(data = cbind(C,D), statistic = myCor, R = 2000)

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 *

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(

Dear R-experts, Here below, my R code working BUT I get a strange result I was not expecting! Indeed, the?95% percentile bootstrap CIs is (-54.81, -54.81 ). Is anything going wrong? Best, ########################################## Score=c(345,564,467,675,432,346,476,512,567,543,234,435,654,411,356,658,432,345,432,345, 345,456,543,501) ? Country=c("Italy", "Italy",

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

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, ] <-

R-Experts, Here below my R code working without error message but I don't get the results I am expecting. Here is the result I get: [1] "All values of t are equal to 0.28611928397257 \n Cannot calculate confidence intervals" NULL If someone knows how to solve my problem, really appreciate. Best, S ######################################################### # Difference in Spearman

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

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()

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 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. ########## Y=c(2,4,5,4,3,4,2,3,56,5,4,3,4,5,6,5,4,5,34,21,12,13,12,8,9,7,43,12,19,21)

Dear R-experts, I can not install the package "matie". If somebody can tell me where the problem is. > install.packages("matie") Installing package into ?/Users/Caro/Library/R/3.3/library? (as ?lib? is unspecified) essai de l'URL 'https://cran.rstudio.com/bin/macosx/mavericks/contrib/3.3/matie_1.2.tgz' Content type 'application/x-gzip' length 80151

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

Dear R-experts, I am trying to fit a GAM with 2 binary predictors (variables coded 0,1). I guess I cannot just smooth binary variables. By the way I code them as 0=no,1=yes, then mgcv will think those variables are numeric.? I have tried to change 0 and 1 in no and yes. It does not work. How to solve my problem. Here below my toy example. Many thanks. Best, Sacha ? ########################