Dear Rui,
I was wondering whether we have to square root of SD to find SE, right?
bootprop <- function(data, index){
d <- data[index, ]
sum(d[["BothTimes"]], na.rm = TRUE)/sum(d[["Time1"]], na.rm = TRUE)
}
R <- 1e3
set.seed(2020)
b <- boot(daT, bootprop, R)
b
b$t0 # original
sd(b$t) #
Buen día estimada comunidad,
Actualmente tengo un problema cuando carga mi tabla de metadatos, una vez
importada de excel, escribo el siguiente código para preparar mis metadatos
para crear mi objeto phyloseq, sin embargo a la hora de realizar esto me
elimina la primera columna con los datos de
OK, but that's not the point of my comment.
Bert Gunter
"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Fri, Jan 22, 2021 at 5:03 PM Abby Spurdle wrote:
> Sorry, Bert.
>
Sorry, Bert.
The fitdistr function estimates parameters via maximum likelihood.
(i.e. The "lognormal" part of this, is not a kernel).
On Fri, Jan 22, 2021 at 5:14 AM Bert Gunter wrote:
>
> In future, you should try to search before posting. I realize that getting
> good search terms can
Hi Jean,
kripp.alpha expects a classifier by object (in your case, score)
matrix as the first argument. If I read your code correctly you are
getting the scores zigzagging across six columns when you want a 6 x
180 matrix. My guess is that you want:
Trainers<-matrix(c(Trainer_1,Trainer_2,
Well,
"Appreciate the help!"
Indeed, you should for any that you receive. But do note per the posting
guide linked below:
"For questions about functions in standard packages distributed with R (see
the FAQ Add-on packages in R
Hello!
I am requesting help for a data set in which I have 6 evaluators (I've called
them 'Trainers'), each of which took a questionnaire with 180 questions. The
possible answers to the questions are ordinal numbers. I am looking to test
inter-rater reliability of the questionnaire using
Hello,
Something like this, using base package boot?
library(boot)
bootprop <- function(data, index){
d <- data[index, ]
sum(d[["BothTimes"]], na.rm = TRUE)/sum(d[["Time1"]], na.rm = TRUE)
}
R <- 1e3
set.seed(2020)
b <- boot(daT, bootprop, R)
b
b$t0 # original
sd(b$t) # bootstrapped
Hi All,
I was trying to estimate standard error (SE) for the proportion value using
some kind of randomization process (bootstrapping or jackknifing) in R, but
I could not figure it out.
Is there any way to generate SE for the proportion?
The example of the data and the code I am using is
1. Please read the posting guide. Statistical questions and questions on
non-standard (not shipped with standard R distro) packages are largely off
topic here.
2. This CRAN task view might be of interest to you:
https://cran.r-project.org/web/views/Econometrics.html
3. But as a general comment,
Thanks all of you for your answers.
I managed to solve it after all. I was looking at the wrong place. It
was an argument on the server.R.
On Fri, 2021-01-22 at 09:17 -0600, Yihui Xie wrote:
> I don't know the answer (there are multiple possible reasons for the
> file not being found), but as
Hi,
I’m currently using R studio to analyse data - I have 3 data sets consisting of
months beginning in 02/2005-02/2020 with numbers of internet searches for 3
different countries. I have used a GAM to analyse this data and see if there
are any significant differences or trends, but I am unsure
Hello everyone, My name is Vadym, I am a last-year student at Stockholm
School of Economics in Riga, currently working on my Bachelor's Thesis (so
still learning R). For my research, I need a GMM estimator (preferable
Arellano and Bond, 1991). I have been searching for a solution to my issue
for
Gracias Jesús, sí, haré remuestreos. Yo suelo usar una función que me pasó
un colega:
# A function to generate the equal number of samples (n) for different
classes
bsample <- function(data,cname,n) {
d <- data[-c(1:nrow(data)),]
u <- unique(data[,cname])
for (uu in u) {
w <-
Gracias Carlos, la verdad es que no me lo solucionó, seguramente por falta
mía de conocimientos. Haré remuestreo como dice Jesús.
El vie, 22 ene 2021 a las 12:46, Carlos Ortega ()
escribió:
> Hola Manuel,
>
> No dan una respuesta concluyente, pero sí algunas pistas...
>
>
>
I think it's always difficult and sometimes impossible to take an
existing session and convert it to the vanilla state, but it's very easy
to run a new instance of R from an existing one.
So instead of a clearws() function, I'd suggest a "runInVanilla"
function, that takes some code as input,
Thanks Duncan for a clear argument about the "why".
The suggestion of R --vanilla started a train of thought that one could do
something like
clearws <- function(){ # Try to clear workspace
tmp <- readline("Are you sure you want to clear the workspace? ")
print(tmp)
if (
I don't know the answer (there are multiple possible reasons for the
file not being found), but as the first step to debug the problem, you
may replace this chunk
```{r, echo=FALSE}
library(knitr)
source("helper.R", local = knitr::knit_global())
Has probado en usar en vez de algoritmos sensibles al coste, usar t�cnicas de
remuestreo? Hay un paquete de la UGR llamado imbalance que funciona muy bien.
De: R-help-es en nombre de Manuel Mendoza
Enviado: viernes, 22 de enero de 2021 11:43
Para: Lista R
You would be better to try https://community.rstudio.com. Shiny is
basically Tidyverse/Rstudio item and there will be a lot more expertise
there.
On Fri, 22 Jan 2021 at 02:02, Georgios via R-help
wrote:
> Hi!
> I'm new in R and this list.
> I made a shiny app using R studio.
> my files are:
>
Hola Manuel,
No dan una respuesta concluyente, pero sí algunas pistas...
https://stackoverflow.com/questions/57076570/how-to-calculate-class-weights-for-random-forests
Gracias,
Carlos Ortega
www.qualityexcellence.es
El vie, 22 ene 2021 a las 11:43, Manuel Mendoza ()
escribió:
> Buenos días,
Buenos días, tengo una base de datos desequilibrados (unbalanced) en la que
las ausencias son 9 veces más abundantes que las presencias (*ratio *= 9).
Para árboles de clasificación utilizo una matriz de pérdidas
parms=list(loss=matrix(c(0,
FP, *ratio *,0)))o un vector de ponderación que le da 9
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