the following error , at the following point in the code ..." )
On 2025-04-16 12:11 p.m., varin sacha via R-help wrote:
> R-experts,
>
> The R script here below.
>
> How to solve my problem ?
>
> Best,
>
>
> ##
> # Load packa
R-experts,
The R script here below.
How to solve my problem ?
Best,
##
# Load packages
install.packages("remotes")
remotes::install_github("bd2kccd/r-causal")
# Install 'gssr' from 'ropensci' universe
install.packages('gssr', repos =
c('https://kjhealy.r-unive
Dear R-experts,
Here below a toy example not working. After some researches on the Net, still
don't get it !
Many thanks for your precious help.
#
library(knitr)
library(ggplot2)
library(pollster)
library(dplyr)
statut=c("married","not married","not
married","mar
Dear R-experts,
I write to you to know if somebody is aware of a R package (or function) able
to plot graphs for extrapolation.
I need to be clear on what extrapolation really is to me. It is when we use the
model for X variables outside the range of X variables that were used to
construct the
Dear R-experts,
I really thank you all for your responses.
Best,
Le dimanche 14 janvier 2024 à 10:22:12 UTC+1, Duncan Murdoch
a écrit :
On 13/01/2024 8:58 p.m., Rolf Turner wrote:
> On Sat, 13 Jan 2024 17:59:16 -0500
> Duncan Murdoch wrote:
>
>
>
>> My guess is that one of the boot
Time)
library(boot)
func= function(data, idx) {
coef(lm(Score~ Time + factor(Country),data=data[idx,]))
}
B= boot(e, func, R=1000)
boot.ci(B, index=2, type="perc")
#
Le samedi 13 janvier 2024 à 21:56:58 UTC+1, Ivan Kr
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
; On 12/10/2023 2:50 PM, Rui Barradas wrote:
>> Às 22:35 de 10/12/2023, varin sacha via R-help escreveu:
>>>
>>> Dear R-experts,
>>>
>>> Here below my R code, as my X-axis is "year", I must be missing one
>>> or more steps! I am trying to
Dear R-experts,
Here below my R code, as my X-axis is "year", I must be missing one or more
steps! I am trying to get the regression line with the 95% confidence bands
around the regression line. Any help would be appreciated.
Best,
S.
#
library(g
y's are not looked for in data (i.e. NSE) but in the environment
where the function was defined, which is standard evaluation. Change the above
to:
cor1 <- with(d, cor(x1, y1, method="spearman"))
cor2 <- with(d, cor(x2, y2, method="spearman"))
and all should be fi
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.
Bes
Dear Rui,
I really thank you a lot for your precious R help. It is exactly what I was
trying to do! Once more, many thanks!
Best,
Sacha
Le vendredi 27 octobre 2023 à 09:36:18 UTC+2, Rui Barradas
a écrit :
Às 19:23 de 26/10/2023, varin sacha via R-help escreveu:
> Dear R-Expe
Dear R-Experts,
Here below my R code working but I don't know how to complete/finish my R code
to get the final plot with the extrapolation for the10 more years.
Indeed, I try to extrapolate my data with a linear fit over the next 10 years.
So I create a date sequence for the next 10 years and
Dear Rui,
I really thank you a lot for your response and your R code.
Best,
Sacha
Le mardi 24 octobre 2023 à 16:37:56 UTC+2, Rui Barradas
a écrit :
Às 20:12 de 23/10/2023, varin sacha via R-help escreveu:
> Dear R-experts,
>
> I really thank you all a lot for your responses.
o 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.)
> >>
> >> And it's not necessarily true
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)
x1=c(34,35,12,13,15,37,65,45,47,67,87,
R-experts,
I am trying to find the exact 95% confidence intervals of the mean (and the
median) for the weibull distribution. Here below I have the exact 95% CIs of
the meanlog and the median for a lognormal using EnvStats library (ready-to-run
function in library) but I don't find how to get th
, May 9, 2023 at 6:40 AM Eric Berger wrote:
> Hi,
> One problem you have is with the command:
> regr<-randomForest(y~x1+x2, data=X_train, proximity=TRUE)
>
> What you need is something like this:
>
> X2 <- cbind(X,y)
> regr<-randomForest(y~x1+x2, data=X2, proxi
Dear R-experts,
Here below a toy example with some error messages, especially at the end of the
code (Tuning the parameters). Your help to correct my R code would be highly
appreciated.
###
#libraries
library(lattice)
library(ggplot2)
library(caret)
library(
Dear R-experts,
Here below my R code (toy example) working! The only thing missing is in my GAM
plot: I would like to get on the same graph the 2 bands (prediction and
confidence bands) like in my lm model graph!
Is it possible? How to get that?
Dear Simon,
Thanks ! It works !
Best,
Le lundi 1 mai 2023 à 11:19:26 UTC+2, Simon Wood a écrit
:
try...
sum(residuals(model1)^2)
On 30/04/2023 22:03, varin sacha via R-help wrote:
> Dear R-experts,
>
> Here below my R code. I get a NaN response for gam with mgcv librar
Dear R-experts,
Here below my R code. I get a NaN response for gam with mgcv library. How to
solve that problem?
Many thanks.
#
library(mgcv)
y=c(23,24,34,40,42,43,54,34,52,54,23,32,35,45,46,54,34,36,37,48)
x1=c(0.1,0.3,0.5,0.7,0.8,0.9,0.
Dear R-experts,
Here below a toy example to calculate the MSE (mean squared error).
Starting from this equation : MSE = bias^2 + variance + irreducible error
I am trying to get the bias and the variance in addition to the MSE. How to get
them both?
Many thanks for your help.
###
Me again ! How to solve this?
At the end of this page there is the installation command :
https://gitlab.science.ru.nl/gbucur/RUcausal/-/blob/master/README.Rmd
Working with a MAC, I have tried to install the RUcausal library (copy and
paste the installation command).
It is written that the pack
Bert,
Thanks ! It works !
Best,
Le samedi 22 avril 2023 à 19:42:18 UTC+2, Bert Gunter
a écrit :
Is lvida.R in your working directory?
Try using the full path name to the file instead in source()
-- Bert
On Sat, Apr 22, 2023 at 9:38 AM varin sacha via R-help
wrote:
>
&g
Best,
Le samedi 22 avril 2023 à 18:30:48 UTC+2, Eric Berger a
écrit :
looks fine.
what's the problem?
On Sat, Apr 22, 2023 at 7:27 PM varin sacha wrote:
> Eric,
>
> Here it is :
>
>
> library(RBGL)
> Loading required package: graph
> Loading required
’
The following objects are masked from ‘package:igraph’:
bfs, dfs, transitivity
Le samedi 22 avril 2023 à 18:12:56 UTC+2, Eric Berger a
écrit :
What happens with the command
> library(RBGL)
On Sat, Apr 22, 2023 at 7:08 PM varin sacha via R-help
wrote:
> Dear R-e
Dear R-experts,
How to solve that problem?
My R version is 4.2.1
Here below trying to install RGBL library found here :
https://bioconductor.org/packages/release/bioc/html/RBGL.html
So, I run this R code :
if (!require("BiocManager", quietly = TRUE))
install.packages("BiocManager")
BiocM
Dear R-experts,
Here below my R code working but I would like to plot (get the graph) of the
Causal additive model exactly the same way I get the graph (CPDAG) from the
bnlearn R package. Is it possible? If yes, how?
Many thanks
###
# libraries
library(devtools)
this is quite clearly not the case.
I suppose you may have intended to fit the models _with_ the intercept and then
_ignore_ the intercept for plotting purposes, i.e. lm(y~x11+x12,
Dataset)$coef[-1], etc.?
(Also, I suspect that you don't actually have y=7 and y=867 in the dataset.)
-pd
11 feb. 2023 om 18:35 schreef varin sacha via R-help
:
> 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
Dear R-experts,
Here below my R code working with quite a few warnings.
x11 and x12 are dichotomous variable (0=no and 1=yes). I substract 1 to ignore
intercept.
I would like not to ignore intercept. How to modify my R code because if I just
remove -1 it does not work?
y= c(32,45,65,34,23,43,
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
ersions. Probably simple ... in function definition and print(plot(res, axes
> = axes, ...)) addition could do the trick, but I am not sure.
>
> Cheers
> Petr
>
>> -Original Message-
>> From: R-help On Behalf Of varin sacha via R-
>> help
>>
Dear R-experts,
Here below the R code working (page 8
http://www2.uaem.mx/r-mirror/web/packages/FactoMineR/FactoMineR.pdf).
But I am trying to get all the labels (the writes) : comfort, university,
economic, world, ... smaller.
How could I do that ?
Many thanks.
library(FactoMineR)
data(chi
https://fentechsolutions.github.io/CausalDiscoveryToolbox/html/index.html
>
> HTH,
> Eric
>
>
>> On Mon, Nov 21, 2022 at 9:00 PM varin sacha via R-help
>> wrote:
>> Dear R experts,
>>
>> Google is very often my friend but this time it does not !
>> Are you
Dear R experts,
Google is very often my friend but this time it does not !
Are you aware of an R package in which the directed causal discovery algorithm
called the Information Geometric Causal Inference (IGCI) of (Daniusis et al.,
2010) is implemented ?
Best,
Sacha
Envoyé de mon iPhone
_
Dear R-experts,
While comparing groups, it is better to assess confidence intervals of those
differences rather than comparing confidence intervals for each group.
I am trying to calculate the CIs of the difference between the two Cramer's V
and not the CI to the estimate of each group’s Cramer'
Dear R-experts,
Here below my R code working but I would like to get the names of the variables
to appear on the graph instead of 1 ; 2 ; 3 and 4.
Is it possible ? I have tried something with colnames(x) but I get an error
message.
Many thanks for your help.
#
librar
Dear R-experts,
Here below my R code working but I don't know how I can get the graph
(estimated DAG).
If you could help me to get the graph, many thanks.
###
library(pcalg)
x1<-c(508,413,426,500,568,372,484,512,529,322,544,586,480,561,567,488,450,548,526,561,435,567,537,521,5
djust(r$p.value, method = "holm")) :
# no non-missing arguments to min; returning Inf
p.dagitty.param.correct
#[1] Inf
Hope this helps,
Rui Barradas
Às 10:58 de 31/03/2022, varin sacha via R-help escreveu:
> Dear R-experts,
>
> Here below my R code (toy example). No
icate cannot be authenticated with given CA certificates'
Le jeudi 24 mars 2022, 14:39:03 UTC+1, Ivan Krylov a
écrit :
On Thu, 24 Mar 2022 13:29:45 + (UTC)
varin sacha wrote:
> I have changed my CRANmirror to the Switzerland (ETHZ) one using the
> function you sent to
Many thanks for your response.
Le mercredi 16 mars 2022, 21:11:06 UTC+1, varin sacha via R-help
a écrit :
Dear R-experts,
I have used the EnvStats package and the elnorm function (p. 248).
I would like to calculate the exact 95% confidence intervals around the mean,
not around
Dear R-experts,
Here below my sessionInfo( ). I cannot download the pcalg package/library. Is
my R version too old ?
sessionInfo()
R version 4.1.2 (2021-11-01)
Platform: x86_64-apple-darwin17.0 (64-bit)
Running under: macOS High Sierra 10.13.6
Matrix products: default
BLAS:
/Library/Framewo
Dear R-experts,
I have used the EnvStats package and the elnorm function (p. 248).
I would like to calculate the exact 95% confidence intervals around the mean,
not around the meanlog.
Here below my R code, how can I get the exact 95% CIs around the mean ?
Many thanks.
library(EnvStats)
x=rlnor
kages that appear to provide that
information. Have you tried any of them?
On Sun, Feb 13, 2022 at 12:44 PM varin sacha via R-help
wrote:
>
> Dear R-experts,
>
> Here below my R code. I am trying to get the exact 95% confidence intervals
> around the meanlog for lognormal, around the
John Fox, Professor Emeritus
McMaster University
Hamilton, Ontario, Canada
Web: http::/socserv.mcmaster.ca/jfox
> On Jan 8, 2022, at 12:04 PM, varin sacha via R-help
> wrote:
>
> Dear R-experts,
>
> Here below my R code for the percentile bootstr
Dear R-experts,
Here below my R code for the percentile bootstrap confidence intervals with an
error message.
Is there a way to make my R code work ?
Many thanks for your help and time.
library(boot)
s=rnorm(10,0,1)
(m<-median(s))
N <- 100
n<-
Dear John, Dear Rui,
I really thank you a lot for your R code.
Best,
SV
Le jeudi 30 décembre 2021, 05:25:11 UTC+1, Fox, John a écrit
:
Dear varin sacha,
You didn't correctly adapt the code to the median. The outer call to mean() in
the last line shouldn't be replaced w
;- sample(s, size = 5)
boot.out <- boot(data = a, statistic = med, R = 1)
boot.ci(boot.out, type = "stud")$stud[, 4:5]
})
mean(out[1, ] < mean(s) & mean(s) < out[2, ])
#[1] 0.952
Hope this helps,
Rui Barradas
Às 11:45 de 19/12/21, varin sacha via R-help escreveu:
Dear R-experts,
Here below my R code working but really really slowly ! I need 2 hours with my
computer to finally get an answer ! Is there a way to improve my R code to
speed it up ? At least to win 1 hour ;=)
Many thanks
library(boot)
ue in ?boot.ci.
So just change
$norm[, 4:5]
to
$norm[, 2:3]
Hope this helps,
Rui Barradas
Às 22:24 de 16/12/21, varin sacha via R-help escreveu:
> Dear R-experts,
>
> Here below my R code. Using "bca" in the boot.ci function at the end of my R
> code works perfect
Dear R-experts,
Here below my R code. Using "bca" in the boot.ci function at the end of my R
code works perfectly. Using "perc" and "basic" perfectly works as well. But
using "norm", my R code does not work anymore.
Is it possible to solve that problem and to make my R code work ?
#
rowMeans(arr) ))
#[1] 111
Now, the equal values. The return value of 1 means that all BCa
intervals include mean(s). The outer mean(.) is a mean of an indicator
function given by a logical conjunction. And there are only TRUE's.
Hope this helps,
Rui Barradas
Às 15:37 de 11/12/21, va
;- 1e3, not 1e4.
set.seed(2021)
N <- 1e3
out <- replicate(N, {
boot.out <- boot(data = dat, statistic = med, R = 1)
boot.ci(boot.out, type = "bca")$bca[, 4:5]
})
mean(out[1,] < mean(s) & mean(s) < out[2,])
Hope this helps,
Rui Barradas
Às 10:47 de 11/12/21, vari
Dear R-experts,
Here below my R code. I am trying to do 2 things :
1) I would like to repeat this R code 1 times
2) Out of the 1 repetitions I would have liked R to tell me how many times
the "true" mean value of the population called "s" in my R example here below
is included in the 1
Hi,
Many thanks for your responses.
Le samedi 6 novembre 2021, 08:39:22 UTC+1, Rui Barradas
a écrit :
Hello,
Às 01:36 de 06/11/21, David Winsemius escreveu:
>
> On 11/5/21 1:16 PM, varin sacha via R-help wrote:
>> Dear R-experts,
>>
>> Here is a toy ex
Hi,
I really thank you a lot for your response.
Le samedi 6 novembre 2021, 02:37:46 UTC+1, David Winsemius
a écrit :
On 11/5/21 1:16 PM, varin sacha via R-help wrote:
> Dear R-experts,
>
> Here is a toy example. How can I get the bootstrap confidence intervals
> workin
Dear R-experts,
Here is a toy example. How can I get the bootstrap confidence intervals working
?
Many thanks for your help
library(DescTools)
library(boot)
A=c(488,437,500,449,364)
dat<-data.frame(A)
med<-function(d,i) {
temp<-d[i,]
HodgesLehmann(A)
}
boot.out<-
"Appenzell Rhodes Extérieures","Genève","Zoug","Tessin",
"Neuchâtel","Vaud","Uri","Nidwald","Berne(d)","Zurich","Obwald",
"Saint-Gall","Soleure","Lucerne","Glari
ot;Schaffhouse","Berne(f)",
"Thurgovie","Valais(f)","Argovie","Appenzell Rhodes Extérieures",
"Genève","Zoug","Tessin","Neuchâtel","Vaud","Uri","Nidwald",
Dear R-experts,
Here below my R code. I would need your help to improve my graph/plot.
- The x-axis to be longer not to stop at 500 value
- All the name on the y-axis to appear not only a few of them and the name
(Fribourg(f), Appenzell Rhodes Intérieures,...) to appear entire, not to be cut
Ma
Dear R-experts,
Here below a toy R code example. I would like some countries (not all of them)
"Italy", "Canada", "Greece" and "Norway" to appear in red color. The others
remaining black. How can I do that without big changes in my R code ? Indeed, I
would like my R code to remain like this as
gt; Were you expecting image files? I don't see any plot device e.g. pdf()
> in your code.
>
> Jim
>
>> On Wed, May 12, 2021 at 6:34 PM varin sacha via R-help
>> wrote:
>>
>> Dear Experts,
>>
>> My R code was perfectly working since I decide to
", type = 'b')
points((1:30)/10, power.cors[6,], pch = 2, col = "green", type = 'b')
points((1:30)/10, power.cork[6,], pch = 3, col = "blue", type = 'b')
points((1:30)/10, power.dcor[6,], pch = 4, col = "red", type = 'b')
po
ng 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 Sun, May 9, 2021 at 2:59 PM varin sacha via R-help
> wrote:
>
>> Rui,
>>
>> The cr
The comments only line is your last code line.
The result is attached.
Hope this helps,
Rui Barradas
Às 19:39 de 09/05/21, varin sacha via R-help escreveu:
> Dear R-experts,
>
> I am trying to get the 8 graphs like the ones in this paper :
> https://statweb.stanford.edu/~tibs/
Rui,
Angelo,
I found it :=)
Many thanks
S.
Le dimanche 9 mai 2021 à 23:49:41 UTC+2, Angelo Canty
a écrit :
Have you looked in the pdf file (power.pdf) to which you instructed R to
send the plots?
On 2021-05-09 5:27 p.m., varin sacha via R-help wrote:
> Dear Rui,
>
> I
(1,2,3), col = c("black","green","blue","red"))
plot((1:30)/10, power.cor[7,], ylim = c(0,1), main = "Circle", xlab = "Noise
Level", ylab = "Power", pch = 1, col = "black", type = 'b')
points((1:30)/10,
Dear R-experts,
I am trying to get the 8 graphs like the ones in this paper :
https://statweb.stanford.edu/~tibs/reshef/comment.pdf
My R code does not show any error message neither warnings but I d'on't get
what I would like to get (I mean the 8 graphs), so I am missing something.
What's it ?
Dear R-experts,
Here below my R code working but I would like to get 3 values not only 1. The
value I get is, according to my R code, the variance value. My goal is to get 3
values : the bias value, the variance value and the MSE value. How to solve my
problem ?
Many thanks.
#
thing like
> ypred = pred.robustgam( fit18, data.frame(X=Testing)
>
> 2. your logic is wrong. You define the vectors x and y at the top. They
> should remain untouched during your program.
> However in the
at the top. They should
remain untouched during your program.
However in the loop you redefine y and then use the redefined y as an
argument to robustgam() the next time through
the loop. This looks like a serious error.
HTH,
Eric
On Sun, Jan 17, 2021 at 12:20 PM varin sacha via R-help
w
Dear R-experts,
Here below my reproducible R code. I get an error message (end of code) I can't
solve.
Many thanks for your help.
##
#Data
y=c(34000,45000,19000,48900,65000,67000,78000,9,51000,32000,54000,85000,38000,76345,87654,90990,78654,67894,56789,65432,18998,789
Murdoch wrote:
>
> On 12/01/2021 2:43 p.m., varin sacha via R-help wrote:
>> Dear R-experts,
>> I have tried to reach the maintainer of the rgam package. Until now, no
>> response.
>> Since I'm in a bit of a hurry, I try to reach you because as I try to
&g
y to respond to these kinds of questions.
Also, I note the most obvious choice for GAM-based models in R, is
mgcv, which ships with R.
But there are other packages, if mgcv doesn't meet your needs.
On Wed, Jan 13, 2021 at 8:44 AM varin sacha via R-help
wrote:
>
> Dear R-experts,
>
>
Dear R-experts,
I have tried to reach the maintainer of the rgam package. Until now, no
response.
Since I'm in a bit of a hurry, I try to reach you because as I try to install
the rgam package using the command :
install.packages("rgam")
I get this warning message :
Warning message:
package
Dear R-experts,
I dont get any error message but I can't get the green curve/line on the plot.
What is going on ? How to solve my problem ?
Here is the R code :
PIB.hab<-c(12000,34000,25000,43000,12500,32400,76320,45890,76345,90565,76580,45670,23450,34560,65430,65435,56755,8
ry
o <- order(x)
lines(x[o], yfit[o], col="green", lwd=2)
Duncan Murdoch
On 09/01/2021 2:58 p.m., varin sacha via R-help wrote:
>
Dear R-experts,
Here below my R code. What is happening with my green curve ? How to solve the
problem ?
Many thanks,
##
#DATA
y<-c(499,491,500,517,438,495,501,525,516,494,500,453,479,481,505,465,477,520,520,480,477,416,502,503,497,513,492,469,504,482,502,498,463,504
cking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Wed, Jan 6, 2021 at 12:17 PM varin sacha via R-help
wrote:
> Dear R-Experts,
>
> Here below my reproducible R code.
> How can I get the
Dear R-Experts,
Here below my reproducible R code.
How can I get the AIC of my model (robust GAM) ?
Best Regards,
y<-c(499,491,500,517,438,495,501,525,516,494,500,453,479,481,505,465,477,520,520,480,477,416,502,503,497,513,492,469,504,482,502,498,463,504,495)
x<-c(499,496,424,537,480,484,503,57
rs. However this would probably break some existing
code, since there are various functions that return NULL instead of numeric(0).
-Bill
On Fri, Nov 13, 2020 at 2:05 PM varin sacha via R-help
wrote:
> Dear R-experts,
>
> Here below my reproducible example. No error message but I can not g
Dear R-experts,
Here below my reproducible example. No error message but I can not get a
result. I get "NaN" as a result. I don't understand what is going on. Many
thanks for your precious help, as usual.
# # # # # # # # # # # # # # # # # # # # # # # # #
x<-c(499,491,500,517,438,495,501,525,5
Dear R-helpers,
Here below my R code showing warnings and error messages I don't understand.
What is going wrong ?
install.packages("devtools")
library(devtools)
install_github("ProcessMiner/nlcor")
library(nlcor)
A=c(505, 530, 419, 486, 608, 468, 519, 486, 532, 289, 5
Many thanks Duncan,
It works !
Best.
Le mardi 27 octobre 2020 à 20:49:25 UTC+1, Duncan Murdoch
a écrit :
On 27/10/2020 3:06 p.m., varin sacha via R-help wrote:
> Dear R-experts,
>
> Here below my R code. The warning message is not a problem to me but there is
&
bs)
Thanks.
Md
On Tue, Oct 27, 2020 at 7:21 PM Sarah Goslee wrote:
> 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
>
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 <-
Dear R-experts,
I have fitted an orthogonal regression and have some difficulties to get the
adjusted R^2 and R^2, the AICc, the coefficients and R^2 bootstrap confidence
intervals. Here below my R codes.
Many thanks for your precious help.
y=c(231,212,112,143,154,165,1
Dear Matthias,
Many thanks for your response.
Best,
SV
Le mardi 4 août 2020 à 16:22:41 UTC+2, Prof. Dr. Matthias Kohl
a écrit :
you could try:
library(MKinfer)
meanDiffCI(a, b, boot = TRUE)
Best
Matthias
Am 04.08.20 um 16:08 schrieb varin sacha via R-help:
> Dear R-expe
Dear R-experts,
Using the bootES package I can easily calculate the bootstrap confidence
intervals of the means like in the toy example here below. Now, I am looking
for the confidence intervals for the difference between group means. In my
case, the point estimate of the mean difference is 64.
))
}
out
}
double.mad(x)
# lo hi
#0.81543 0.44478
double.mad(c(x, 1))
# lo hi
#2.29803 0.44478
double.mad(c(x, 1), include.right = TRUE)
# lo hi
#1.03782 1.63086
Hope this helps,
Rui Barradas
Às 15:22 de 03/08/2020, varin sacha via R-help escreveu:
> Dear R-Exper
Dear R-Experts,
Is there an all-ready function to calculate the Double MAD (Median absolute
deviation) as there is an easy function to calculate the MAD "mad function". Or
I have to write my own function for Double MAD ?
To calculate the double MAD, the idea is the following : for the obtained
Dear R-experts,
I am trying to do a permutation test for the Ramsey RESET test. More precisely,
I am interested in the "exact" p-value of the test.
I have checked the coin package and all the functions (oneway_test; ...). There
are plenty of functions but no one is helping me for RESET test. I
in `colnames<-`(`*tmp*`, value = "MSE_OLS") :
# attempt to set 'colnames' on an object with less than two dimensions
mean(my.data)
#[1] 105.6951
Hope this helps,
Rui Barradas
Às 22:25 de 19/05/20, varin sacha escreveu:
> Hi Rui,
>
> If I don
C+2, Rui Barradas a
écrit :
Hello,
Inline.
Às 21:38 de 19/05/20, varin sacha via R-help escreveu:
>
> Hi Richard,
>
> Thanks for your response.
> However, how can I correct my R code knowing that I want, as a result, only
> one value : the mean of the 500 MSE_OLS values ?
0
On Tue, May 19, 2020 at 3:51 PM varin sacha via R-help
wrote:
> Dear R-experts,
>
> Here is my R code, I get a result but I also get an error message so I doubt
> I can trust the result I get.
> What is going wrong ? Many thanks.
>
>
Dear R-experts,
Here is my R code, I get a result but I also get an error message so I doubt I
can trust the result I get.
What is going wrong ? Many thanks.
a<-c(2,4,3,4,6,5,3,1,2,3,4,3,4,5,65)
b<-c(23,45,32,12,23,43,56,44,33,11,12,54,23,34,54)
d<-c(9,4
Dear R-experts,
My goal is to get only 1 value : the average/ the mean of the 100 MSE values.
How can I finish my R code ?
###
my.experiment <- function() {
n<-500
x<-runif(n, 0, 5)
z <- rnorm(n, 2, 3)
a <- runif(n, 0, 5)
y_mode
till could be a function representing a probability
distribution...
On Sun, Apr 26, 2020 at 7:09 AM varin sacha via R-help
wrote:
>
> Dear R-experts,
>
> I am trying to fit a gaussian density curve. More precisely, I would like to
> obtain the fitted "Gaussian curve".
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