Estimado Javier.
A parte de los modos concretos que hay en R para trabajar con texto y codigo,
en emacs tienes el modo .org que permite integrar codigo de diferentes
lenguajes.
Aqui dejo un enlace donde se pueden descargar versiones actualizadas de emacs
que vienen ya con el modo ess para
> On Mar 21, 2017, at 5:04 AM, Michael Dayan wrote:
>
> Hi,
>
> I would like to fit a mixture of two beta distributions with parameters
> (alpha1, beta1) for the first component, (alpha2, beta2) for the second
> component, and lambda for the mixing parameter. I also
> On Mar 21, 2017, at 6:31 PM, Bert Gunter wrote:
>
> It is not clear to me what you mean, but:
>
>> grep ("x1 \\+.* \\+ x3",test, value = TRUE)
> [1] "x1 + x2 + x3"
>
> ## This will miss "x1 + x3" though.
So then this might be acceptable:
grep ("x1\\ \\+.* x3",
Estimado Fernando Arce
Hace mucho que no uso Emacs, recuerdo que lo vi por primera vez cuándo compre
una versión de suse 5.3, que instale en un Pentium III. Podría explorarlo, creo
que está XEmacs con algo de entorno gráfico, la experiencia de usuario es nada
que ver, lo que más me gusto
It is not clear to me what you mean, but:
> grep ("x1 \\+.* \\+ x3",test, value = TRUE)
[1] "x1 + x2 + x3"
## This will miss "x1 + x3" though.
seems to do what you want, maybe. Perhaps you need to read up about
regular expressions and/or clarify what you want to do.
Cheers,
Bert
Bert Gunter
Según entiendo, te hace falta una distribución de látex para la conversión
del documento o la tienes instalada pero no encuentra la ruta de los
ejecutables. Si no la tienes puedes instalar miktek que es una distribución
de látex para Windows
El 21 mar. 2017 6:44 PM, "Mauricio Mardones Inostroza"
Hi Joe,
This may help you:
test <- c("x1", "x2", "x3", "x1 + x2 + x3")
multigrep<-function(x1,x2) {
xbits<-unlist(strsplit(x1," "))
nbits<-length(xbits)
xans<-rep(FALSE,nbits)
for(i in 1:nbits) if(length(grep(xbits[i],x2))) xans[i]<-TRUE
return(all(xans))
}
multigrep("x1 + x3","x1 + x2 +
Hi Folks,
Is there a way to find "x1 + x2 + x3" given "x1 + x3" as the pattern?
Or is that a ridiculous question, since I'm trying to find something
based on a pattern that doesn't exist?
test <- c("x1", "x2", "x3", "x1 + x2 + x3")
test
[1] "x1" "x2" "x3" "x1 + x2 +
Estimado Fernando
Si, cerré desde el administrador del sistema operativo, reinicié, reinstalé.
Javier Rubén Marcuzzi
De: Fernando Macedo
Enviado: martes, 21 de marzo de 2017 20:16
Para: Marcuzzi, Javier
CC: R-help-es
Asunto: RE: [R-es] Alternativa a RStudio
Probaste matando los procesos de R
El 21/03/17 a las 23:48, javier.ruben.marcu...@gmail.com escribió:
> Estimados
>
> Alguno utiliza una alternativa a RStudio, últimamente no me gusta como
> funciona,
> por ejemplo, al cargar una archivo (abrirlo) se coloca como a ejecutar algo,
> la consola no marca
> nada, pero pasa el
Estimado Fernando Macedo
Encontré algo que dice sobre la indexación, como que iría creando un índice en
algún lado para encontrar partes del código utilizado en autocompletar, o algo
por el estilo.
Estoy cambiando desde un archivo R notebook a R Markdown, sospecho que ese
archivo hacía
Buscaste en Internet por posibles problemas? Entiendo que lo usas en
Windows. En Linux no he tenido problemas de ese tipo.
El 21 mar. 2017 19:48, escribió:
> Estimados
>
> Alguno utiliza una alternativa a RStudio, últimamente no me gusta como
> funciona, por
Estimados
Alguno utiliza una alternativa a RStudio, últimamente no me gusta como
funciona, por ejemplo, al cargar una archivo (abrirlo) se coloca como a
ejecutar algo, la consola no marca nada, pero pasa el tiempo y el administrador
de tareas de Windows 10 informa como va aumentando los megas
He offered two solutions, and I want to second the vote against the first one.
I often put large numbers of configuration variables in a few CSV files
organized by topic area and read them in. The resulting data frames are capable
of holding multiple cases if desired and I just specify which
That worked perfectly!
This makes using a large number of values for programming and their
documentation significantly easier.
Thank you
Shawn Way, PE
-Original Message-
From: Enrico Schumann [mailto:e...@enricoschumann.net]
Sent: Tuesday, March 21, 2017 4:40 PM
To: Shawn Way
On Tue, 21 Mar 2017, Shawn Way writes:
> I have an org-mode table with the following structure
> that I am pulling into an R data.frame, using the
> sfsmisc package and using xtable to print in org-mode
>
> | Symbol | Value | Units |
> |--+---+---|
> | A
Note that the recently released package nlsr by Duncan Murdoch and I has a slight update in the functions nlxb() and
nlfb() as well as a lot of features for symbolic derivative calculation.
JN
On 2017-03-21 06:57 AM, DANIEL PRECIADO wrote:
Dear list,
I want to use nlxb (package nlmrt) to fit
I have an org-mode table with the following structure that I am pulling into an
R data.frame, using the sfsmisc package and using xtable to print in org-mode
| Symbol | Value | Units |
|--+---+---|
| A | 1 | kg/hr|
Thanks, Bill.
Sooo... taking the error literally,
temp1 = array(NA, c(3,2,1))
dimnames(temp1)[[3]] = list( "test" )
works, even though
mode( dimnames(temp1)[[3]] )
yields
"character".
Setting all the dimension names simultaneously still feels way better.
--
Sent from my phone. Please
It happens because dimnames(originalArray) is NULL and when [[<- extends
NULL it turns it into a list if the size of the new element is not one
but into a vector with the type of new element if the new element's
size is one.
> str( `[[<-`(NULL, 3, value="One") )
chr [1:3] NA NA "One"
> str(
While I generally agree that it is better to design code that sets all of the
dimension names simultaneously, the discrepancy between behavior when the
dimensions are longer than 1 and when they are equal to 1 seems irregular.
Someone went to some lengths to make it possible to set dimnames
When you use the data= argument in glm(), the function looks in the data.frame
for a variable first. You have created two versions of stype, one in the
data.frame and one outside it. So your first glm() selects all the cases
apistrat since apistrat$stype always equals apistrat$stype. You can
This one was actually clear enough to those exposed to the econometric lingo:
Panel = "repeated measurements", VAR = Vector AutoRegression (not to be
confused with var() or VaR).
-pd
> On 20 Mar 2017, at 17:44 , Bert Gunter wrote:
>
> Excuse my denseness, but huh??
>
Hello,
I am experiencing odd behavior with the subset parameter for glm. It appears
that the parameter uses non-standard evaluation, but only in some cases. Below
is a reproducible example.
library(survey) # for example dataset
data(api)
stype <- "E"
(a <- glm(api00~ell+meals+mobility, data =
Hi,
This mail is in regards with a query we have in usage of Cubist package in R.
We are using the CRAN package - "Cubist" for predicting sales. We have a large
training dataset which in turn gives a huge specification result file.
(Note: Independent variables consists of both character and
Dear R users,
I am pleased to announce the release on CRAN of brant 0.1-1:
https://cran.r-project.org/package=brant. I implemented the brant test
(Brant, Rollin 1990) in R together with Prof. Marco Steenbergen at our
Institute.
The function brant() tests the parallel regression assumption for
> On Mar 20, 2017, at 4:46 PM, Douglas Ezra Morrison
> wrote:
>
> Dear R-Help readers,
>
> I am writing to ask about some behavior of base::dimnames() that surprised
> me. If I create an array with one of its dimensions = 1, and then try to
> assign names to that
Estimada Miriam Alzate
No se entiende que desea, la primer parte de la pregunta es utilizar tres
criterios para ordenación, primero uno, luego los siguientes, y a cada criterio
le puede asignar el orden como descendiente o ascendente. Pero usted dice de
asignar a una misma fecha y un mismo
You can use wrapnls from the nlmrt package to get an nls object. Run
it instead of nlxb. It runs nlxb followed by nls so that the output is
an nls object.. Then you can use all of nls' methods.
On occiasion that fails even if nlxb succeeds since the nls
optimization can fail independently of
Buenas,
Necesito hacer lo siguiente:
Tengo una lista de reviews (opiniones). Quiero ordenarlas primero por "Id
del producto", 2º por "Mas votos recibidos" y 3º por "Fecha más nueva". El
problema está en el tercer criterio, que hay varias que coinciden en
fecha. Por eso quiero tomar este método:
Hi,
I would like to fit a mixture of two beta distributions with parameters
(alpha1, beta1) for the first component, (alpha2, beta2) for the second
component, and lambda for the mixing parameter. I also would like to set a
maximum of 200 iterations and a tolerance of 1e-08.
My question is: how
Dear list,
I want to use nlxb (package nlmrt) to fit different datasets to a gaussian,
obtain parameters (including standard error, t-and p-value) and confidence
intervals.
nlxb generates the parameters, but very often results in NA standard
error,t-and p-values. Furthermore, using confint()
Below are samples from a kernel density estimated "data" with gaussian
kernel.
I really like this solution of estimation of a kernel because it is nice
and elegant.
fit<-density(data)
rnorm(N, sample(data, size = N, replace = TRUE), fit$bw) #samples from
kernel density estimation
I am however
> Gerrit Eichner
> on Tue, 21 Mar 2017 10:45:13 +0100 writes:
> Hi, Tal,
> in print.default it says:
> digits:
> a non-null value for digits specifies the __minimum__
> number of __significant__ digits to be printed in values.
I'd say it's about equivalent to suggesting to Vladimir Putin that he
stop trying to take over territories bordering Russia.
Jim
On Tue, Mar 21, 2017 at 8:30 PM, Tal Galili wrote:
> This may have been asked before, but I don't understand the behavior
> of options(digits =
Hi, Tal,
in print.default it says:
digits:
a non-null value for digits specifies the __minimum__
number of __significant__ digits to be printed in values.
Maybe this clarifies your observation.
Hth -- Gerrit
-
Dr. Gerrit
This may have been asked before, but I don't understand the behavior
of options(digits = 1) for vectors containing values such as 0.01
For example, why is this happening:
options(digits = 1)
> 0.01
[1] 0.01
More examples:
> options(digits = 7)
> 0.1
[1] 0.1
> 0.01
[1] 0.01
> 0.11
[1] 0.11
>
Hi Alicia,
If I understand this, perhaps:
df<-data.frame(VALUE=c("<60","Positive","Negative","Less than 0.30",
"12%","<0.2","Unknown"),
DESCRIPTION = c("A","A", "B","C","D","E","E"),
NUMERIC_VALUE=c(9,9,9,9,9,9,9))
df$NUMERIC_VALUE[df$VALUE == "Positive" & df$DESCRIPTION == "A"]<-
9
The solution proposed below does not accomplish my goal. In the column
labeled NEW_VALUE, there are two NAs where the value in NUMERIC_VALUE for
that row should be.
Can anyone else suggest a solution to my problem?
Thanks!
On Sat, Mar 18, 2017 at 1:33 PM, David Winsemius
Hello, I am having some troubles extracting pixels from a raster using
polygons. When I attempt to do so, pixels which are only partially
intersected by polygons are not included.
In the example below the number of pixels returned is less than the number
of pixels which can be seen intersecting
Dear R-Help readers,
I am writing to ask about some behavior of base::dimnames() that surprised
me. If I create an array with one of its dimensions = 1, and then try to
assign names to that dimension, I get an error unless I name one of the
other dimensions first. For example:
> temp1 =
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