Hi, I have a question about how to use the mlogit package in R to do analysis
of discrete choice survey data. Our survey is about asking people to choose
from different insurance policies(with two attributes of deductible and
premium).
The code I used to fit mixed logit is:
[1] ml - mlogit.data
On Thu, Jul 23, 2015 at 3:56 PM, maxbre mbres...@arpa.veneto.it wrote:
Hi
thank you for your reply: it's a neat solution but unfortunately not
applicable to my specific case;
I'm going to assume you're replying to me, although there's no context
whatsoever in your response (this is the R-help
Turns out my column and row names were too long to fit in the console and
when I widened the window the table didn't automatically adjust and widen
as well. I had to widen the console first and then retype the same command
so it fit. Silly mistake. I usually keep the console pretty narrow on my
Marie-Louise,
As long sa I know you have to gie googleVis the ISO code for provinces.
GR - data.frame( ISO = c(DE-BW,
DE-BY,
DE-BE,
DE-BB,
DE-HB,
DE-HH,
DE-HE,
DE-MV,
DE-NI,
DE-NW,
DE-RP,
DE-SL,
DE-SN,
DE-ST,
DE-SH,
DE-TH),
name = c(Baden-Württemberg,
Bayern,
Berlin,
Brandenburg,
Bremen,
Estimados
Pude encontrar uno de mis errores, el cuál está en el XPath, en rvest funciona
y comparto el código, pero por alguna causa solucionando el XPath RSelenium
falla informando el mismo error.
library(rvest)
# Cargo la página
Pagina.R - html(x = http://www.r-project.org;)
## mirar bien
Do you have legal advice that suing the University (if that is the right
context)
would actually be a fruitful way forwards, that it would achieve anything
useful within reasonable time and without causing the student severe
financial risk?
What may work in that context is for students to
François Collin stxfc at nottingham.ac.uk writes:
Dear all,
I would like to run a linear model which includes two factors:
- The first one has two levels, including a reference level. Thus I
have to use the treatment contrast (contr.treatment, reference level
effect = 0, then the
You could do something like the following
rowsToShiftLeft - c(2,4,5) # 4, not the 3 that was in the original post
mat - as.matrix(df_start)
mat[rowsToShiftLeft, 1:3] - mat[rowsToShiftLeft, 2:4]
result - data.frame(mat[, 1:3], stringsAsFactors=FALSE)
str(result)
'data.frame': 5
Oops, Bill's reply and mine crossed in the email. His is essentially
the same as mine except probably more efficient.
-- Bert
Bert Gunter
Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom.
-- Clifford Stoll
On Thu, Jul 23, 2015 at 2:44 PM, William
Hey, I can't seem to import my csv file in such a way that the table looks
normal. The dimensions are correct, 4x7, and I set my header =TRUE, but
it still looks weird. Attached is the picture. Any ideas??
Thanks!!
-Marissa
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R-help@r-project.org
Dear All,
Does anyone know of any R functions that compute partial sums of series?
Thanks in advance!
Janh
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Jeff NewmillerThe . . Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#..
On 24/07/15 07:07, Therneau, Terry M., Ph.D. wrote:
The following are in parody (but like all good parody correct wrt the
salient features). The musings of
Guernsey McPhearson
http://www.senns.demon.co.uk/wprose.html#Mixed
http://www.senns.demon.co.uk/wprose.html#FDA
In formal
Dear R-Experts,
I am trying to calculate the confidence intervals of the Goodman Kruskal
gamma statistics using bootstrap. There is no gamma function in the boot
package. There is a gamma function in the base package, but it is the usual
mathematical function.
So, I decide to try to
Hi
thank you for your reply: it's a neat solution but unfortunately not
applicable to my specific case;
in fact as I specified in my first post (I may have been not enough clear,
sorry for that!) I can not rely on any search method grep-like because the
value xxx in the rows of df_start can be
Some shapefiles for Germany can be found here
http://www.statsilk.com/maps/download-free-shapefile-maps
http://www.statsilk.com/maps/download-free-shapefile-maps
On Jul 23, 2015, at 21:36, Erin Hodgess erinm.hodg...@gmail.com wrote:
Hello Erich:
I just looked at your leaflet package and
Ah, so apparently you require some sort of psychic abilities...
For how else would one choose which three values to keep in a row that was:
a 2 b 5
based on your specification that xxx could be anything.
Cheers,
Bert
Bert Gunter
Data is not information. Information is not knowledge. And
sorry but honestly I do not get your point
I need to shift to left by one position (i.e. one column) the entire rows
2,4,5 of df_start so that to obtain as final result the structure
indicated in df_end
I know in advance the rows that I need to shift
hope it clears a bit, now
--
View this
Hola,
Estoy tratando de crear un método S3 llamado anthr dentro del paquete
que estoy desarrollando, cuyo argumento principal es res que
básicamente es una lista con un solo componente. Pero si el segundo
argumento llamado oneSize es FALSE, res es una lista de listas.
Lo que he escrito hasta el
Dear Sir,
Thanks for your great guidance. Made me realize that I need to think out of
box.
As regards the low losses, BASEL guidelines do say to get rid of such low
losses which create noise in analysing the losses caused by Operational Loss
events.
Its the right tail events do matter which
You might be interested in the HWidentify and HTKidentify functions in
the TeachingDemos package. They currently don't do maps, but since
the functions are pure R code it would not be hard to modify them.
On Wed, Jul 22, 2015 at 5:35 PM, Marie-Louise tim...@hotmail.de wrote:
Hello,
I am trying
Hi Jim,
I tried it and it while it does make the diagram look more like what I want,
there are a few categories still out of order. Thank you for your help!
-Angela
On Thu, 7/23/15, Jim Lemon drjimle...@gmail.com wrote:
Subject: Re: [R] Ordering
by considering the following reproducible example:
v0-c(a,xxx,c,rep(xxx,2))
v1-c(1,b,3,d,e)
v2-c(6,2,8,4,5)
v3-c(xxx,7,xxx,9,10)
df_start-data.frame(v0,v1,v2,v3)
df_start
v0-letters[1:5]
v1-1:5
v2-6:10
df_end-data.frame(v0,v1,v2)
df_end
I need to shift by one column some given rows in the
I am quite happy with that package leaflet which is not yet on CRAN
but available on Githib.
https://github.com/rstudio/leaflet https://github.com/rstudio/leaflet
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With one minor change to your reproducible example (thank you!):
df_start - data.frame(v0,v1,v2,v3, stringsAsFactors=FALSE)
data.frame(t(apply(df_start, 1, function(i)i[!grepl(xxx, i)])),
stringsAsFactors=FALSE)
I'll leave it to you to deal with columns that you'd like to have
numeric. (You
The following are in parody (but like all good parody correct wrt the salient features).
The musings of
Guernsey McPhearson
http://www.senns.demon.co.uk/wprose.html#Mixed
http://www.senns.demon.co.uk/wprose.html#FDA
In formal publication:
Senn, Statistical Issues in Drug Development,
Amelia Marsh via R-help r-help at r-project.org writes:
Dear Sir,
[snip]
Lastly using the command rsnorm(1, mean = m, sd = s, xi = x)
where m, s and x are the estimated parameters obtained from loss
data. The usual procedure is to arrange these simulated values in
descending
Thank you both very much for your tips, and excuse my horrible english
skills.
R is new to me so I tried to work with code that was already there.
I managed to get close to what I wanted with
require(googleVis)
G4 = gvisGeoChart(con2,
locationvar = DE,
Dear all,
I would like to run a linear model which includes two factors:
- The first one has two levels, including a reference level. Thus I have to use
the treatment contrast (contr.treatment, reference level effect = 0, then the
intercept).
- The second is a 6-level factor without reference
Hi,
I am using four spatial regression models (SAR, SEM, SAC and SDM) to
evaluate the spillover effect of some factors. LeSage and Pace (2009)
pointed out that when the spatial lag of the dependent variable is included
in the model, parameter estimates lose their conventional interpretation as
Hi Antony,
Except if you have good reasons to use R2.15, it is generally a good
idea to upgrade to the latest version.
In any case, the solutions that were proposed on the list will
definitely work fine.
There are however advantages of using readxl, in my opinion. readxl does
not use
Hi Ivan,
This way i would've tried but i am using R 2.15 - ReadXL package will
support R =3 versions.
Thanks,
Antony.
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View this message in context:
http://r.789695.n4.nabble.com/Read-xlsx-and-convert-date-column-value-into-Dataframe-tp4710192p4710232.html
Sent from the R help mailing list
Hi Jim,
My requirement is simple. I have to read date-values from the excel file
into dataframe, that's all.
and i tried using the way you mentioned and it works. Thank you very much !
--
View this message in context:
Apologies for cross-posting
There are various remaining seats on the following two statistics courses:
Course 1:
Introduction to Generalized Linear Models with R. -Bayesian and
frequentist approaches -
Dates: 11 - 14 August, 2015
Location: Hotel Grand Chancellor, Coral Coast Drive, Palm
Hi Rich,
Thanks to the link provided by Boris, I now realize that the third
example in the radial.plot function help page is almost a Tickell
diagram. Another plotting function that is close to the illustrations
in that paper is starPie. Learn something every day. Hope this is
helpful.
Jim
On
Hi Angela,
Assuming that your reformatted data is named data, have you tried:
data[order(data$count,data$before,decreasing=TRUE),]
Jim
On Thu, Jul 23, 2015 at 3:15 AM, Angela via R-help r-help@r-project.org wrote:
Hello,
I am trying to figure out if there is a way to order the left side of a
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