[R] “Error in if (abs(x - oldx) ftol)” when using “lognormal” distribution in mixed logit
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 (mydata, choice=choice, shape = wide, id =
individual,
opposite =c ('deductible', 'premium'),varying = 5:10)
[2] ml.w5 - mlogit (choice~deductible+premium|0, ml, panel = TRUE,
rpar = c(deductible='ln', premium='ln'),
R = 100, halton = NA, print.level=0)
I try to use lognormal because we hope the coefficients for both deductible
and premium are negative. And I use opposite in [1] to reverse the sign
because lognormal is always positive.
But I always get the error warning:
Error in if (abs(x - oldx) ftol) { : missing value where TRUE/FALSE
needed
In addition: Warning message: In log(start[ln]) : NaNs produced
I double check the data and am sure there isn't any missing data. And if I
change the lognormal ln to n or cn, it will work without any warning.
Does anyone know how to deal with this? Thank you for your help.
--
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Re: [R] shift by one column given rows in a dataframe
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 email list, not Nabble). 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 anything (string or numeric and always different) so that I necessarely need to apply by row index position (i.e. in my reproducible example rows: 2, 3, 5); Then how do you know which positions? Do you have another R object that specifies row and column number? Or do you guess? I can easily remove a random element from each row... If you aren't removing based on value, then you didn't provide a reproducible example after all, and you need to supply the index for removal. Always different? Different within a single data frame? Different between data frames? (Then you can simply change xxx to whatever.) Is telepathy required? Or perhaps precognition? thank you again for your kind help but still searching for a solution... best -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] shift by one column given rows in a dataframe
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 laptop and never considered that would change the format. Thanks for all the help - I'm sure I'll be back again very soon with more questions!! On Jul 23, 2015 4:53 PM, Bert Gunter bgunter.4...@gmail.com wrote: 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 Dunlap wdun...@tibco.com wrote: 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 obs. of 3 variables: $ v0: chr a b c d ... $ v1: chr 1 2 3 4 ... $ v2: chr 6 7 8 9 ... You will then have to convert the columns which ought to be numeric to numeric. (All the columns in df_start were factors because of the extra xxx that offset some of them.) Bill Dunlap TIBCO Software wdunlap tibco.com On Thu, Jul 23, 2015 at 1:19 PM, maxbre mbres...@arpa.veneto.it wrote: 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 message in context: http://r.789695.n4.nabble.com/shift-by-one-column-given-rows-in-a-dataframe-tp4710256p4710276.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] interactive Map: Popups
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, Hamburg, Hessen, Mecklenburg-Vorpommern, Niedersachsen, Nordrhein-Westfalen, Rheinland-Pfalz, Saarland, Sachsen, Sachsen-Anhalt, Schleswig-Holstein, Thüringen), value = 1:16) G3 - gvisGeoMap(GR, locationvar='ISO', numvar='value', options=list(region=DE, displayMode=regions, resolution=provinces,)) plot(G3) Daniel Merino 2015-07-23 17:00 GMT-03:00 Erich Neuwirth erich.neuwi...@univie.ac.at: 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 its examples. It is awesome! Thanks, Erin On Thu, Jul 23, 2015 at 6:28 AM, Erich Neuwirth erich.neuwi...@univie.ac.at mailto:erich.neuwi...@univie.ac.at wrote: 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 https://github.com/rstudio/leaflet https://github.com/rstudio/leaflet __ R-help@r-project.org mailto:R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html http://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Erin Hodgess Associate Professor Department of Mathematical and Statistics University of Houston - Downtown mailto: erinm.hodg...@gmail.com mailto:erinm.hodg...@gmail.com __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Daniel [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R-es] Fw: web RSelenium
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 los XPath- los modifico, quitar el [1], no me interesa el
primero, quiero todos
# /html/body/div/div[1]/div[1]/div/div[1]/ul/li[1]
# /html/body/div/div[1]/div[1]/div/div[1]/ul/li[1]/a
# tomar link URLs
urls - Pagina.R %% # desde Pagina.R realizar el siguiente paso
html_nodes(xpath='/html/body/div/div[1]/div[1]/div/div[1]/ul/li/a') %%
html_attr(href) # extraer URLs
# Get link text
links - Pagina.R %% # desde Pagina.R realizar el siguiente paso
html_nodes(xpath='/html/body/div/div[1]/div[1]/div/div[1]/ul/li') %%
html_text() # extraer link texto
# Combinar `links` y `urls` en un data.frame
Resultado - data.frame(links = links, urls = urls, stringsAsFactors = FALSE)
Resultado
Javier Rubén Marcuzzi
Técnico en Industrias Lácteas
Veterinario
De: Javier Ruben Marcuzzi
Enviado el: miércoles, 22 de julio de 2015 11:56 p.m.
Para: R-help-es@r-project.org
Estimados, les envío un código, el cuál me da error, aunque dudo si está
escrito correctamente puesto que muchos ejemplos me están fallando.
Agradezco si alguno me informa un error al igual que el informado por mi
computadora, no creo que el inconveniente sea por la opción de R que instalé en
Windows 8.1, ya me perdí buscando mi error.
Revolution R Open 8.0.3
Using CRAN snapshot taken on 2015-04-01
library(RSelenium)
checkForServer()
startServer()
remDr - remoteDriver()
remDr$open()
remDr$navigate(http://www.r-project.org;)
webElems - remDr$findElement(using = 'css selector', body div div.row
div.col-xs-12.col-sm-offset-1.col-sm-2.sidebar div div:nth-child(1) ul)
resHeaders - unlist(lapply(webElems, function(x){x$getElementText()}))
#Error in x$getElementText : object of type 'closure' is not subsettable
Javier Rubén Marcuzzi
Técnico en Industrias Lácteas
Veterinario
[[alternative HTML version deleted]]
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[R] R: Re: Differences in output of lme() when introducing interactions
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 collectively complain that important aspects of their training and support are being neglected. With the rapidity of recent technological change, the issue is widespread. To an extent, able post-docs and PhDs have to lead the charge in getting training and support updated and brought into the modern world. John Maindonald email: john.maindon...@anu.edu.aumailto:john.maindon...@anu.edu.au On 22/07/2015, at 22:00, r-help-requ...@r-project.orgmailto:r-help-requ...@r-project.org wrote: Da: li...@dewey.myzen.co.ukmailto:li...@dewey.myzen.co.uk Data: 21-lug-2015 11.58 A: angelo.arc...@virgilio.itmailto:angelo.arc...@virgilio.itangelo.arc...@virgilio.itmailto:angelo.arc...@virgilio.it, bgunter.4...@gmail.commailto:bgunter.4...@gmail.com Cc: r-help@r-project.orgmailto:r-help@r-project.org Ogg: Re: R: Re: [R] R: Re: Differences in output of lme() when introducing interactions Dear Angelo I suggest you do an online search for marginality which may help to explain the relationship between main effects and interactions. As I said in my original email this is a complicated subject which we are not going to retype for you. If you are doing this as a student I suggest you sue your university for failing to train you appropriately and if it is part of your employment I suggest you find a better employer. On 21/07/2015 10:04, angelo.arc...@virgilio.itmailto:angelo.arc...@virgilio.it wrote: Dear Bert, thank you for your feedback. Can you please provide some references online so I can improve my ignorance? Anyways, please notice that it is not true that I do not know statistics and regressions at all, and I am strongly convinced that my question can be of interest for some one else in the future. This is what forums serve for, isn't it? This is why people help each other, isn't it? Moreover, don't you think that I would not have asked to this R forum if I had the possibility to ask or pay a statician? Don't you think I have done already my best to study and learn before posting this message? Trust me, I have read different online tutorials on lme and lmer, and I am confident that I have got the basic concepts. Still I have not found the answer to solve my problem, so if you know the answer can you please give me some suggestions that can help me? I do not have a book where to learn and unfortunately I have to analyze the results soon. Any help? Any online reference to-the-point that can help me in solving this problem? Thank you in advance Best regards Angelo [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Contr.sum and coefficient tests
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 intercept). - The second is a 6-level factor without reference contrast nor order. So, I would like to use sum contrat: sum of the effects = 0. The problem arises when it comes to the coefficient test. I understand it is not relevant to test the reference level for the first factor as the reference level is set to 0. However, using sum contrast for the second factor, I would have expected the test of each level to be included in the classical summary print of the lm function result but it is not. And here is my problem, how can I have every coefficients tested and printed in the summary output when my factor is studied from this sum.contrast standpoint? [some context snipped to make gmane happy -- sorry ] I think you should look at the effects package or the lsmeans package (or possibly the multcomp package, if you want to be careful about the number of (non-orthogonal) tests) -- the issue is that summary.lm always reports on the *parameters* estimated. If some parameters are not independently estimable (e.g. the effect corresponding to the last level can be reconstructed from the parameter values for all of the preceding levels), then summary.lm() won't give them to you. In fact, these packages will (probably) give you the results you want even if the original model uses default treatment contrasts. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] shift by one column given rows in a dataframe
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 obs. of 3 variables: $ v0: chr a b c d ... $ v1: chr 1 2 3 4 ... $ v2: chr 6 7 8 9 ... You will then have to convert the columns which ought to be numeric to numeric. (All the columns in df_start were factors because of the extra xxx that offset some of them.) Bill Dunlap TIBCO Software wdunlap tibco.com On Thu, Jul 23, 2015 at 1:19 PM, maxbre mbres...@arpa.veneto.it wrote: 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 message in context: http://r.789695.n4.nabble.com/shift-by-one-column-given-rows-in-a-dataframe-tp4710256p4710276.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] shift by one column given rows in a dataframe
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 Dunlap wdun...@tibco.com wrote: 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 obs. of 3 variables: $ v0: chr a b c d ... $ v1: chr 1 2 3 4 ... $ v2: chr 6 7 8 9 ... You will then have to convert the columns which ought to be numeric to numeric. (All the columns in df_start were factors because of the extra xxx that offset some of them.) Bill Dunlap TIBCO Software wdunlap tibco.com On Thu, Jul 23, 2015 at 1:19 PM, maxbre mbres...@arpa.veneto.it wrote: 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 message in context: http://r.789695.n4.nabble.com/shift-by-one-column-given-rows-in-a-dataframe-tp4710256p4710276.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Table Looks Funny
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 __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Infinite Series
Dear All, Does anyone know of any R functions that compute partial sums of series? Thanks in advance! Janh [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Infinite Series
?cumsum --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. On July 23, 2015 8:23:39 PM PDT, Janh Anni annij...@gmail.com wrote: Dear All, Does anyone know of any R functions that compute partial sums of series? Thanks in advance! Janh [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Differences in output of lme() when introducing interactions
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 publication: Senn, Statistical Issues in Drug Development, second edition, Chapter 14: Multicentre Trials Senn, The many modes of meta, Drug information journal, 34:535-549, 2000. The second points out that in a meta analysis no one would ever consider giving both large and small trials equal weights, and relates that to several other bits of standard practice. The 'equal weights' notion embedded in a fixed effects model + SAS type 3 is an isolated backwater. Terry T. PS. The Devils' Drug Development Dictionary at the same source has some gems. Three rather random choices: Bayesian - One who, vaguely expecting a horse and catching a glimpse of a donkey, strongly concludes he has seen a mule. Medical Statistician - One who won't accept that Columbus discovered America because he said he was looking for India in the trial Plan. Trend Towards Significance - An ever present help in times of trouble. On 07/22/2015 06:02 PM, Rolf Turner wrote: On 23/07/15 01:15, Therneau, Terry M., Ph.D. wrote: SNIP 3. Should you ever use it [i.e. Type III SS]? No. There is a very strong inverse correlation between understand what it really is and recommend its use. Stephen Senn has written very intellgently on the issues. Terry --- can you please supply an explicit citation? Ta. Thanks Terry! cheers, Rolf -- Technical Editor ANZJS Department of Statistics University of Auckland Phone: +64-9-373-7599 ext. 88276 __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Confidence intervals of GK gamma statistics using bootstrap
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 calculate the confidence intervals using this site :
http://www.statmethods.net/advstats/bootstrapping.html
I get a normal warning message, but my R code is not working and I don't
understand where my mistake(s) is (are).
Here is the reproducible example with imaginary/fake data.
install.packages(ryouready)
library(ryouready)
a=c(satisfait, pas satisfait, tres satisfait,satisfait,tres
satisfait,pas satisfait,satisfait,satisfait,tres satisfait,pas
satisfait)
b=c(grand, petit, petit, grand, petit, grand, grand, petit,
petit, grand)
x=table(a,b)
ord.gamma(x)
# calculate Goodman Kruskal gamma using bootstrap
library(boot)
GK - function(formula, data, indices) {
d - data[indices,] # allows boot to select sample
tab - xtabs(formula, data=d)
stat - ord.gamma(tab)
return(stat)
}
# bootstrapping with 2000 replications
results - boot(data=d, statistic=GK,R=2000)
# view results
results
plot(results)
# get 95% confidence interval
boot.ci(results, type=all)
Best Regards, thanks for your precious help!
SV
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Re: [R] shift by one column given rows in a dataframe
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 anything (string or numeric and always different) so that I necessarely need to apply by row index position (i.e. in my reproducible example rows: 2, 3, 5); thank you again for your kind help but still searching for a solution... best -- View this message in context: http://r.789695.n4.nabble.com/shift-by-one-column-given-rows-in-a-dataframe-tp4710256p4710271.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] interactive Map: Popups
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 its examples. It is awesome! Thanks, Erin On Thu, Jul 23, 2015 at 6:28 AM, Erich Neuwirth erich.neuwi...@univie.ac.at mailto:erich.neuwi...@univie.ac.at wrote: 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 https://github.com/rstudio/leaflet https://github.com/rstudio/leaflet __ R-help@r-project.org mailto:R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html http://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Erin Hodgess Associate Professor Department of Mathematical and Statistics University of Houston - Downtown mailto: erinm.hodg...@gmail.com mailto:erinm.hodg...@gmail.com signature.asc Description: Message signed with OpenPGP using GPGMail __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] shift by one column given rows in a dataframe
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 knowledge is certainly not wisdom. -- Clifford Stoll On Thu, Jul 23, 2015 at 12:56 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; 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 anything (string or numeric and always different) so that I necessarely need to apply by row index position (i.e. in my reproducible example rows: 2, 3, 5); thank you again for your kind help but still searching for a solution... best -- View this message in context: http://r.789695.n4.nabble.com/shift-by-one-column-given-rows-in-a-dataframe-tp4710256p4710271.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] shift by one column given rows in a dataframe
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 message in context: http://r.789695.n4.nabble.com/shift-by-one-column-given-rows-in-a-dataframe-tp4710256p4710276.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R-es] Método S3 paquete
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 momento es lo siguiente:
anthr - function(res, oneSize, nsizes){
UseMethod(anthr)
}
anthr.tri - function(res, oneSize, nsizes){
if(oneSize){
cases - c()
cases - res$meds
}else{
cases - list()
for (i in 1 : (nsizes - 1)){
cases[[i]] - res[[i]]$meds
}
}
return(cases)
}
El problema cuando instalo el paquete y utilizo este método, es que R no
me reconoce que res sea una lista. En concreto, me aparece este error:
Error in UseMethod(anthr) :
no applicable method for 'anthr' applied to an object of class list
He tratado de añadir esto:
tri - function(x){
value - list(meds = x$meds)
attr(value, class) - tri
value
}
pero sigue sin funcionarme. ¿Alguien puede ofrecerme alguna ayuda?.
Muchas gracias de antemano.
Un saludo,
Guillermo
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Re: [R] Statistical distribution not fitting
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 represent low frequency high magnitude nature losses. But my client is so adamant about it, that although we have shown them research papers about threshold limits which need to apply to arrive at some meaningful analyses, he is insisting that we do include these low losses too and fit some distribution. 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 order and select an observation representing (say 99.9%) and this is Value at Risk (VaR) which is say 'p'. My understanding is to this value 'p', I need to apply the transformation 10^p to arrive at the value which is in line with my original loss data. Am I right? Thanks again sir for your great help. I have something to look ahead now. Regards Amelia _ On Thursday, 23 July 2015 2:20 AM, Boris Steipe boris.ste...@utoronto.ca wrote: So - as you can see, your data can be modelled. Now the interesting question is: what do you do with that knowledge. I know nearly nothing about your domain, but given that the data looks log-normal, I am curious abut the following: - Most of the events are in the small-loss category. But most of the damage is done by the rare large losses. Is it even meaningful to guard against a single 1/1000 event? Shouldn't you be saying: my contingency funds need to be large enough to allow survival of, say, a fiscal year with 99.9 % probability? This is a very different question. - If a loss occurs, in what time do the funds need to be replenished? Do you need to take series of events into account? - The model assumes that the data are independent. This is probably a poor (and dangerous) assumption. Cheers, B. On Jul 22, 2015, at 3:56 PM, Ben Bolker bbol...@gmail.com wrote: Amelia Marsh amelia_marsh08 at yahoo.com writes: Hello! (I dont know if I can raise this query here on this forum, but I had already raised on teh finance forum, but have not received any sugegstion, so now raising on this list. Sorry for the same. The query is about what to do, if no statistical distribution is fitting to data). I am into risk management and deal with Operatioanl risk. As a part of BASEL II guidelines, we need to arrive at the capital charge the banks must set aside to counter any operational risk, if it happens. As a part of Loss Distribution Approach (LDA), we need to collate past loss events and use these loss amounts. The usual process as being practised in the industry is as follows - Using these historical loss amounts and using the various statistical tests like KS test, AD test, PP plot, QQ plot etc, we try to identify best statistical (continuous) distribution fitting this historical loss data. Then using these estimated parameters w.r.t. the statistical distribution, we simulate say 1 miliion loss anounts and then taking appropriate percentile (say 99.9%), we arrive at the capital charge. However, many a times, loss data is such that fitting of distribution to loss data is not possible. May be loss data is multimodal or has significant variability, making the fitting of distribution impossible. Can someone guide me how to deal with such data and what can be done to simulate losses using this historical loss data in R. A skew-(log)-normal fit doesn't look too bad ... (whenever you have positive data that are this strongly skewed, log-transforming is a good step) hist(log10(mydat),col=gray,breaks=FD,freq=FALSE) ## default breaks are much coarser: ## hist(log10(mydat),col=gray,breaks=Sturges,freq=FALSE) lines(density(log10(mydat)),col=2,lwd=2) library(fGarch) ss - snormFit(log10(mydat)) xvec - seq(2,6.5,length=101) lines(xvec,do.call(dsnorm,c(list(x=xvec),as.list(ss$par))), col=blue,lwd=2) ## or try a skew-Student-t: not very different: ss2 - sstdFit(log10(mydat)) lines(xvec,do.call(dsstd,c(list(x=xvec),as.list(ss2$estimate))), col=purple,lwd=2) There are more flexible distributional families (Johnson, log-spline ...) Multimodal data are a different can of worms -- consider fitting a finite mixture model ... __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To
Re: [R] interactive Map: Popups
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 to build a map of a country which shows informations to its regions in a popup window as soon as someone clicks on a region. Thank you -- View this message in context: http://r.789695.n4.nabble.com/interactive-Map-Popups-tp4710226.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Gregory (Greg) L. Snow Ph.D. 538...@gmail.com __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Ordering in Sankey diagram using R and googleVis
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 in Sankey diagram using R and googleVis
Cc: r-help mailing list r-help@r-project.org
Date: Thursday, July 23, 2015, 6:43 AM
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 Sankey diagram from most
frequent to least frequent. I am using R version 3.2.1 and
using googleVis version 0.5.9 for the Sankey. I've tried
sorting, but that does not work. Is there anyway to force it
to arrange the left (before) side in decreasing
frequency? Something I am missing? Does not have to be using
[[elided Yahoo spam]]
-Angela
Example of the data I have, in a csv
file:
before
after
A B
A
B
A B
A
C
A A
A
A
A B
D
E
F B
F
B
F F
G
H
G A
I reformat the data in R so it looks like
this:
before
after count
A B 4
A C 1
A
A 2
D E 1
F B 2
F
F 1
G H 1
G A 1
Then plot using this:
plot( gvisSankey (data, from=before,
to=after, weight=freq,
options=list(width=600, height=800,
sankey={iterations: 2})))
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[R] shift by one column given rows in a dataframe
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 initial data frame called df_start so that to get the final structure as in df_end; please consider that the value xxx in the rows of df_start can be anything so that I necessarly need to apply by row index position (in my reproducible example rows: 2, 3, 5); I'm really stuck with that problem and I can not conceive any viable solution up to now any hints? best regards m __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] interactive Map: Popups
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 signature.asc Description: Message signed with OpenPGP using GPGMail __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] shift by one column given rows in a dataframe
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 might also try str(df_start)). Sarah On Thu, Jul 23, 2015 at 12:14 PM, Massimo Bressan mbres...@arpa.veneto.it wrote: 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 initial data frame called df_start so that to get the final structure as in df_end; please consider that the value xxx in the rows of df_start can be anything so that I necessarly need to apply by row index position (in my reproducible example rows: 2, 3, 5); I'm really stuck with that problem and I can not conceive any viable solution up to now any hints? best regards m -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Differences in output of lme() when introducing interactions
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, second edition, Chapter 14: Multicentre Trials Senn, The many modes of meta, Drug information journal, 34:535-549, 2000. The second points out that in a meta analysis no one would ever consider giving both large and small trials equal weights, and relates that to several other bits of standard practice. The 'equal weights' notion embedded in a fixed effects model + SAS type 3 is an isolated backwater. Terry T. PS. The Devils' Drug Development Dictionary at the same source has some gems. Three rather random choices: Bayesian - One who, vaguely expecting a horse and catching a glimpse of a donkey, strongly concludes he has seen a mule. Medical Statistician - One who won't accept that Columbus discovered America because he said he was looking for India in the trial Plan. Trend Towards Significance - An ever present help in times of trouble. On 07/22/2015 06:02 PM, Rolf Turner wrote: On 23/07/15 01:15, Therneau, Terry M., Ph.D. wrote: SNIP 3. Should you ever use it [i.e. Type III SS]? No. There is a very strong inverse correlation between understand what it really is and recommend its use. Stephen Senn has written very intellgently on the issues. Terry --- can you please supply an explicit citation? Ta. cheers, Rolf __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Statistical distribution not fitting
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 order and select an observation representing (say 99.9%) and this is Value at Risk (VaR) which is say 'p'. My understanding is to this value 'p', I need to apply the transformation 10^p to arrive at the value which is in line with my original loss data. Am I right? [snip; sorry to remove context, but Gmane doesn't like it] (1) you can probably calculate the 0.999 quantile directly from qsnorm(0.999, [params]) rather than by simulating ... (2) ... I believe that my original example used log(), so you would need to use exp() (not 10^x) to get back to the original scale ... (3) ... if you're concerned about extreme events it would be a very good idea to use the skew-t rather than the skew-Normal (4) you should certainly consider Boris Steipe's concerns about non-independence (although I have to admit that without more information and further time/effort/thought I don't have any simple suggestions how ...) cheers Ben Bolker __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] interactive Map: Popups
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, colorvar = data_7, hovervar = data_6, options=list( width=800, height=600, region=DE, displayMode=regions, resolution=provinces)) plot(G4) but it seems like googleVis does not support resolution=provinces for Germany :( Do you know anything that simple as this code to get to what I wanted? Thank you very much -- View this message in context: http://r.789695.n4.nabble.com/interactive-Map-Popups-tp4710226p4710257.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Contr.sum and coefficient tests
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 contrast nor order. So, I would like to use sum contrat: sum of the effects = 0. The problem arises when it comes to the coefficient test. I understand it is not relevant to test the reference level for the first factor as the reference level is set to 0. However, using sum contrast for the second factor, I would have expected the test of each level to be included in the classical summary print of the lm function result but it is not. And here is my problem, how can I have every coefficients tested and printed in the summary output when my factor is studied from this sum.contrast standpoint? Here is an example from a previous thread: (http://comments.gmane.org/gmane.comp.lang.r.general/258886) == x - as.factor(c(1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3)) y - c(1.1,1.15,1.2,1.1,1.1,1.1,1.2,1.2,1.2,2.1,2.2,2.3,2.4,2.5, + 2.6,2.7,2.8,2.9,3,3.1) test - data.frame(x,y) reg1 - lm(y~C(x,contr.sum),data=test) summary(reg1) Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 1.60.06577 24.834 8.48e-15 *** C(x, contr.sum)1 -0.483330.10792 -4.479 0.00033 *** C(x, contr.sum)2 -0.483330.08936 -5.409 4.70e-05 *** == How can I include the third factor level in this table? I fill the answer provided in this previous thread is relevant for contr.treatment but not for contr.sum. Am I right? Or can you explain me? Many thanks, Fanch This message and any attachment are intended solely for the addressee and may contain confidential information. If you have received this message in error, please send it back to me, and immediately delete it. Please do not use, copy or disclose the information contained in this message or in any attachment. Any views or opinions expressed by the author of this email do not necessarily reflect the views of the University of Nottingham. This message has been checked for viruses but the contents of an attachment may still contain software viruses which could damage your computer system, you are advised to perform your own checks. Email communications with the University of Nottingham may be monitored as permitted by UK legislation. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to calculate the average direct effect, average total effect and average indirect effect for spatial regression models with spatial lag of dependent variable
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 marginal effects, because the spatial lag gives rise to a series of feedback loops and spillover effects across regions. Therefore, I need to calculate the three different marginal effects: average direct effect, average total effect and average indirect (spillover) effect, and this is what I don't know how to perform in R. Can you teach me about how to calculate the average direct effect, average total effect and average indirect (spillover) effect for the spatial regression models in R and tell me the related R code? I would like to express my great appreciation to you! Thank you very much and best regards. Yours sincerely, Wenyue Yang -- View this message in context: http://r.789695.n4.nabble.com/How-to-calculate-the-average-direct-effect-average-total-effect-and-average-indirect-effect-for-spate-tp4710253.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Read .xlsx and convert date-column value into Dataframe
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 dependencies; it correctly converts to dates; and if you have empty cells in numeric columns, xlsx will convert to NaN and not NA as would be expected (correct me if this behavior has been modified in the latest version). The problem with readxl is that it only reads! A similar write function would be nice! Ivan -- Ivan Calandra, ATER University of Reims Champagne-Ardenne GEGENAA - EA 3795 CREA - 2 esplanade Roland Garros 51100 Reims, France +33(0)3 26 77 36 89 ivan.calan...@univ-reims.fr https://www.researchgate.net/profile/Ivan_Calandra Le 23/07/15 08:50, R_Antony a écrit : Hi Ivan, This way i would've tried but i am using R 2.15 - ReadXL package will support R =3 versions. Thanks, Antony. -- 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 archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Read .xlsx and convert date-column value into Dataframe
Hi Ivan, This way i would've tried but i am using R 2.15 - ReadXL package will support R =3 versions. Thanks, Antony. -- 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 archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Read .xlsx and convert date-column value into Dataframe
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: http://r.789695.n4.nabble.com/Read-xlsx-and-convert-date-column-value-into-Dataframe-tp4710192p4710233.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Stats courses in Queensland, Australia
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 Cove. Australia Course 2: Introduction to Zero Inflated Models with R. - Frequentist and Bayesian approaches - Dates: 17 - 21 August, 2015 Location: Hotel Grand Chancellor, Coral Coast Drive, Palm Cove. Australia Course website: http://highstat.com/statscourse.htm Course flyers: http://highstat.com/Courses/Flyers/Flyer2015_08PalmCoveI.pdf http://highstat.com/Courses/Flyers/Flyer2015_08PalmCoveII.pdf Both courses are pre-required knowledge for our 'Introduction to GLMs with spatial and temporal correlation'. Kind regards, Alain Zuur -- Dr. Alain F. Zuur First author of: 1. Beginner's Guide to GAMM with R (2014). 2. Beginner's Guide to GLM and GLMM with R (2013). 3. Beginner's Guide to GAM with R (2012). 4. Zero Inflated Models and GLMM with R (2012). 5. A Beginner's Guide to R (2009). 6. Mixed effects models and extensions in ecology with R (2009). 7. Analysing Ecological Data (2007). Highland Statistics Ltd. 9 St Clair Wynd UK - AB41 6DZ Newburgh Tel: 0044 1358 788177 Email: highs...@highstat.com URL: www.highstat.com __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Langelier-Ludwig Plots
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 Thu, Jul 23, 2015 at 2:49 AM, Rich Shepard rshep...@appl-ecosys.com wrote: On Wed, 22 Jul 2015, Boris Steipe wrote: According to ... http://info.ngwa.org/gwol/pdf/721000139.PDF (Graphical Interpretation of Water Quality Data) ... a Langelier-Ludwig plot is simply a scatterplot of cations vs. anions (in percent). Surely that would be beyond trivial to produce in R. Or am I missing a subtle something? Boris, Yes, that's what it is. I'll see just how trivial it is to produce it in the four-quadrant format I've seen used. It's a different layout from the basic two-variable scatterplot. Thanks for the URL. I worked from a reference in a 2003 paper to the original 1942 paper. Rich __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Ordering in Sankey diagram using R and googleVis
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
Sankey diagram from most frequent to least frequent. I am using R version
3.2.1 and using googleVis version 0.5.9 for the Sankey. I've tried sorting,
but that does not work. Is there anyway to force it to arrange the left
(before) side in decreasing frequency? Something I am missing? Does not
have to be using googleVis. Thank you!
-Angela
Example of the data I have, in a csv file:
beforeafter
AB
AB
AB
AC
AA
AA
AB
DE
FB
FB
FF
GH
GA
I reformat the data in R so it looks like this:
beforeaftercount
AB4
AC1
AA2
DE1
FB2
FF1
GH1
GA1
Then plot using this:
plot( gvisSankey (data, from=before, to=after, weight=freq,
options=list(width=600, height=800,
sankey={iterations: 2})))
__
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