[R] Sampling
I have 1000 data points. i want to take 30 samples and find mean. I also want to repeat this process 100 times. How to go about it? Regards Parth __ 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-es] Resumen de R-help-es, Vol 73, Envío 40
Por lo que comentas parece que tuvieras Java de 32 bits y que quisieras hacerlo funcionar con R de 64 bits. Lo mejor es que tengas las dos versiones de Java instaladas puesto que hay software que lo requiere en 32bits y otro en 64. Ejemplo: Aunque tu sistema operativo sea de 64 bits, si usas un navegador de 32 bits, necesitarás Java de 32bits. Un Saludo, Miguel. El 27/03/2015 a las 17:11, Our Utopy escribió: Muchísimas gracias, lo he probado y ya me funciona la interconectividad con Excel, pero con el R de 32 bits, ya que con el de 64 bits me sigue dando problemas. Os cuento tan pronto lo resuelva. Creo sospechar que el Java está involucrado y que si uso 64 bits todo tiene que ser de 64 bits... o algo así. Saludos Nota: A información contida nesta mensaxe e os seus posibles documentos adxuntos é privada e confidencial e está dirixida únicamente ó seu destinatario/a. Se vostede non é o/a destinatario/a orixinal desta mensaxe, por favor elimínea. A distribución ou copia desta mensaxe non está autorizada. Nota: La información contenida en este mensaje y sus posibles documentos adjuntos es privada y confidencial y está dirigida únicamente a su destinatario/a. Si usted no es el/la destinatario/a original de este mensaje, por favor elimínelo. La distribución o copia de este mensaje no está autorizada. See more languages: http://www.sergas.es/aviso_confidencialidad.htm ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
Re: [R] generating phi using function()
The argument 'K' is missing since you are only passing four arguments to
the phi() function, but you defined it with five formal parameters. It
looks like the argument 'j' is not necessary in the function. It is an
unnecessary carry-over from the summation notation and it is never used
in the function.
Dan
On 3/29/2015 4:08 PM, Jim Lemon wrote:
Hi T.,
Your translation of the formula looks okay, and the error message is about
a missing argument. Perhaps you have not included the necessary arguments
to phi in the call to mls.
Jim
On Sun, Mar 29, 2015 at 11:59 PM, T.Riedle tr...@kent.ac.uk wrote:
Hi everybody,
I am trying to generate the formula shown in the attachment. My formula so
far looks as follows:
phi - function(w1, w2, j, k, K){
zaehler - (k/K)^(w1-1)*(1-k/K)^(w2-1)
nenner - sum( ((1:K)/K)^(w1-1)*(1-(1:K)/K)^(w2-1))
return( zaehler/nenner )
}
Unfortunately something must be wrong here as I get the following message
when running a midas regression
m22.phi- midas_r(rv~mls(rvh,1:max.lag+h1,1,phi), start = list(rvh=c(1,1)))
Error in phi(c(1, 1), 44L, 1) : argument K is missing, with no default
Called from: .rs.breakOnError(TRUE)
Browse[1] K-125
Browse[1] 125
Could anybody look into my phi formula and tell me what is wrong with it?
Thanks in advance.
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--
Daniel Nordlund
Bothell, WA USA
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[R] temporal autocorrelation in MCMCglmm
Hi, For a number of individuals, I have measured several behavioral traits in the wild. Those traits (e.g. home range) can be estimated on different temporal scales, for example daily, weekly or monthly. I want to estimate repeatability of those traits, assuming that the daily/weekly/monthly measurements represent replicates. I have 3 months (90 days) of data for each trait. Two questions: 1) How can assess if there is temporal autocorrelation in my model? I guess that if I consider daily measurements as replicates (90 replicates), I will have some autocorrelation, but if I use just monthly measurements (3 replicates) maybe I avoid it. 2) How can account for temporal autocorrelation in MCMCglmm? Sorry for this pretty basic questions but I haven't found an answer so far. Thanks! David [[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] Sampling
On 3/29/2015 11:10 PM, Partha Sinha wrote: I have 1000 data points. i want to take 30 samples and find mean. I also want to repeat this process 100 times. How to go about it? Regards Parth __ 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. see ?replicate and ?sample. Simple example where yourdata is a simple vector of values, and assuming you want to sample without replacement. Generalizing it to other data structures is left as an exercise for the reader. replicate(100,mean(sample(yourdata,30, replace=FALSE))) hope this is helpful, Dan -- Daniel Nordlund Bothell, WA USA __ 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] Nueva publicación en blog en r-es.org
Nueva publicación en blog: Cursos, CURSO NO PRESENCIAL CONTROL DE CALIDAD CON R, por vaamonde en 30/03/15 10:27h Ver el blog en: http://r-es.org/tiki-view_blog_post.php?blogId=4postId=98 Si no desea recibir estas notificaciones siga este enlace: http://r-es.org/tiki-user_watches.php?id=49 ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
Re: [R] matrix manipulation question
On 30-03-2015, at 09:59, Stéphane Adamowicz stephane.adamow...@avignon.inra.fr wrote: Le 27 mars 2015 à 18:01, David Winsemius dwinsem...@comcast.net a écrit : On Mar 27, 2015, at 3:41 AM, Stéphane Adamowicz wrote: Well, it seems to work with me. No one is doubting that it worked for you in this instance. What Peter D. was criticizing was the construction : complete.cases(t(Y))==T ... and it was on two bases that it is wrong. The first is that `T` is not guaranteed to be TRUE. The second is that the test ==T (or similarly ==TRUE) is completely unnecessary because `complete.cases` returns a logical vector and so that expression is a waste of time. Indeed, You are right, the following code was enough : « Z - Y[, complete.cases(t(Y) ] » However, in order to help me understand, would you be so kind as to give me a matrix or data.frame example where « complete.cases(X)== T » or « complete.cases(X)== TRUE » would give some unwanted result ? T can be redefined. Try this in your example with airquality: T - hello Z - Y[,complete.cases(t(Y))==T] Z TRUE is a reserved word and cannot be changed. But why use ==TRUE if not necessary? All of this mentioned already by David Winsemius in a previous reply. Berend __ 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] matrix manipulation question
On 30 Mar 2015, at 09:59 , Stéphane Adamowicz stephane.adamow...@avignon.inra.fr wrote: However, in order to help me understand, would you be so kind as to give me a matrix or data.frame example where « complete.cases(X)== T » or « complete.cases(X)== TRUE » would give some unwanted result ? The standard problem with T for TRUE is if T has been used for some other purpose, like a time variable. E.g., T - 0 ; complete.cases(X)==T. complete.cases()==TRUE is just silly, like (x==0)==TRUE or ((x==0)==TRUE)==TRUE). (However, notice that x==TRUE is different from as.logical(x) if x is numeric, so ifelse(x,y,z) may differ from ifelse(x==TRUE,y,z).) -- Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@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.
[R] POSIX and ecdf()
Below is some working code that, generally speaking, accomplishes why I want,
but am looking for a necessary improvement in the final step. The code below
scrapes data from a website (thousands of pages actually) and organizes
athlete�s scores in a data frame. The final variable, called Workout05 in the
original data is a timed event. So, I use strplit() to pull out the data I want
in that column and format it using as.POSIXct() as you can see in the code
below (using a regular expression I�m sure would improve on how to pull out
those data in the column, but that is not my primary question).
After I have all data, I want to find the empirical CDF of the data, so I use
ecdf() on those data just as I would on other variables. Now, the main issue
I�m interested is in the final step where you plug in a specific time to find
its percentile
## These are below in context of the real problem as well
fn - ecdf(dat$score5)
fn(dat$score5[1])
This works, but not in the way I want. What I want is for a user to easily be
able to enter their time in �lay� terms such as 5:35 and from that it would
return the percentile rank.
So, I�d like something like the following to be able to work
fn(5:35)
The larger context for this problem for why I want this can be seen if you
visit my web app built using shiny. I�ve built a site where athletes can build
customized reports based on their performance on certain events by entering in
data. This specific issue would be found on the �get my percentile� tab where a
user can use the text input box to enter their time in a way humans typically
understand it and then it gets passed to the R fn() function that runs in the
background and builds the plot for them.
https://hdoran.shinyapps.io/openAnalysis/
So, my question is how can I structure this such that a time can be expressed
as simply minute:seconds (e.g., 4:52) in a text box so that it would still work
to return a percentile rank as I�ve described here.
Thanks
library(XML)
i = 1; j = 0; division = 1
url -
paste(paste('http://games.crossfit.com/scores/leaderboard.php?stage=5sort=0page=',
i, sep=''), paste('division=1region=', j, sep=''),
'numberperpage=100competition=0frontpage=0expanded=1year=15full=1showtoggles=0hidedropdowns=0showathleteac=1=is_mobile=0',
sep='')
tmp - try(readHTMLTable(readLines(url), which=1, header=TRUE))
if(!is.null(dim(tmp))){ # new part here
names(tmp) - gsub(\\n, , names(tmp))
names(tmp) - gsub( +, , names(tmp))
tmp[] - lapply(tmp, function(x) gsub(\\n, , x))
tmp$region - j
}
dat - tmp
aa - strsplit(dat$Workout05, split = '\\(')
bb - sapply(aa, function(x) x[2])
aa - strsplit(bb, split = '\\)')
dat$score5 - as.character(sapply(strsplit(bb, split = '\\)'), function(x) x))
dat$score5 - as.POSIXct(dat$score5, format=%M:%S)
fn - ecdf(dat$score5)
fn(dat$score5[1])
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Re: [R-es] Comparaciones múltiples
Hola, Quizas puedas considerar tus contrastes como comparaciones o contrastes planificados (planned contrasts o creo que tambi�n llamados por algunos a-priori contrasts). Una b�squeda r�pida da algunos buenos ejemplos: https://www.google.es/search?q=r+anova+planned+contrasts Por ejemplo http://faculty.smu.edu/kyler/courses/7311/planned_4up.pdf Saludos, Pedro El 30/03/2015 a las 12:00, r-help-es-requ...@r-project.orgmailto:r-help-es-requ...@r-project.org escribi�: Message: 2 Date: Sat, 28 Mar 2015 23:10:04 +0100 From: Carlos Hern�ndez-Castellano carlos.hernandezcastell...@gmail.commailto:carlos.hernandezcastell...@gmail.com To: r-help-es@r-project.orgmailto:r-help-es@r-project.org Subject: [R-es] Comparaciones m�ltiples Message-ID: cafqssgjvun3tssut8gyo0ggaktd8mymszld2stnbwfw6o9f...@mail.gmail.commailto:cafqssgjvun3tssut8gyo0ggaktd8mymszld2stnbwfw6o9f...@mail.gmail.com Content-Type: text/plain; charset=UTF-8 Saludos, tengo una base de datos con tres variables: especie, tratamiento y abundancia. Quiero hacer comparaciones m�ltiples con Tukey. Ya hab�a usado el comando TukeyHSD(aov(x~y)). La cuesti�n esque ahora tengo varios niveles del factor especie y varios niveles del factor tratamiento. Si hago TukeyHSD(aov(abundancia~especie:tratamiento)) me salen todas las comparaciones posibles, pero yo s�lo estoy interesado en que para cada especie (i.e. para cada nivel del factor especie) me salgan las comparaciones de los datos de abundancia para cada par de tratamientos (niveles del factor tratamiento). Una soluci�n laboriosa ser�a partir el documento original en tantos como especies, pero seguro que alguien me puede sugerir una soluci�n m�s inteligente. Saludos y gracias, -- *?* *Carlos Hern�ndez-Castellano* Environmental Scientist Student, MSc Terrestrial Ecology and Biodiversity Management Research Collaborator at Centre of Ecological Research and Forestry Applications (CREAF) Ecology Unit - Department of Animal Biology, Plant Biology and Ecology; Faculty of Sciences, Autonomous University of Barcelona (UAB) 08193 Bellaterra, Spain Email: carlos.hernandezcastell...@gmail.commailto:carlos.hernandezcastell...@gmail.com [[alternative HTML version deleted]] -- Message: 3 Date: Sat, 28 Mar 2015 23:31:44 + From: V�ctor Granda Garc�a victorgrandagar...@gmail.commailto:victorgrandagar...@gmail.com To: Carlos Hern�ndez-Castellano carlos.hernandezcastell...@gmail.commailto:carlos.hernandezcastell...@gmail.com, r-help-es@r-project.orgmailto:r-help-es@r-project.org Subject: Re: [R-es] Comparaciones m�ltiples Message-ID: caopgseatessjj0kju0y7dumtuomd8zr+a3bov3hy3tmppn4...@mail.gmail.commailto:caopgseatessjj0kju0y7dumtuomd8zr+a3bov3hy3tmppn4...@mail.gmail.com Content-Type: text/plain; charset=UTF-8 Hola Carlos. La orden TukeyHSD tiene un par�metro which que te permite indicar para que variable explicativa quieres las comparaciones. Mira la ayuda de TukeyHSD con '?TukeyHSD' donde tienes un ejemplo, aunque en tu caso ser�a algo parecido a esto: modelo - aov(abundancia~especie:tratamiento) TukeyHSD(modelo, tratamiento) Espero que te sirva, un saludo. El s�b., 28 de marzo de 2015 a las 23:10, Carlos Hern�ndez-Castellano ( carlos.hernandezcastell...@gmail.commailto:carlos.hernandezcastell...@gmail.com) escribi�: Saludos, tengo una base de datos con tres variables: especie, tratamiento y abundancia. Quiero hacer comparaciones m�ltiples con Tukey. Ya hab�a usado el comando TukeyHSD(aov(x~y)). La cuesti�n esque ahora tengo varios niveles del factor especie y varios niveles del factor tratamiento. Si hago TukeyHSD(aov(abundancia~especie:tratamiento)) me salen todas las comparaciones posibles, pero yo s�lo estoy interesado en que para cada especie (i.e. para cada nivel del factor especie) me salgan las comparaciones de los datos de abundancia para cada par de tratamientos (niveles del factor tratamiento). Una soluci�n laboriosa ser�a partir el documento original en tantos como especies, pero seguro que alguien me puede sugerir una soluci�n m�s inteligente. Saludos y gracias, -- ** *Carlos Hern�ndez-Castellano* Environmental Scientist Student, MSc Terrestrial Ecology and Biodiversity Management Research Collaborator at Centre of Ecological Research and Forestry Applications (CREAF) Ecology Unit - Department of Animal Biology, Plant Biology and Ecology; Faculty of Sciences, Autonomous University of Barcelona (UAB) 08193 Bellaterra, Spain Email: carlos.hernandezcastell...@gmail.commailto:carlos.hernandezcastell...@gmail.com [[alternative HTML version deleted]] ___ R-help-es mailing list R-help-es@r-project.orgmailto:R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es [[alternative
Re: [R] matrix manipulation question
Le 27 mars 2015 � 18:01, David Winsemius dwinsem...@comcast.net a �crit : On Mar 27, 2015, at 3:41 AM, St�phane Adamowicz wrote: Well, it seems to work with me. No one is doubting that it worked for you in this instance. What Peter D. was criticizing was the construction : complete.cases(t(Y))==T ... and it was on two bases that it is wrong. The first is that `T` is not guaranteed to be TRUE. The second is that the test ==T (or similarly ==TRUE) is completely unnecessary because `complete.cases` returns a logical vector and so that expression is a waste of time. Indeed, You are right, the following code was enough : � Z - Y[, complete.cases(t(Y) ] � However, in order to help me understand, would you be so kind as to give me a matrix or data.frame example where � complete.cases(X)== T � or � complete.cases(X)== TRUE � would give some unwanted result ? St�phane [[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] Fwd: non-conformable arguments
Dear Thierry Onkelinx, Thank you so much to answer me. But I do not know what information I should send for you I want to run neural network on my data.I run these codes and I saw this Error in the last line #load mydata dim(mydata) # 20 3111 library(neuralnet) fm - as.formula(paste(resp ~, paste(colnames(mydata)[1:3110], collapse=+))) out - neuralnet(fm,data=mydata, hidden = 4, lifesign = minimal, linear.output = FALSE, threshold = 0.1) Call: neuralnet(formula = fm, data = mydata, hidden = 4, threshold = 0.1, lifesign = minimal, linear.output = FALSE) 1 repetition was calculated. Error Reached Threshold Steps 1 2.402157856 0.06958374995 9 #load testset dim(testset) # 20 3111 out.results - compute(out, testset) Error in neurons[[i]] %*% weights[[i]] : non-conformable arguments Regards, Soheila [[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] non-conformable arguments
We need enough information to run your code on our computer and get the same error as you. In this case we are missing the data. See e.g. http://adv-r.had.co.nz/Reproducibility.html If you can't provide the original data, then try to make a (small) example which reproduces your problem. ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest team Biometrie Kwaliteitszorg / team Biometrics Quality Assurance Kliniekstraat 25 1070 Anderlecht Belgium To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey 2015-03-30 14:47 GMT+02:00 Soheila Khodakarim lkhodaka...@gmail.com: Dear Thierry Onkelinx, Thank you so much to answer me. But I do not know what information I should send for you I want to run neural network on my data.I run these codes and I saw this Error in the last line #load mydata dim(mydata) # 20 3111 library(neuralnet) fm - as.formula(paste(resp ~, paste(colnames(mydata)[1:3110], collapse=+))) out - neuralnet(fm,data=mydata, hidden = 4, lifesign = minimal, linear.output = FALSE, threshold = 0.1) Call: neuralnet(formula = fm, data = mydata, hidden = 4, threshold = 0.1, lifesign = minimal, linear.output = FALSE) 1 repetition was calculated. Error Reached Threshold Steps 1 2.402157856 0.06958374995 9 #load testset dim(testset) # 20 3111 out.results - compute(out, testset) Error in neurons[[i]] %*% weights[[i]] : non-conformable arguments Regards, Soheila [[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] A problem someone should know about
Rich, You’ve probably reported the error to the wrong group. A quick search suggests this is not an R issue, but an RStudio issue. The error message is unique enough. Google returns this as the first link: https://support.rstudio.com/hc/communities/public/questions/200807456-Error-when-plotting-graphics-on-Mac-OSX-Mavericks Peter On Mar 29, 2015, at 9:21 PM, Richard M. Heiberger r...@temple.edu wrote: This looks like a specific Macintosh error that appears at random intervals. I get it at random, and unreproducible times. I reported it (or perhaps a close relative) to the r-sig-mac list in September 2014. Rich On Sun, Mar 29, 2015 at 9:59 PM, Rolf Turner r.tur...@auckland.ac.nz wrote: On 30/03/15 11:52, Ian Lester wrote: I’m a novice and this message looks like it shouldn’t be ignored. Someone who knows what they’re doing should probably take a look. Thanks Ian Lester logfat.lm-(lm(body.fat~log(BMI))) plot(logfat) Error in plot(logfat) : object 'logfat' not found plot(logfat.lm) Hit Return to see next plot: Hit Return to see next plot: Hit Return to see next plot: Mar 29 18:10:18 iansimac.gateway rsession[69550] Error: Error: this application, or a library it uses, has passed an invalid numeric value (NaN, or not-a-number) to CoreGraphics API. This is a serious error and contributes to an overall degradation of system stability and reliability. This notice is a courtesy: please fix this problem. It will become a fatal error in an upcoming update. Please make your examples *reproducible* as the posting guide requests. I *presume* that your data are the fat data from the UsingR package, which you did not mention. After installing and loading UsingR I did logfat.lm - lm(body.fat~log(BMI),data=fat) plot(logfat.lm) and got a sequence of plots, with no error thrown. It would appear that whatever is causing the error that you saw is peculiar to your system. cheers, Rolf Turner -- Rolf Turner Technical Editor ANZJS Department of Statistics University of Auckland Phone: +64-9-373-7599 ext. 88276 Home phone: +64-9-480-4619 __ 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-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] changing column labels for data frames inside a list
summary(mygenfreqt)
Length Class Mode
dat1.str 59220 -none- numeric
dat2.str 59220 -none- numeric
dat3.str 59220 -none- numeric
head(mylist[[1]])
1 2 3 4 5 6 7 8 91011
12
L0001.1 0.60 0.500 0.325 0.675 0.600 0.500 0.500 0.375 0.550 0.475 0.350
0.275
L0001.2 0.40 0.500 0.675 0.325 0.400 0.500 0.500 0.625 0.450 0.525 0.650
0.725
I want to change 1:12 to pop1:pop12
mylist- lapply(mylist, function(e) colnames(e) - paste0('pop',1:12))
What this is doing is replacing the data frames with just names
pop1:pop12. I just want to replace the column labels.
Thanks for any suggestions.
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[R] Joint distribution of wind data
* Hi, ** I'm dealing with wind data and I'd like to model their** distribution.** Wind** direction * * are typically following a vonmises distribution and wind speeds **follow a weibull distribution. I'd like to build a joint** distribution** of directions and speeds as a VonMises-Weibull bivariate* * distribution. **An alternative in such cases (i.e., when marginals are available but** the** joint is difficult to postulate) is to use copulas, which can** construct* * multivariate distributions from univariate marginals. Therefore, if anyone has an idea of how to do whether joint distribution or copula forwind data in R,Please, contact with me.* *Best Regards * *Nasr * [[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] vif in package car: there are aliased coefficients in the model
Thanks a lot for the answer and I'm sorry for the silly question! Also thanks for the conceptual advice! It was also a doubt of me and my advisors. Best! 2015-03-28 15:17 GMT-03:00 John Fox j...@mcmaster.ca: Dear Rodolfo, Sending the data helps, though if you had done what I suggested, you would have seen what's going on: snip -- dim(data) [1] 8 8 summary(lm(response_variable ~ predictor_1 + predictor_2 + predictor_3 + predictor_4 + + predictor_5 + predictor_6 + predictor_7, data = data)) Call: lm(formula = response_variable ~ predictor_1 + predictor_2 + predictor_3 + predictor_4 + predictor_5 + predictor_6 + predictor_7, data = data) Residuals: ALL 8 residuals are 0: no residual degrees of freedom! Coefficients: (1 not defined because of singularities) Estimate Std. Error t value Pr(|t|) (Intercept) -5.1905 NA NA NA predictor_1yellow 2.4477 NA NA NA predictor_2fora 6.5056 NA NA NA predictor_2interior 6.0769 NA NA NA predictor_3 0.6750 NA NA NA predictor_4 3.0742 NA NA NA predictor_5 0.6715 NA NA NA predictor_6 -0.9850 NA NA NA predictor_7 NA NA NA NA Residual standard error: NaN on 0 degrees of freedom Multiple R-squared: 1, Adjusted R-squared:NaN F-statistic: NaN on 7 and 0 DF, p-value: NA snip -- So the data set that you're using has 8 cases and 8 variables, one of which is a factor with 3 levels. Consequently, the model you're fitting my LS has 9 coefficients. Necessarily the rank of the model matrix is deficient. When you eliminate a coefficient, you get a perfect fit: 8 coefficients fit to 8 cases with 0 df for error. This is of course nonsense: You don't have enough data to fit a model of this complexity. In fact, you might not have enough data to reasonably fit a model with just 1 predictor. I'm cc'ing this response to the r-help email list, where you started this thread. Best, John On Sat, 28 Mar 2015 12:04:05 -0300 Rodolfo Pelinson rodolfopelin...@gmail.com wrote: Thanks a lot for your answer and your time! But Im still having the same problem. That's the script I am using: library(car) data -read.table(data_vif.txt, header = T, sep = \t, row.names = 1) data vif(lm(response_variable ~ predictor_1 + predictor_2 + predictor_3 + predictor_4 + predictor_5 + predictor_6 + predictor_7, data = data)) vif(lm(response_variable ~ predictor_1 + predictor_2 + predictor_3 + predictor_4 + predictor_5 + predictor_6, data = data)) the first vif function above returns me the following error: Error in vif.default(lm(response_variable ~ predictor_1 + predictor_2 + : there are aliased coefficients in the model Then if I remove any one of the predictors (in the script I removed predictor_7 as an example), it returns this: GVIF Df GVIF^(1/(2*Df)) predictor_1 NaN 1 NaN predictor_2 NaN 2 NaN predictor_3 NaN 1 NaN predictor_4 NaN 1 NaN predictor_5 NaN 1 NaN predictor_6 NaN 1 NaN Warning message: In cov2cor(v) : diag(.) had 0 or NA entries; non-finite result is doubtful Can you help me with this? I even attached to this e-mail my data set. It's a small table. Sorry for the question. 2015-03-27 21:51 GMT-03:00 John Fox j...@mcmaster.ca: Dear Rodolfo, It's apparently the case that at least one of the columns of the model matrix for your model is perfectly collinear with others. There's not nearly enough information here to figure out exactly what the problem is, and the information that you provided certainly falls short of allowing me or anyone else to reproduce your problem and diagnose it properly. It's not even clear from your message exactly what the structure of the model is, although localizacao is apparently a factor with 3 levels. If you look at the summary() output for your model or just print it, you should at least see which coefficients are aliased, and that might help you understand what went wrong. I hope this helps, John --- John Fox, Professor McMaster University Hamilton, Ontario, Canada http://socserv.mcmaster.ca/jfox/ -Original Message- From: R-help [mailto:r-help-boun...@r-project.org] On Behalf
[R] Multiple Plots using ggplot
Dear All, I want to plot multiple using ggplot function from a data frame of many columns. I want to plot only str1, str2 and str3 and I failed to make it. What I want is to compare str1, str2 and str3 by plotting vertical line. I also need to add points to the plot to be able to separate them. Here is how the data look like and how I tried to make it. Date NumberofRaindays TotalRains str1 str2 str3 1/1/1952 86 1360.5 92 120 112 1/1/1953 96 1100 98 100 110 ... ... df1 -data.frame(data) df1 df2 - melt(df1 , id = 'Date', variable_name = 'start of Rains') df2 ggplot(df2, aes(Date,value)) + geom_line(aes(colour =red),type = h) Kindly any help is welcome. Thanks Regards, Frederic. Frederic Ntirenganya Maseno University, African Maths Initiative, Kenya. Mobile:(+254)718492836 Email: fr...@aims.ac.za https://sites.google.com/a/aims.ac.za/fredo/ [[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] Ayuda con un árbol de regresión
Miguel, Sin tus datos es imposible brindarte una respuesta precisa pero ten en cuenta que la descripción del argumento 'fórmula' en la función tree del paquete del mismo nombre dice que la variable dependiente -PRECIO en tu caso- debe ser un vector numérico si quieres estimar un árbol de regressión o un factor si deseas producir un árbol de clasificación. No veo términos de interacción en tu fórmula, cuya presencia no es permitida, por lo que te sugiero revises qué clase de objeto es PRECIO, para lo cual puedes escribir str(CARROST$PRECIO). Puede que necesites modificar la clase a vector numérico. Without any of your data we can't give you a precise answer, but take into account that the description of the formula argument of the tree function (package tree) reads: The left-hand-side (response) should be either a numerical vector when a regression tree will be fitted or a factor, when a classification tree is produced. The right-hand-side should be a series of numeric or factor variables separated by +; there should be no interaction terms. Your formula contains no interaction terms, so I'd suggest you should check what class of object PRECIO is -type str(CARROST$PRECIO). You may need to change it to a numerical vector. José -Original Message- From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Michael Dewey Sent: 24 March 2015 18:16 To: Miguel angel Lopez Martinez; r-help@r-project.org Subject: Re: [R] Ayuda con un árbol de regresión Miguel, si prefieres escribir en espanol https://stat.ethz.ch/mailman/listinfo/r-help-es (This is the address of the Spanish language mailing list) On 23/03/2015 20:13, Miguel angel Lopez Martinez wrote: ARBOL-tree(PRECIO~CILINDRAJE_DEL_MOTOR+MODELO,data=CARROST) ARBOL plot(ARBOL) text(ARBOL,cex=0.9) este es el código que utilizamos las variables cilindraje y modelo son categóricas transformadas por que si utilizamos las varibales como factor no las reconoce [[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. - No virus found in this message. Checked by AVG - www.avg.com Version: 2015.0.5751 / Virus Database: 4315/9373 - Release Date: 03/24/15 -- Michael http://www.dewey.myzen.co.uk __ 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. Age UK Group Age UK is a registered charity and company limited by guarantee, (registered charity number 1128267, registered company number 6825798) Registered office: Tavis House, 1-6 Tavistock Square, London WC1H 9NA. For the purposes of promoting Age UK Insurance, Age UK is an Appointed Representative of Age UK Enterprises Limited. Age UK Enterprises Limited is authorised and regulated by the Financial Conduct Authority. Charitable Services are offered through Age UK (the Charity) and commercial products and services are offered by the Charity’s subsidiary companies. The Age UK Group comprises of Age UK, and its subsidiary companies and charities, dedicated to improving the lives of people in later life. Our network includes the three national charities Age Cymru, Age NI and Age Scotland and more than 160 local Age UK charities. This email and any files transmitted with it are confidential and intended solely for the use of the individual or entity to whom they are addressed. If you receive a message in error, please advise the sender and delete immediately. Except where this email is sent in the usual course of our business, any opinions expressed in this email are those of the author and do not necessarily reflect the opinions of Age UK or its subsidiaries and associated companies. Age UK monitors all e-mail transmissions passing through its network and may block or modify mails which are deemed to be unsuitable. __ 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] data.frame: data-driven column selections that vary by row??
Sorry if that's confusing: I'm probably confused. :-( I am collecting and trying to analyze data regarding performance of computer systems. After extracting the data from its repository, I have created and used a Perl script to generate a (relatively) simple CSV, each record of which contains: * a POSIXct timestamp * a hostname * a collection of metrics for the interval identified by the timestamp, and specific to the host in question, as well as some factors to group the hosts (e.g., whether it's in a control vs. a test group; a broad categorization of how the host is provisioned; which version of the software it was running at the time...). (Each metric and factor is in a uniquely-named column.) As extracted from the repository, there were several records for each such hostname/timestamp pair -- e.g., there would be separate records for: * Input bandwidth utilization for network interface 1 * Output bandwidth utilization for network interface 1 * Input bandwidth utilization for network interface 2 * Output bandwidth utilization for network interface 2 (And the same field would be used for each of these -- the interpretation being driven by the content of other fields in teh record.) Working with the data as described (immediately) above directly in R seemed... daunting, at best: thus the excursion into Perl. And for some of the data, what I have works well enough. But now I also want to analyze information from disk drives, and things get messy (as far as I can see). First, each disk drive has a collection of 17 metrics (such as busy_pct, kb_per_transfer_read, and transfers_per_second_write), as well as a factor (dev_type). Each also has a device name that is unique within the host where it resides (e.g. da1, da2, da3). (The dev_type factor identifies whether the drive is a solid-state device or a spinning disk.) I have thus made the corresponding columns unique by pasting the drive name and the name of the metric (or factor), separating the two with _ (e.g. da7_busy_pct; ada0_mb_per_second_write; ada4_queue_length). I am not certain that's the best thing I could have done -- and I'm open to changing the approach. The challenge for me is that different (classes of) machines are provisioned differently; some consequennces of that: * While da1 may be a spinning disk on host A, that has no bearing on whether or not the da1 on host B is a spinning disk or an SSD. * Host C may not even have a da1 device. * Host D may be of a type that normally has a da1, but in this case, the drive has failed and has been disabled (so host D won't report anything about da1). (I'm not too bothered about the non-reporting case, but cite it so we all know about it.) I expect I will want to be using groupings: * All disk devices -- this one is easy. * All SSD devices (excluding spinning disks). * All spinning disks (excluding SSDs). I'm having trouble with the latter two (though, certainly, if I solve one, the other is also solved). Also, for some of the metrics, I will want to sum them; for others, I will want to do other things -- find minima or maxima, or average them. So pre-calculating such aggregates in the Perl script isn't something that appeals to me. Finally (as far as complications go), I'm trying to write the code in such a way that if we deploy a new configuration of machine that has (say) twice as many drives as the biggest one we presently deploy, the code Just Works -- I shouldn't need to update the code merely to adapt to another hardware configuration. I have been able to write a function that takes the data.frame obtained by reading the above-cited CSV, and generates a data.frame with a row for each host, and depicts the dev_type for each device for that host; here's an abbreviated (and slightly redacted) copy of its output to illustrate some of the above: ada0 ada1 ada2 ada3 ada4 ada5 da30 da31 da32 da33 da34 da35 da36 da3 host_A ssd ssd hdd hdd hdd hdd hdd hdd hdd hdd hdd hdd hdd hdd host_B ssd ssd hdd hdd hdd hdd hdd hdd hdd hdd hdd hdd hdd hdd host_G ssd ssd ssd ssd ssd ssdssd host_H ssd ssd ssd ssd ssd ssdssd host_M ssd ssd ssd ssd ssd ssdssd host_N ssd ssd ssd ssd ssd ssdssd (That function is written with the explicit assumption(!) that for the period covered by a given set of input data, a given host's configuration remains static: we won't have drives changing type mid-stream.) So the point of this lengthy(!) note is to ask if there's a somewhat-sane way to be able to group the metrics for the ssd devices (for example), given the above. (So far, the least obnoxious way that comes to mind is to actually create 2 columns for each device metric: one for the device if it's an ssd;l the other for hdd -- so instead of columns such as: * da3_busy_pct * da3_dev_type *
Re: [R] generating phi using function()
Your function phi has 5 arguments with no defaults. Your call only has 3
arguments. Hence the error message.
phi - function(w1, w2, j, k, K){
+ zaehler - (k/K)^(w1-1)*(1-k/K)^(w2-1)
+ nenner - sum( ((1:K)/K)^(w1-1)*(1-(1:K)/K)^(w2-1))
+ return( zaehler/nenner )
+ }
phi(c(1, 1), 44L, 1)
Error in phi(c(1, 1), 44L, 1) : argument k is missing, with no default
T.Riedle tr...@kent.ac.uk
Sent by: R-help r-help-boun...@r-project.org
03/29/2015 08:59 AM
To
r-help@r-project.org r-help@r-project.org,
cc
Subject
[R] generating phi using function()
Hi everybody,
I am trying to generate the formula shown in the attachment. My formula so
far looks as follows:
phi - function(w1, w2, j, k, K){
zaehler - (k/K)^(w1-1)*(1-k/K)^(w2-1)
nenner - sum( ((1:K)/K)^(w1-1)*(1-(1:K)/K)^(w2-1))
return( zaehler/nenner )
}
Unfortunately something must be wrong here as I get the following message
when running a midas regression
m22.phi- midas_r(rv~mls(rvh,1:max.lag+h1,1,phi), start =
list(rvh=c(1,1)))
Error in phi(c(1, 1), 44L, 1) : argument K is missing, with no default
Called from: .rs.breakOnError(TRUE)
Browse[1] K-125
Browse[1] 125
Could anybody look into my phi formula and tell me what is wrong with it?
Thanks in advance.
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[R] Again: A problem someone should know about
i have no idea what to do plot(body.fat, BMI,xlab=Body fat,ylab=BMI,main=“Figure 2.1: BMI vs Body fat (n=252)”) Error: unexpected input in plot(body.fat, BMI,xlab=Body fat,ylab=BMI,main=� plot(body.fat, BMI,xlab=Body fat,ylab=BMI) serious error. This application, or a library it uses, is using an invalid context and is thereby contributing to an overall degradation of system stability and reliability. This notice is a courtesy: please fix this problem. It will become a fatal error in an upcoming update. Begin forwarded message: From: Ian Lester ihles...@mensa.org.au Reply-To: ihles...@mensa.org.au Subject: A problem someone should know about Date: 30 March 2015 9:52:54 am AEDT To: r-help@r-project.org I’m a novice and this message looks like it shouldn’t be ignored. Someone who knows what they’re doing should probably take a look. Thanks Ian Lester logfat.lm-(lm(body.fat~log(BMI))) plot(logfat) Error in plot(logfat) : object 'logfat' not found plot(logfat.lm) Hit Return to see next plot: Hit Return to see next plot: Hit Return to see next plot: Mar 29 18:10:18 iansimac.gateway rsession[69550] Error: Error: this application, or a library it uses, has passed an invalid numeric value (NaN, or not-a-number) to CoreGraphics API. This is a serious error and contributes to an overall degradation of system stability and reliability. This notice is a courtesy: please fix this problem. It will become a fatal error in an upcoming update. [[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] non-conformable arguments
Dear All, I want to run neural network on my data. i run these codes: #load mydata dim(mydata) # 20 3111 library(neuralnet) fm - as.formula(paste(resp ~, paste(colnames(mydata)[1:3110], collapse=+))) out - neuralnet(fm,data=mydata, hidden = 4, lifesign = minimal, linear.output = FALSE, threshold = 0.1) #load testset dim(testset) # 20 3111 out.results - compute(out, testset) Error in neurons[[i]] %*% weights[[i]] : non-conformable arguments what should I do now? Regards, Soheila [[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-es] Comparaciones múltiples
Hola Miguel yo tengo una función para hacer comparaciones multiples ( muy
util si hacer medidas repetidas, ya que en el otro caso tienes la función
HSD)
t.test.Comparison.Function.Ch - function( data, StringResponse,
StringFactor){
# Factor
data[,StringFactor ] - factor(data[,StringFactor ] )
# Create data.frame
T.Test - data.frame(Name = NA, LevelOne =NA, LevelTwo= NA,statistic =
NA, df = NA, df = NA, p.value = NA )
T.Test - T.Test[-1,]
NumberOfFactors - length(levels(data[, StringFactor]) )
# t.test computation
for(i in 1:( NumberOfFactors - 1 ) ){
LevelOne - levels(data[, StringFactor])[i]
for(j in (i+1):( NumberOfFactors) ){
LevelTwo - levels(data[, StringFactor])[j]
x -t.test( x =data[data[, StringFactor] == LevelOne, StringResponse]
,
y =data[data[, StringFactor] == LevelTwo, StringResponse]
)
T.Test - rbind( T.Test,
data.frame(Name = paste(LevelOne,LevelTwo, sep= .),
LevelOne = LevelOne,
LevelTwo= LevelTwo,
statistic = x$statistic, df =
x$parameter, p.value = x$p.value )
)
}
}
# Significacion
T.Test$Sign - n.s.
if(dim( T.Test[ T.Test$p.value = 0.1 T.Test$p.value 0.05 ,] )[1]
0 ){
T.Test[ T.Test$p.value = 0.1 T.Test$p.value 0.05 ,]$Sign - ·
}
if(dim( T.Test[ T.Test$p.value = 0.05 T.Test$p.value 0.01 ,] )[1]
0 ){
T.Test[ T.Test$p.value = 0.05 T.Test$p.value 0.01 ,]$Sign - *
}
if(dim( T.Test[ T.Test$p.value = 0.01 T.Test$p.value 0.001 ,] )[1]
0 ){
T.Test[ T.Test$p.value = 0.01 T.Test$p.value 0.001 ,]$Sign - **
}
if(dim( T.Test[ T.Test$p.value = 0.001 ,] )[1] 0 ){
T.Test[ T.Test$p.value = 0.001 ,]$Sign - ***
}
# Bonferroni p.value
T.Test$p.value.Adjusted - p.adjust(T.Test$p.value, method = bonferroni)
# Significacion
T.Test$Sign.Adjusted - n.s.
if(dim( T.Test[ T.Test$p.value.Adjusted = 0.1 T.Test$p.value.Adjusted
0.05 ,] )[1] 0 ){
T.Test[ T.Test$p.value.Adjusted = 0.1 T.Test$p.value.Adjusted
0.05 ,]$Sign.Adjusted - ·
}
if(dim( T.Test[ T.Test$p.value.Adjusted = 0.05
T.Test$p.value.Adjusted 0.01 ,] )[1] 0 ){
T.Test[ T.Test$p.value.Adjusted = 0.05 T.Test$p.value.Adjusted
0.01 ,]$Sign.Adjusted - *
}
if(dim( T.Test[ T.Test$p.value.Adjusted = 0.01
T.Test$p.value.Adjusted 0.001 ,] )[1] 0 ){
T.Test[ T.Test$p.value.Adjusted = 0.01 T.Test$p.value.Adjusted
0.001 ,]$Sign.Adjusted - **
}
if(dim( T.Test[ T.Test$p.value.Adjusted = 0.001 ,] )[1] 0 ){
T.Test[ T.Test$p.value.Adjusted = 0.001 ,]$Sign.Adjusted - ***
}
# Effect Size
t - T.Test$statistic
df -T.Test$df
T.Test$Effect.Size - sqrt( t^2/(t^2+df) )
T.Test$statistic - round( T.Test$statistic, digits = 3)
T.Test$df - round( T.Test$df, digits = 3)
T.Test$p.value - round( T.Test$p.value, digits = 3)
T.Test$p.value.Adjusted - round( T.Test$p.value.Adjusted, digits = 3)
T.Test$Effect.Size - round( T.Test$Effect.Size, digits = 3)
return(T.Test)
}
--
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Re: [R] non-conformable arguments
You should sent us a commented, minimal, self-contained, reproducible code as the posting guide tells you to do. Your code is not self-contained. So we can only speculate what when wrong. ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest team Biometrie Kwaliteitszorg / team Biometrics Quality Assurance Kliniekstraat 25 1070 Anderlecht Belgium To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey 2015-03-30 14:10 GMT+02:00 Soheila Khodakarim lkhodaka...@gmail.com: Dear All, I want to run neural network on my data. i run these codes: #load mydata dim(mydata) # 20 3111 library(neuralnet) fm - as.formula(paste(resp ~, paste(colnames(mydata)[1:3110], collapse=+))) out - neuralnet(fm,data=mydata, hidden = 4, lifesign = minimal, linear.output = FALSE, threshold = 0.1) #load testset dim(testset) # 20 3111 out.results - compute(out, testset) Error in neurons[[i]] %*% weights[[i]] : non-conformable arguments what should I do now? Regards, Soheila [[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. [[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] Multiple Plots using ggplot
Hi Frederic, Can you provide a minimal reproducible example including either real data (dput), or simulated data that mimics your situation? This will allow more people to help. Stephen On Mon, Mar 30, 2015 at 8:39 AM, Frederic Ntirenganya ntfr...@gmail.com wrote: Dear All, I want to plot multiple using ggplot function from a data frame of many columns. I want to plot only str1, str2 and str3 and I failed to make it. What I want is to compare str1, str2 and str3 by plotting vertical line. I also need to add points to the plot to be able to separate them. Here is how the data look like and how I tried to make it. Date NumberofRaindays TotalRains str1 str2 str3 1/1/1952 86 1360.5 92 120 112 1/1/1953 96 1100 98 100 110 ... ... df1 -data.frame(data) df1 df2 - melt(df1 , id = 'Date', variable_name = 'start of Rains') df2 ggplot(df2, aes(Date,value)) + geom_line(aes(colour =red),type = h) Kindly any help is welcome. Thanks Regards, Frederic. Frederic Ntirenganya Maseno University, African Maths Initiative, Kenya. Mobile:(+254)718492836 Email: fr...@aims.ac.za https://sites.google.com/a/aims.ac.za/fredo/ [[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. -- Stephen Sefick ** Auburn University Biological Sciences 331 Funchess Hall Auburn, Alabama 36849 ** sas0...@auburn.edu http://www.auburn.edu/~sas0025 ** Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis A big computer, a complex algorithm and a long time does not equal science. -Robert Gentleman [[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] Again: A problem someone should know about
On Mon, Mar 30, 2015 at 1:59 AM, Ian Lester ianleste...@gmail.com wrote: i have no idea what to do One of the other responses already told you it was probably an Rstudio problem, so not using Rstudip seems like a good place to start. Sarah plot(body.fat, BMI,xlab=Body fat,ylab=BMI,main=“Figure 2.1: BMI vs Body fat (n=252)”) Error: unexpected input in plot(body.fat, BMI,xlab=Body fat,ylab=BMI,main=� plot(body.fat, BMI,xlab=Body fat,ylab=BMI) serious error. This application, or a library it uses, is using an invalid context and is thereby contributing to an overall degradation of system stability and reliability. This notice is a courtesy: please fix this problem. It will become a fatal error in an upcoming update. Begin forwarded message: __ 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] changing column labels for data frames inside a list
Assuming that the elements of mylist are data frames, try this:
mylist - lapply(mylist, function(e) { names(e) - paste0('pop',1:12) ; e})
With certain exceptions, the result of a function is the result of the
last expression in the function body. As you defined it, the last
expression was
colnames(e) - paste0('pop',1:12)
that is, the column names (not labels, but names).
If the elements really are data frames, then names() can be used instead
of colnames(), but colnames() is ok. I don't know if one of them is better
than the other for data frames.
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 3/30/15, 6:54 AM, Vikram Chhatre crypticline...@gmail.com wrote:
summary(mygenfreqt)
Length Class Mode
dat1.str 59220 -none- numeric
dat2.str 59220 -none- numeric
dat3.str 59220 -none- numeric
head(mylist[[1]])
1 2 3 4 5 6 7 8 91011
12
L0001.1 0.60 0.500 0.325 0.675 0.600 0.500 0.500 0.375 0.550 0.475 0.350
0.275
L0001.2 0.40 0.500 0.675 0.325 0.400 0.500 0.500 0.625 0.450 0.525 0.650
0.725
I want to change 1:12 to pop1:pop12
mylist- lapply(mylist, function(e) colnames(e) - paste0('pop',1:12))
What this is doing is replacing the data frames with just names
pop1:pop12. I just want to replace the column labels.
Thanks for any suggestions.
[[alternative HTML version deleted]]
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Re: [R] changing column labels for data frames inside a list
I am not sure it would do it since there is no reproducible example, but
try names() instead of colnames().
HTH,
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 30/03/15 15:54, Vikram Chhatre a écrit :
summary(mygenfreqt)
Length Class Mode
dat1.str 59220 -none- numeric
dat2.str 59220 -none- numeric
dat3.str 59220 -none- numeric
head(mylist[[1]])
1 2 3 4 5 6 7 8 91011
12
L0001.1 0.60 0.500 0.325 0.675 0.600 0.500 0.500 0.375 0.550 0.475 0.350
0.275
L0001.2 0.40 0.500 0.675 0.325 0.400 0.500 0.500 0.625 0.450 0.525 0.650
0.725
I want to change 1:12 to pop1:pop12
mylist- lapply(mylist, function(e) colnames(e) - paste0('pop',1:12))
What this is doing is replacing the data frames with just names
pop1:pop12. I just want to replace the column labels.
Thanks for any suggestions.
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] error MANOVA in R
Dear R-usrs, I am trying to perform a MANOVA on a data frame with 31 columns about soil parameters and 1 column containing the explanatory variable (Fraction) that have three levels. my code is the following: datam - read.table(data_manova2.csv, header=T, sep=,) names(datam) manova_fraction2 - manova(cbind(pH, AWC, WEOC, WEN, C.mic, CO2.C, Ca, Mg, K, Na, sol.exch.Fe, easily.reducible.Fe, amourphou.Fe.oxide...Fe.OM, crystalline.Fe.oxides, TN, TOC, NH4.N, NO3.N, N.org, organic.P, avaiable.P, Total.PLFA, Tot.Bat, Gram., Gram..1, Funghi, AMF, protozoa, actinomiceti, non.specifici) ~ as.factor(Fraction), data= datam) summary(manova_fraction2) when I did the summary I got this error summary(manova_fraction2) Error in summary.manova(manova_fraction2) : residuals have rank 18 30 Is this error possibly due to high correlation between my variables? Many thanks in advance, -- Gian [[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] changing column labels for data frames inside a list
Sarah's statement is correct.
So is yours. They are not contradictory, and I believe Sarah's point
was that the OP needed to learn the appropriate syntax.
-- Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom.
Clifford Stoll
On Mon, Mar 30, 2015 at 7:56 AM, Sven E. Templer sven.temp...@gmail.com wrote:
On 30 March 2015 at 16:47, Sarah Goslee sarah.gos...@gmail.com wrote:
colnames(e) - paste0('pop',1:12)
isn't a function and doesn't return anything.
But
function(e){colnames(e) - paste0('pop', 1:2)}
is a function and it returns something (the last evaluated expression! -
here the paste0 return):
mylist2 - lapply(mylist, function(e){colnames(e) - paste0('pop', 1:2)})
mylist2
[[1]]
[1] pop1 pop2
[[2]]
[1] pop1 pop2
[[3]]
[1] pop1 pop2
from ?return:
If the end of a function is reached without calling return, the value of
the last evaluated expression is returned.
mylist - list(
+ data.frame(a = runif(10), b = runif(10)),
+ data.frame(c = runif(10), d = runif(10)),
+ data.frame(e = runif(10), f = runif(10)))
mylist2 - lapply(mylist, function(e){colnames(e) - paste0('pop', 1:2);
e})
colnames(mylist2[[1]])
[1] pop1 pop2
Sarah
On Mon, Mar 30, 2015 at 9:54 AM, Vikram Chhatre
crypticline...@gmail.com wrote:
summary(mygenfreqt)
Length Class Mode
dat1.str 59220 -none- numeric
dat2.str 59220 -none- numeric
dat3.str 59220 -none- numeric
head(mylist[[1]])
1 2 3 4 5 6 7 8 91011
12
L0001.1 0.60 0.500 0.325 0.675 0.600 0.500 0.500 0.375 0.550 0.475 0.350
0.275
L0001.2 0.40 0.500 0.675 0.325 0.400 0.500 0.500 0.625 0.450 0.525 0.650
0.725
I want to change 1:12 to pop1:pop12
mylist- lapply(mylist, function(e) colnames(e) - paste0('pop',1:12))
What this is doing is replacing the data frames with just names
pop1:pop12. I just want to replace the column labels.
Thanks for any suggestions.
--
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.
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Again: A problem someone should know about
In your first example you created logfat.lm and then tried to plot logfat so you got an error indicating that logfat did not exist. In your second example we have no idea what body.fat. You must make your examples reproducible so that we can reproduce your error. It looks like you could also benefit from spending a little time learning about R using a free tutorial. - David L Carlson Department of Anthropology Texas AM University College Station, TX 77840-4352 -Original Message- From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Ian Lester Sent: Monday, March 30, 2015 12:59 AM To: r-help@r-project.org Subject: [R] Again: A problem someone should know about i have no idea what to do plot(body.fat, BMI,xlab=Body fat,ylab=BMI,main=“Figure 2.1: BMI vs Body fat (n=252)”) Error: unexpected input in plot(body.fat, BMI,xlab=Body fat,ylab=BMI,main=� plot(body.fat, BMI,xlab=Body fat,ylab=BMI) serious error. This application, or a library it uses, is using an invalid context and is thereby contributing to an overall degradation of system stability and reliability. This notice is a courtesy: please fix this problem. It will become a fatal error in an upcoming update. Begin forwarded message: From: Ian Lester ihles...@mensa.org.au Reply-To: ihles...@mensa.org.au Subject: A problem someone should know about Date: 30 March 2015 9:52:54 am AEDT To: r-help@r-project.org I’m a novice and this message looks like it shouldn’t be ignored. Someone who knows what they’re doing should probably take a look. Thanks Ian Lester logfat.lm-(lm(body.fat~log(BMI))) plot(logfat) Error in plot(logfat) : object 'logfat' not found plot(logfat.lm) Hit Return to see next plot: Hit Return to see next plot: Hit Return to see next plot: Mar 29 18:10:18 iansimac.gateway rsession[69550] Error: Error: this application, or a library it uses, has passed an invalid numeric value (NaN, or not-a-number) to CoreGraphics API. This is a serious error and contributes to an overall degradation of system stability and reliability. This notice is a courtesy: please fix this problem. It will become a fatal error in an upcoming update. [[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] changing column labels for data frames inside a list
colnames(e) - paste0('pop',1:12)
isn't a function and doesn't return anything.
mylist - list(
+ data.frame(a = runif(10), b = runif(10)),
+ data.frame(c = runif(10), d = runif(10)),
+ data.frame(e = runif(10), f = runif(10)))
mylist2 - lapply(mylist, function(e){colnames(e) - paste0('pop', 1:2); e})
colnames(mylist2[[1]])
[1] pop1 pop2
Sarah
On Mon, Mar 30, 2015 at 9:54 AM, Vikram Chhatre
crypticline...@gmail.com wrote:
summary(mygenfreqt)
Length Class Mode
dat1.str 59220 -none- numeric
dat2.str 59220 -none- numeric
dat3.str 59220 -none- numeric
head(mylist[[1]])
1 2 3 4 5 6 7 8 91011
12
L0001.1 0.60 0.500 0.325 0.675 0.600 0.500 0.500 0.375 0.550 0.475 0.350
0.275
L0001.2 0.40 0.500 0.675 0.325 0.400 0.500 0.500 0.625 0.450 0.525 0.650
0.725
I want to change 1:12 to pop1:pop12
mylist- lapply(mylist, function(e) colnames(e) - paste0('pop',1:12))
What this is doing is replacing the data frames with just names
pop1:pop12. I just want to replace the column labels.
Thanks for any suggestions.
--
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] changing column labels for data frames inside a list
On 30 March 2015 at 16:47, Sarah Goslee sarah.gos...@gmail.com wrote:
colnames(e) - paste0('pop',1:12)
isn't a function and doesn't return anything.
But
function(e){colnames(e) - paste0('pop', 1:2)}
is a function and it returns something (the last evaluated expression! -
here the paste0 return):
mylist2 - lapply(mylist, function(e){colnames(e) - paste0('pop', 1:2)})
mylist2
[[1]]
[1] pop1 pop2
[[2]]
[1] pop1 pop2
[[3]]
[1] pop1 pop2
from ?return:
If the end of a function is reached without calling return, the value of
the last evaluated expression is returned.
mylist - list(
+ data.frame(a = runif(10), b = runif(10)),
+ data.frame(c = runif(10), d = runif(10)),
+ data.frame(e = runif(10), f = runif(10)))
mylist2 - lapply(mylist, function(e){colnames(e) - paste0('pop', 1:2);
e})
colnames(mylist2[[1]])
[1] pop1 pop2
Sarah
On Mon, Mar 30, 2015 at 9:54 AM, Vikram Chhatre
crypticline...@gmail.com wrote:
summary(mygenfreqt)
Length Class Mode
dat1.str 59220 -none- numeric
dat2.str 59220 -none- numeric
dat3.str 59220 -none- numeric
head(mylist[[1]])
1 2 3 4 5 6 7 8 91011
12
L0001.1 0.60 0.500 0.325 0.675 0.600 0.500 0.500 0.375 0.550 0.475 0.350
0.275
L0001.2 0.40 0.500 0.675 0.325 0.400 0.500 0.500 0.625 0.450 0.525 0.650
0.725
I want to change 1:12 to pop1:pop12
mylist- lapply(mylist, function(e) colnames(e) - paste0('pop',1:12))
What this is doing is replacing the data frames with just names
pop1:pop12. I just want to replace the column labels.
Thanks for any suggestions.
--
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.
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Transformation of the circular variable into linear variable
I am working with a circular data ( wind direction ) and i have problem in transforming the circular variable into linear variable. I have found an equation that can be used to convert the circular variable into linear variable which is included in this paper * ARMA based approaches for forecasting the tuple of wind speed and direction. *The equation is called as the inverse of a link function and i have attached the summery of it.T herefore,Please, if you have an idea of how to conduct it in R programming or if you have another method for transformation, let me know it. Your cooperation is highly appreciated. My Best Regards All The Best Nasr [[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] changing column labels for data frames inside a list
First of all, thank you for all the quick replies. Here is a solution that
worked for me.
mylist2 - lapply(mylist, function(e){colnames(e) - paste0('pop',1:12);
return(e)})
head(mylist2[[1]])
pop1 pop2 pop3 pop4 pop5 pop6 pop7 pop8 pop9 pop10 pop11
pop12
L0001.1 0.60 0.500 0.325 0.675 0.600 0.500 0.500 0.375 0.550 0.475 0.350
0.275
L0001.2 0.40 0.500 0.675 0.325 0.400 0.500 0.500 0.625 0.450 0.525 0.650
0.725
While we are at this, I wanted to create a 13th column in each data frame
for average of each row.
# Calculate average
myavg - lapply(mylist2, function(e) rowSums(mylist2)/12)
# Attach to the main data frame
mylist3 - cbind(mylist2, myavg)
This does not work the way I imagined it would. The myavg vector is
attached directly to mylist2, not to individual dataframes within.
p.s. Is it a standard convention to always copy the reply to the last
person who responded?
On Mon, Mar 30, 2015 at 10:56 AM, Sven E. Templer sven.temp...@gmail.com
wrote:
On 30 March 2015 at 16:47, Sarah Goslee sarah.gos...@gmail.com wrote:
colnames(e) - paste0('pop',1:12)
isn't a function and doesn't return anything.
But
function(e){colnames(e) - paste0('pop', 1:2)}
is a function and it returns something (the last evaluated expression! -
here the paste0 return):
mylist2 - lapply(mylist, function(e){colnames(e) - paste0('pop', 1:2)})
mylist2
[[1]]
[1] pop1 pop2
[[2]]
[1] pop1 pop2
[[3]]
[1] pop1 pop2
from ?return:
If the end of a function is reached without calling return, the value of
the last evaluated expression is returned.
mylist - list(
+ data.frame(a = runif(10), b = runif(10)),
+ data.frame(c = runif(10), d = runif(10)),
+ data.frame(e = runif(10), f = runif(10)))
mylist2 - lapply(mylist, function(e){colnames(e) - paste0('pop',
1:2);
e})
colnames(mylist2[[1]])
[1] pop1 pop2
Sarah
On Mon, Mar 30, 2015 at 9:54 AM, Vikram Chhatre
crypticline...@gmail.com wrote:
summary(mygenfreqt)
Length Class Mode
dat1.str 59220 -none- numeric
dat2.str 59220 -none- numeric
dat3.str 59220 -none- numeric
head(mylist[[1]])
1 2 3 4 5 6 7 8 910
11
12
L0001.1 0.60 0.500 0.325 0.675 0.600 0.500 0.500 0.375 0.550 0.475
0.350
0.275
L0001.2 0.40 0.500 0.675 0.325 0.400 0.500 0.500 0.625 0.450 0.525
0.650
0.725
I want to change 1:12 to pop1:pop12
mylist- lapply(mylist, function(e) colnames(e) - paste0('pop',1:12))
What this is doing is replacing the data frames with just names
pop1:pop12. I just want to replace the column labels.
Thanks for any suggestions.
--
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.
[[alternative HTML version deleted]]
__
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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]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] changing column labels for data frames inside a list
On 30 March 2015 at 17:19, Vikram Chhatre crypticline...@gmail.com wrote:
First of all, thank you for all the quick replies. Here is a solution that
worked for me.
mylist2 - lapply(mylist, function(e){colnames(e) - paste0('pop',1:12);
return(e)})
head(mylist2[[1]])
pop1 pop2 pop3 pop4 pop5 pop6 pop7 pop8 pop9 pop10 pop11
pop12
L0001.1 0.60 0.500 0.325 0.675 0.600 0.500 0.500 0.375 0.550 0.475 0.350
0.275
L0001.2 0.40 0.500 0.675 0.325 0.400 0.500 0.500 0.625 0.450 0.525 0.650
0.725
While we are at this, I wanted to create a 13th column in each data frame
for average of each row.
For new problems you should use a new topic.
# Calculate average
myavg - lapply(mylist2, function(e) rowSums(mylist2)/12)
# Attach to the main data frame
mylist3 - cbind(mylist2, myavg)
This does not work the way I imagined it would. The myavg vector is
attached directly to mylist2, not to individual dataframes within.
But it works as expected (read ?cbind).
You try to cbind two lists (myavg and mylist2).
You want to cbind each list object (the data.frames) with each rowSums
output.
So, use cbind within your first lapply.
p.s. Is it a standard convention to always copy the reply to the last
person who responded?
I guess it depends on which answer you refer to.
On Mon, Mar 30, 2015 at 10:56 AM, Sven E. Templer sven.temp...@gmail.com
wrote:
On 30 March 2015 at 16:47, Sarah Goslee sarah.gos...@gmail.com wrote:
colnames(e) - paste0('pop',1:12)
isn't a function and doesn't return anything.
But
function(e){colnames(e) - paste0('pop', 1:2)}
is a function and it returns something (the last evaluated expression! -
here the paste0 return):
mylist2 - lapply(mylist, function(e){colnames(e) - paste0('pop',
1:2)})
mylist2
[[1]]
[1] pop1 pop2
[[2]]
[1] pop1 pop2
[[3]]
[1] pop1 pop2
from ?return:
If the end of a function is reached without calling return, the value of
the last evaluated expression is returned.
mylist - list(
+ data.frame(a = runif(10), b = runif(10)),
+ data.frame(c = runif(10), d = runif(10)),
+ data.frame(e = runif(10), f = runif(10)))
mylist2 - lapply(mylist, function(e){colnames(e) - paste0('pop',
1:2);
e})
colnames(mylist2[[1]])
[1] pop1 pop2
Sarah
On Mon, Mar 30, 2015 at 9:54 AM, Vikram Chhatre
crypticline...@gmail.com wrote:
summary(mygenfreqt)
Length Class Mode
dat1.str 59220 -none- numeric
dat2.str 59220 -none- numeric
dat3.str 59220 -none- numeric
head(mylist[[1]])
1 2 3 4 5 6 7 8 910
11
12
L0001.1 0.60 0.500 0.325 0.675 0.600 0.500 0.500 0.375 0.550 0.475
0.350
0.275
L0001.2 0.40 0.500 0.675 0.325 0.400 0.500 0.500 0.625 0.450 0.525
0.650
0.725
I want to change 1:12 to pop1:pop12
mylist- lapply(mylist, function(e) colnames(e) -
paste0('pop',1:12))
What this is doing is replacing the data frames with just names
pop1:pop12. I just want to replace the column labels.
Thanks for any suggestions.
--
Sarah Goslee
http://www.functionaldiversity.org
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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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]]
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and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] changing column labels for data frames inside a list
On 30 March 2015 at 17:31, Bert Gunter gunter.ber...@gene.com wrote:
Sarah's statement is correct.
So is yours. They are not contradictory, and I believe Sarah's point
was that the OP needed to learn the appropriate syntax.
That's why I pointed to ?return.
Sarah's statement was not so clear (and might have been misleading) for me
regarding the R expertise of the OP.
-- Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom.
Clifford Stoll
On Mon, Mar 30, 2015 at 7:56 AM, Sven E. Templer sven.temp...@gmail.com
wrote:
On 30 March 2015 at 16:47, Sarah Goslee sarah.gos...@gmail.com wrote:
colnames(e) - paste0('pop',1:12)
isn't a function and doesn't return anything.
But
function(e){colnames(e) - paste0('pop', 1:2)}
is a function and it returns something (the last evaluated expression! -
here the paste0 return):
mylist2 - lapply(mylist, function(e){colnames(e) - paste0('pop',
1:2)})
mylist2
[[1]]
[1] pop1 pop2
[[2]]
[1] pop1 pop2
[[3]]
[1] pop1 pop2
from ?return:
If the end of a function is reached without calling return, the value of
the last evaluated expression is returned.
mylist - list(
+ data.frame(a = runif(10), b = runif(10)),
+ data.frame(c = runif(10), d = runif(10)),
+ data.frame(e = runif(10), f = runif(10)))
mylist2 - lapply(mylist, function(e){colnames(e) - paste0('pop',
1:2);
e})
colnames(mylist2[[1]])
[1] pop1 pop2
Sarah
On Mon, Mar 30, 2015 at 9:54 AM, Vikram Chhatre
crypticline...@gmail.com wrote:
summary(mygenfreqt)
Length Class Mode
dat1.str 59220 -none- numeric
dat2.str 59220 -none- numeric
dat3.str 59220 -none- numeric
head(mylist[[1]])
1 2 3 4 5 6 7 8 910
11
12
L0001.1 0.60 0.500 0.325 0.675 0.600 0.500 0.500 0.375 0.550 0.475
0.350
0.275
L0001.2 0.40 0.500 0.675 0.325 0.400 0.500 0.500 0.625 0.450 0.525
0.650
0.725
I want to change 1:12 to pop1:pop12
mylist- lapply(mylist, function(e) colnames(e) - paste0('pop',1:12))
What this is doing is replacing the data frames with just names
pop1:pop12. I just want to replace the column labels.
Thanks for any suggestions.
--
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.
[[alternative HTML version deleted]]
__
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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]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Vennerable plot
Hi all,
I frequently use Vennerable to design diagrams of my data.
However, when I want to make a weighed diagram for a set of 4 samples, type
ChowRuskey, I get a nice view, but the font size of both numbers and sample
names is too small. Furthermore, the place of the sample names isn't always
that good. It places some names in the diagram (always on the same place).
Does anyone know how to
1) Change font size
2) Place labels somewhere else
This is an example of a plot I make:
library(Vennerable)
Vdemo3 - Venn(SetNames = c(KP2,KP15,KP43, KP53), Weight = c(''
= 0, '1000' = 40, '0100' = 54, '0010' = 38, '0001' = 68, '1100' = 0, '1010'
= 0, '0110' = 0, '0011' =1, '0101' = 0, '1001' = 1, '1110' = 0, '1101' = 0,
'1011' = 0, '0111' = 0, '' = 0))
plot(Vdemo3, doWeight = TRUE, type = ChowRuskey)
Many thanks in advance,
Guillaume
--
View this message in context:
http://r.789695.n4.nabble.com/Vennerable-plot-tp4705278.html
Sent from the R help mailing list archive at Nabble.com.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Fwd: changing column labels for data frames inside a list
Sorry, I failed to cc the list.
-- Bert
-- Forwarded message --
From: Bert Gunter bgun...@gene.com
Date: Mon, Mar 30, 2015 at 8:36 AM
Subject: Re: [R] changing column labels for data frames inside a list
To: Vikram Chhatre crypticline...@gmail.com
You really really need to spend (more?) time with a good R tutorial
before posting further. Your questions are entirely due to your
ignorance of standard language syntax that you should learn if you
intend to use R. Of course, feel free to ask for help if there are
places in the tutorials where, after an honest effort, you remain
flummoxed.
Cheers,
Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom.
Clifford Stoll
On Mon, Mar 30, 2015 at 8:19 AM, Vikram Chhatre
crypticline...@gmail.com wrote:
First of all, thank you for all the quick replies. Here is a solution that
worked for me.
mylist2 - lapply(mylist, function(e){colnames(e) - paste0('pop',1:12);
return(e)})
head(mylist2[[1]])
pop1 pop2 pop3 pop4 pop5 pop6 pop7 pop8 pop9 pop10 pop11
pop12
L0001.1 0.60 0.500 0.325 0.675 0.600 0.500 0.500 0.375 0.550 0.475 0.350
0.275
L0001.2 0.40 0.500 0.675 0.325 0.400 0.500 0.500 0.625 0.450 0.525 0.650
0.725
While we are at this, I wanted to create a 13th column in each data frame
for average of each row.
# Calculate average
myavg - lapply(mylist2, function(e) rowSums(mylist2)/12)
# Attach to the main data frame
mylist3 - cbind(mylist2, myavg)
This does not work the way I imagined it would. The myavg vector is
attached directly to mylist2, not to individual dataframes within.
p.s. Is it a standard convention to always copy the reply to the last
person who responded?
On Mon, Mar 30, 2015 at 10:56 AM, Sven E. Templer sven.temp...@gmail.com
wrote:
On 30 March 2015 at 16:47, Sarah Goslee sarah.gos...@gmail.com wrote:
colnames(e) - paste0('pop',1:12)
isn't a function and doesn't return anything.
But
function(e){colnames(e) - paste0('pop', 1:2)}
is a function and it returns something (the last evaluated expression! -
here the paste0 return):
mylist2 - lapply(mylist, function(e){colnames(e) - paste0('pop', 1:2)})
mylist2
[[1]]
[1] pop1 pop2
[[2]]
[1] pop1 pop2
[[3]]
[1] pop1 pop2
from ?return:
If the end of a function is reached without calling return, the value of
the last evaluated expression is returned.
mylist - list(
+ data.frame(a = runif(10), b = runif(10)),
+ data.frame(c = runif(10), d = runif(10)),
+ data.frame(e = runif(10), f = runif(10)))
mylist2 - lapply(mylist, function(e){colnames(e) - paste0('pop',
1:2);
e})
colnames(mylist2[[1]])
[1] pop1 pop2
Sarah
On Mon, Mar 30, 2015 at 9:54 AM, Vikram Chhatre
crypticline...@gmail.com wrote:
summary(mygenfreqt)
Length Class Mode
dat1.str 59220 -none- numeric
dat2.str 59220 -none- numeric
dat3.str 59220 -none- numeric
head(mylist[[1]])
1 2 3 4 5 6 7 8 910
11
12
L0001.1 0.60 0.500 0.325 0.675 0.600 0.500 0.500 0.375 0.550 0.475
0.350
0.275
L0001.2 0.40 0.500 0.675 0.325 0.400 0.500 0.500 0.625 0.450 0.525
0.650
0.725
I want to change 1:12 to pop1:pop12
mylist- lapply(mylist, function(e) colnames(e) - paste0('pop',1:12))
What this is doing is replacing the data frames with just names
pop1:pop12. I just want to replace the column labels.
Thanks for any suggestions.
--
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.
[[alternative HTML version deleted]]
__
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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
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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
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] changing column labels for data frames inside a list
On Mon, Mar 30, 2015 at 11:43 AM, Sven E. Templer
sven.temp...@gmail.com wrote:
On 30 March 2015 at 17:31, Bert Gunter gunter.ber...@gene.com wrote:
Sarah's statement is correct.
So is yours. They are not contradictory, and I believe Sarah's point
was that the OP needed to learn the appropriate syntax.
That's why I pointed to ?return.
Sarah's statement was not so clear (and might have been misleading) for me
regarding the R expertise of the OP.
You're right: I totally wasn't clear. I got involved making a working
reproducible example and didn't explain what I was doing. And you are
totally correct, the last expression will be returned. Mea culpa.
Sarah
-- Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom.
Clifford Stoll
On Mon, Mar 30, 2015 at 7:56 AM, Sven E. Templer sven.temp...@gmail.com
wrote:
On 30 March 2015 at 16:47, Sarah Goslee sarah.gos...@gmail.com wrote:
colnames(e) - paste0('pop',1:12)
isn't a function and doesn't return anything.
But
function(e){colnames(e) - paste0('pop', 1:2)}
is a function and it returns something (the last evaluated expression! -
here the paste0 return):
mylist2 - lapply(mylist, function(e){colnames(e) - paste0('pop',
1:2)})
mylist2
[[1]]
[1] pop1 pop2
[[2]]
[1] pop1 pop2
[[3]]
[1] pop1 pop2
from ?return:
If the end of a function is reached without calling return, the value of
the last evaluated expression is returned.
mylist - list(
+ data.frame(a = runif(10), b = runif(10)),
+ data.frame(c = runif(10), d = runif(10)),
+ data.frame(e = runif(10), f = runif(10)))
mylist2 - lapply(mylist, function(e){colnames(e) - paste0('pop',
1:2);
e})
colnames(mylist2[[1]])
[1] pop1 pop2
Sarah
On Mon, Mar 30, 2015 at 9:54 AM, Vikram Chhatre
crypticline...@gmail.com wrote:
summary(mygenfreqt)
Length Class Mode
dat1.str 59220 -none- numeric
dat2.str 59220 -none- numeric
dat3.str 59220 -none- numeric
head(mylist[[1]])
1 2 3 4 5 6 7 8 910
11
12
L0001.1 0.60 0.500 0.325 0.675 0.600 0.500 0.500 0.375 0.550 0.475
0.350
0.275
L0001.2 0.40 0.500 0.675 0.325 0.400 0.500 0.500 0.625 0.450 0.525
0.650
0.725
I want to change 1:12 to pop1:pop12
mylist- lapply(mylist, function(e) colnames(e) -
paste0('pop',1:12))
What this is doing is replacing the data frames with just names
pop1:pop12. I just want to replace the column labels.
Thanks for any suggestions.
--
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] Debug package options
On 30/03/2015 1:50 PM, Keith S Weintraub wrote: Folks, I would like change some of the options for the Tk window that pops up when using the debug package. I know how to change the options: e.g. options(debug.font = Courier 12 italic”). Is there a way to “preset” these in my environment so when debug starts up I have all the options set up the way I want them? Do I do this in a .First file? Does the .First file have to load the debug package every time I start up R? No need to do my work for me. Just point me to the right doc. See the ?Startup help topic. You probably want to use one of the profile files rather than .First, because .First needs to be in a workspace, and you shouldn't be loading a workspace every time. Duncan Murdoch __ 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-es] comparaciones múltiples
Recomiendo que veas el paquete lsmeans. Podes especificar comparaciones dentro factores. Saludos Luciano El 29 de marzo de 2015, 11:36, Miguel Lázaro lazar...@yahoo.es escribió: Hola a todos, yo he tenido el mismo problema y después de hablar con mucha gente que a su vez habló con mucha gente, he optado por la solución laboriosa y parto la matriz general en tantas matrices como niveles tenga el factor -una vez la interacción es significativa. Es un procedimiento laborioso pero incontestable para cualquier revisor, creo. Saludos Miguel El dom, 29/3/15, r-help-es-requ...@r-project.org r-help-es-requ...@r-project.org escribió: Asunto: Resumen de R-help-es, Vol 73, Envío 42 Para: r-help-es@r-project.org Fecha: domingo, 29 de marzo, 2015 11:00 Envíe los mensajes para la lista R-help-es a r-help-es@r-project.org Para subscribirse o anular su subscripción a través de la WEB https://stat.ethz.ch/mailman/listinfo/r-help-es O por correo electrónico, enviando un mensaje con el texto help en el asunto (subject) o en el cuerpo a: r-help-es-requ...@r-project.org Puede contactar con el responsable de la lista escribiendo a: r-help-es-ow...@r-project.org Si responde a algún contenido de este mensaje, por favor, edite la linea del asunto (subject) para que el texto sea mas especifico que: Re: Contents of R-help-es digest Además, por favor, incluya en la respuesta sólo aquellas partes del mensaje a las que está respondiendo. Asuntos del día: 1. Re: Uso de R a través de Telegram (Pedro Herrero Petisco) 2. Comparaciones múltiples (Carlos Hernández-Castellano) 3. Re: Comparaciones múltiples (Víctor Granda García) -- Message: 1 Date: Sat, 28 Mar 2015 12:06:35 +0100 From: Pedro Herrero Petisco pedroherreropeti...@gmail.com To: Ruben Tobalina Ramirez lagrimaescr...@gmail.com Cc: Lista R r-help-es@r-project.org Subject: Re: [R-es] Uso de R a través de Telegram Message-ID: CAM7H6U_EyWwvWKb4STnXWbNys_Z170zeN2+f6dqvz=me6m+...@mail.gmail.com Content-Type: text/plain; charset=UTF-8 Muchas gracias por compartir, la verdad es que está muy chulo :-) El 28/03/2015 09:29, Ruben Tobalina Ramirez lagrimaescr...@gmail.com escribió: Buenas, He descubierto y queria compartirla con vosotros. La verdad es que tiene buena pinta, es curiosa, pero no sé si tiene mucha utilidad. Tele R es una app para usar R a través de Telegram. Básicamente envias el código de R a una cuenta de Telegram que esta conectada a un servidor en la nube con R y te envia a tu telefono los resultados, vamos como si tuvieses R en tu móvil. Os envio su werb por si os interesa: http://telemath.altervista.org/TeleR.html y existe lo mismo para Octave: http://nbviewer.ipython.org/github/CAChemE/lightning-talks/blob/master/2015-02/TeleOctave.ipynb Un saludo Rubén ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es [[alternative HTML version deleted]] -- Message: 2 Date: Sat, 28 Mar 2015 23:10:04 +0100 From: Carlos Hernández-Castellano carlos.hernandezcastell...@gmail.com To: r-help-es@r-project.org Subject: [R-es] Comparaciones múltiples Message-ID: cafqssgjvun3tssut8gyo0ggaktd8mymszld2stnbwfw6o9f...@mail.gmail.com Content-Type: text/plain; charset=UTF-8 Saludos, tengo una base de datos con tres variables: especie, tratamiento y abundancia. Quiero hacer comparaciones múltiples con Tukey. Ya había usado el comando TukeyHSD(aov(x~y)). La cuestión esque ahora tengo varios niveles del factor especie y varios niveles del factor tratamiento. Si hago TukeyHSD(aov(abundancia~especie:tratamiento)) me salen todas las comparaciones posibles, pero yo sólo estoy interesado en que para cada especie (i.e. para cada nivel del factor especie) me salgan las comparaciones de los datos de abundancia para cada par de tratamientos (niveles del factor tratamiento). Una solución laboriosa sería partir el documento original en tantos como especies, pero seguro que alguien me puede sugerir una solución más inteligente. Saludos y gracias, -- *?* *Carlos Hernández-Castellano* Environmental Scientist Student, MSc Terrestrial Ecology and Biodiversity Management Research Collaborator at Centre of Ecological Research and Forestry Applications (CREAF) Ecology Unit - Department of Animal Biology, Plant Biology and Ecology; Faculty of Sciences, Autonomous University of Barcelona (UAB) 08193 Bellaterra, Spain Email: carlos.hernandezcastell...@gmail.com [[alternative HTML version deleted]] --
[R] multinomial probabilities with mlogit
Hello, When fitting a logit multinomial model with mlogit I can retrieve the response probabilities using fit$fitted.values (for a given object fit) However, I am trying to calculate those response probabilities myself using the maximum likelihood estimates (i.e. fit$coefficients) given by mlogit. I have used the model given in Agresti (2002): Prob_j(x) = exp( linearpredictor_j(x) ) / (1 + sum (linearpredictor(x))) Which is for a category j the exponential of the linear predictor for category j divided by 1 + the sum of all logits across categories, aside from the reference category. But I cannot get my fitted probabilities calculated using this equation to match the output of mlogit fit$fitted values. Can anyone tell me how those fitted values are calculated? Or point me to the corresponding documentation (which I cannot seem to find by googling!) Many thanks Ingrid [[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] changing column labels for data frames inside a list
Regarding the averages, someone else mentioned that it's preferred to
start a new question in a new post to the list.
That said, you are confusing inside the list with outside the list.
Try this:
(the following R expression is supposed to be all on one line, but my
email software may cause a line break)
myavgs - lapply(mylist2, function(e) {cbind( e, avgs=rowSums(e)/12)} )
With no example data I can't test it. The brackets {} are not, strictly
speaking, necessary, but I think they help clarify what is inside the
function with what is outside it.
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 3/30/15, 8:19 AM, Vikram Chhatre crypticline...@gmail.com wrote:
First of all, thank you for all the quick replies. Here is a solution
that
worked for me.
mylist2 - lapply(mylist, function(e){colnames(e) - paste0('pop',1:12);
return(e)})
head(mylist2[[1]])
pop1 pop2 pop3 pop4 pop5 pop6 pop7 pop8 pop9 pop10 pop11
pop12
L0001.1 0.60 0.500 0.325 0.675 0.600 0.500 0.500 0.375 0.550 0.475 0.350
0.275
L0001.2 0.40 0.500 0.675 0.325 0.400 0.500 0.500 0.625 0.450 0.525 0.650
0.725
While we are at this, I wanted to create a 13th column in each data frame
for average of each row.
# Calculate average
myavg - lapply(mylist2, function(e) rowSums(mylist2)/12)
# Attach to the main data frame
mylist3 - cbind(mylist2, myavg)
This does not work the way I imagined it would. The myavg vector is
attached directly to mylist2, not to individual dataframes within.
p.s. Is it a standard convention to always copy the reply to the last
person who responded?
On Mon, Mar 30, 2015 at 10:56 AM, Sven E. Templer sven.temp...@gmail.com
wrote:
On 30 March 2015 at 16:47, Sarah Goslee sarah.gos...@gmail.com wrote:
colnames(e) - paste0('pop',1:12)
isn't a function and doesn't return anything.
But
function(e){colnames(e) - paste0('pop', 1:2)}
is a function and it returns something (the last evaluated expression! -
here the paste0 return):
mylist2 - lapply(mylist, function(e){colnames(e) - paste0('pop',
1:2)})
mylist2
[[1]]
[1] pop1 pop2
[[2]]
[1] pop1 pop2
[[3]]
[1] pop1 pop2
from ?return:
If the end of a function is reached without calling return, the value of
the last evaluated expression is returned.
mylist - list(
+ data.frame(a = runif(10), b = runif(10)),
+ data.frame(c = runif(10), d = runif(10)),
+ data.frame(e = runif(10), f = runif(10)))
mylist2 - lapply(mylist, function(e){colnames(e) - paste0('pop',
1:2);
e})
colnames(mylist2[[1]])
[1] pop1 pop2
Sarah
On Mon, Mar 30, 2015 at 9:54 AM, Vikram Chhatre
crypticline...@gmail.com wrote:
summary(mygenfreqt)
Length Class Mode
dat1.str 59220 -none- numeric
dat2.str 59220 -none- numeric
dat3.str 59220 -none- numeric
head(mylist[[1]])
1 2 3 4 5 6 7 8 910
11
12
L0001.1 0.60 0.500 0.325 0.675 0.600 0.500 0.500 0.375 0.550 0.475
0.350
0.275
L0001.2 0.40 0.500 0.675 0.325 0.400 0.500 0.500 0.625 0.450 0.525
0.650
0.725
I want to change 1:12 to pop1:pop12
mylist- lapply(mylist, function(e) colnames(e) -
paste0('pop',1:12))
What this is doing is replacing the data frames with just names
pop1:pop12. I just want to replace the column labels.
Thanks for any suggestions.
--
Sarah Goslee
http://www.functionaldiversity.org
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[R] Debug package options
Folks, I would like change some of the options for the Tk window that pops up when using the debug package. I know how to change the options: e.g. options(debug.font = Courier 12 italic”). Is there a way to “preset” these in my environment so when debug starts up I have all the options set up the way I want them? Do I do this in a .First file? Does the .First file have to load the debug package every time I start up R? No need to do my work for me. Just point me to the right doc. Best, KW __ 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] A problem someone should know about
My error is Mac because I don't use R-Studio. The phrasing of Ian's error is similar to the error I reported and still occasionally get. As I said, it is random and therefore not reproducible. This is consistent with the comments on the rstudio link you pointed us to. Rich On Mon, Mar 30, 2015 at 9:00 AM, Peter Claussen dakotaj...@mac.com wrote: Rich, You’ve probably reported the error to the wrong group. A quick search suggests this is not an R issue, but an RStudio issue. The error message is unique enough. Google returns this as the first link: https://support.rstudio.com/hc/communities/public/questions/200807456-Error-when-plotting-graphics-on-Mac-OSX-Mavericks Peter On Mar 29, 2015, at 9:21 PM, Richard M. Heiberger r...@temple.edu wrote: This looks like a specific Macintosh error that appears at random intervals. I get it at random, and unreproducible times. I reported it (or perhaps a close relative) to the r-sig-mac list in September 2014. Rich On Sun, Mar 29, 2015 at 9:59 PM, Rolf Turner r.tur...@auckland.ac.nz wrote: On 30/03/15 11:52, Ian Lester wrote: I’m a novice and this message looks like it shouldn’t be ignored. Someone who knows what they’re doing should probably take a look. Thanks Ian Lester logfat.lm-(lm(body.fat~log(BMI))) plot(logfat) Error in plot(logfat) : object 'logfat' not found plot(logfat.lm) Hit Return to see next plot: Hit Return to see next plot: Hit Return to see next plot: Mar 29 18:10:18 iansimac.gateway rsession[69550] Error: Error: this application, or a library it uses, has passed an invalid numeric value (NaN, or not-a-number) to CoreGraphics API. This is a serious error and contributes to an overall degradation of system stability and reliability. This notice is a courtesy: please fix this problem. It will become a fatal error in an upcoming update. Please make your examples *reproducible* as the posting guide requests. I *presume* that your data are the fat data from the UsingR package, which you did not mention. After installing and loading UsingR I did logfat.lm - lm(body.fat~log(BMI),data=fat) plot(logfat.lm) and got a sequence of plots, with no error thrown. It would appear that whatever is causing the error that you saw is peculiar to your system. cheers, Rolf Turner -- Rolf Turner Technical Editor ANZJS Department of Statistics University of Auckland Phone: +64-9-373-7599 ext. 88276 Home phone: +64-9-480-4619 __ 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-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] multinomial probabilities with mlogit
Reproducible example??? cheers, Rolf Turner P.S. Where does mlogit come from? Note fortune(182). R. T. On 31/03/15 06:46, Ingrid Charvet wrote: Hello, When fitting a logit multinomial model with mlogit I can retrieve the response probabilities using fit$fitted.values (for a given object fit) However, I am trying to calculate those response probabilities myself using the maximum likelihood estimates (i.e. fit$coefficients) given by mlogit. I have used the model given in Agresti (2002): Prob_j(x) = exp( linearpredictor_j(x) ) / (1 + sum (linearpredictor(x))) Which is for a category j the exponential of the linear predictor for category j divided by 1 + the sum of all logits across categories, aside from the reference category. But I cannot get my fitted probabilities calculated using this equation to match the output of mlogit fit$fitted values. Can anyone tell me how those fitted values are calculated? Or point me to the corresponding documentation (which I cannot seem to find by googling!) -- Rolf Turner Technical Editor ANZJS Department of Statistics University of Auckland Phone: +64-9-373-7599 ext. 88276 Home phone: +64-9-480-4619 __ 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] Help with POSIX
You seem to be trying to make POSIXt into something it isn't, as though this was Excel. First, POSIXt is not the same as numeric. You can convert between them, but they are not the same. If you want to do numeric operations, convert. Second, POSIXt is not time of day only. When you provide only minutes and seconds to strptime, it includes the current date in the result as well. Thus, values converted today will be different than values converted tomorrow. You should probably pick a specific date to include in every time value, and paste it with the actual data before converting. Finally, formatting of all non-character data is always associated with conversion to character. You either need to reconvert on demand or keep a copy of the data around unconverted for use when you need it. Oh, and your use of HTML is leading to corruption of your code. Learn to use your email program so you don't send us garbage to guess at. --- 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 March 30, 2015 6:15:09 PM PDT, Doran, Harold hdo...@air.org wrote: I�m struggling a bit with learning about POSIX objects to do some basic things with objects of this class. Suppose I have the following simple example times - c(03:20, 29:56, 03:30, 21:03, 56:26) aa - strptime(times, %M:%S�) I can do means, and some other basic things, but I cannot correlate the objects with some other variable cor(aa, rnorm(5)) Also, for purposes of a user-interface I have built with shiny, I need for the time to be viewed as simply as minutes:seconds, such as this format(aa, '%M:%S�) But of course after doing this I lose the ability to work with this object as a time variable. Thank you Harold [[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] Help with POSIX
On Tue, Mar 31, 2015 at 01:15:09AM +, Doran, Harold wrote: I?m struggling a bit with learning about POSIX objects to do some basic things with objects of this class. Suppose I have the following simple example times - c(03:20, 29:56, 03:30, 21:03, 56:26) aa - strptime(times, %M:%S?) I can do means, and some other basic things, but I cannot correlate the objects with some other variable cor(aa, rnorm(5)) I suspect you will find that manipulating times as numeric quantities (so you can do arithmetic on them) will be easier if you convert to seconds (possibly using a different vector, leaving times as it was). Also, for purposes of a user-interface I have built with shiny, I need for the time to be viewed as simply as minutes:seconds, such as this format(aa, '%M:%S?) But of course after doing this I lose the ability to work with this object as a time variable. So do that with a copy -- the output of format() need not destroy its input. ... Peace, david -- David H. Wolfskill r...@catwhisker.org Those who murder in the name of God or prophet are blasphemous cowards. See http://www.catwhisker.org/~david/publickey.gpg for my public key. pgpo9NplNcuzH.pgp Description: PGP signature __ 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] Mann-Kendall test for many independent columns data set at a time
I am analyzing trend test using Mann-kendall monotonic trendtest for 10,368
independent grid cell, each grid has 34 years dataset. I supposed to find
Kendall “tau” for each gridcell (each grid has 34 years data). The data is
arranged in column wise (Iattached part of the grid dataset as a sample).
To find Kendall tau, I wrote R script as:
desta-read.csv(rainfall.csv,header=T, sep=,)
require(Kendall)
MK-function(y) {
nc-ncol(y)
MannKendalltau-numeric(nc)
for(i in 2:nc){
MannKendalltau[i]-MannKendall(y[,i])
}
MannKendalltau
}
MK(desta)
The result displayed both “tau” and “2-sided pvalue”. But, I wantonly “tau”
value that is printed in organized way. Anyone can tell me how can Iget orderly
printed “tau” value for allgrid cells at a time.
Thank you for your help in advance,Desta WodeboTU-DresdenGermany
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[R] Plotting using tapply function output
Hello, I am trying to plot the hourly standard deviation of wind speeds from 13 different measured locations over many years. I imported the data using readLines and into a dataframe called finalData. Using tapply, I determined the standard deviation of the windspeed (ws) for each hour (hour) from every location (stn) using this command line: statHour = tapply(finalData$ws,list(finalData$stn,finalData$hour),sd) I want to plot the standard deviation for each hour of the day, with hours as the x-axis and the standard deviation for the y-axis, and each station as a different color. I've managed to get a boxplot of this, but ideally, I'd like a scatter plot to determine the variations between each instrument throughout the day. The boxplot command is this: boxplot(statHour, names=colnames(statHour),xlab='Hour of the Day',ylab='Standard Deviation of Wind Speed') I also tried to make a dataframe of the tapply output but it ends up using the hours as the column names instead of putting it into the dataframe. Please help!! I have R version 3.1.1 Thanks a lot, Alexandra __ 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] Trying to understand a function passed to lapply
Colleagues, I am trying to understand the syntax of a function passed to apply. The code below generates a matrix, and passes the matrix to a function that is called by apply. I don't understand the syntax of the function. In some way the function computes data[,delta]/data[,SE]. I can't understand how the body of the function, x[c1]/x[c2] refers to the columns data and SE of the matrix data. Can someone help me understand the syntax? Thank you, John myfun - function(x, c1, c2) x[c1]/x[c2] apply(data,1,myfun,c1=delta,c2=SE) CODE: data-matrix(data=c(-0.70 ,-0.90, -0.50, 20, 20, -0.30 ,-0.43, -0.17, 43, 43, -0.50 ,-1.05, 0.05, 16, 18, 0.00 ,-0.21, 0.21, 22, 23, -1.30 ,-1.48, -1.12, 28, 32, -0.90 ,-1.01, -0.79, 18, 15, -0.20 ,-0.47, 0.07, 39, 39, -0.30 ,-0.83, 0.23, 27, 27), nrow=8,ncol=5,byrow=TRUE) dimnames(data) - list(NULL,c(delta,low,high,n1,n2)) data CI - data[,high]-data[,low] data - cbind(data,CI) data data - cbind(data,SE=data[,CI]/(4*1.96)) data data - cbind(data,SD=data[,SE]*sqrt(data[,n1]+data[,n2])) data myfun - function(x, c1, c2) x[c1]/x[c2] apply(data,1,myfun,c1=delta,c2=SE) John David Sorkin M.D., Ph.D. Professor of Medicine Chief, Biostatistics and Informatics University of Maryland School of Medicine Division of Gerontology and Geriatric Medicine Baltimore VA Medical Center 10 North Greene Street GRECC (BT/18/GR) Baltimore, MD 21201-1524 (Phone) 410-605-7119 (Fax) 410-605-7913 (Please call phone number above prior to faxing) Confidentiality Statement: This email message, including any attachments, is for the sole use of the intended recipient(s) and may contain confidential and privileged information. Any unauthorized use, disclosure or distribution is prohibited. If you are not the intended recipient, please contact the sender by reply email and destroy all copies of the original message. __ 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] Trying to understand a function passed to lapply
I can't understand how the body of the function, x[c1]/x[c2] refers to
the columns data and SE of the matrix data.
If you put the line 'str(x)' at the start of myfun(), as in
myfun - function(x, c1, c2) {
str(x)
x[c1]/x[c2]
}
you would start to see why it works - extracting a row from a matrix gives
you a vector with names copied from the column names of the matrix.
Thus subscripting with a single name (or number) works.
apply(data,1,myfun,c1=delta,c2=SE)
Named num [1:8] -0.7 -0.9 -0.5 20 20 ...
- attr(*, names)= chr [1:8] delta low high n1 ...
Named num [1:8] -0.3 -0.43 -0.17 43 43 ...
- attr(*, names)= chr [1:8] delta low high n1 ...
Named num [1:8] -0.5 -1.05 0.05 16 18 ...
...
Named num [1:8] -0.3 -0.83 0.23 27 27 ...
- attr(*, names)= chr [1:8] delta low high n1 ...
[1] -13.72 -9.046154 -3.563636 0.00 -28.31 -32.072727
-2.903704 -2.218868
By the way, that call to apply is just a slow way of dividing two columns
of the matrix
data[, delta]/data[, SE]
[1] -13.72 -9.046154 -3.563636 0.00 -28.31 -32.072727
-2.903704 -2.218868
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Mon, Mar 30, 2015 at 6:48 PM, John Sorkin jsor...@grecc.umaryland.edu
wrote:
Colleagues,
I am trying to understand the syntax of a function passed to apply. The
code below generates a matrix, and passes the matrix to a function that is
called by apply. I don't understand the syntax of the function. In some way
the function computes data[,delta]/data[,SE]. I can't understand how
the body of the function, x[c1]/x[c2] refers to the columns data and SE
of the matrix data. Can someone help me understand the syntax?
Thank you,
John
myfun - function(x, c1, c2) x[c1]/x[c2]
apply(data,1,myfun,c1=delta,c2=SE)
CODE:
data-matrix(data=c(-0.70 ,-0.90, -0.50, 20, 20,
-0.30 ,-0.43, -0.17, 43, 43,
-0.50 ,-1.05, 0.05, 16, 18,
0.00 ,-0.21, 0.21, 22, 23,
-1.30 ,-1.48, -1.12, 28, 32,
-0.90 ,-1.01, -0.79, 18, 15,
-0.20 ,-0.47, 0.07, 39, 39,
-0.30 ,-0.83, 0.23, 27, 27),
nrow=8,ncol=5,byrow=TRUE)
dimnames(data) - list(NULL,c(delta,low,high,n1,n2))
data
CI - data[,high]-data[,low]
data - cbind(data,CI)
data
data - cbind(data,SE=data[,CI]/(4*1.96))
data
data - cbind(data,SD=data[,SE]*sqrt(data[,n1]+data[,n2]))
data
myfun - function(x, c1, c2) x[c1]/x[c2]
apply(data,1,myfun,c1=delta,c2=SE)
John David Sorkin M.D., Ph.D.
Professor of Medicine
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology and
Geriatric Medicine
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
Confidentiality Statement:
This email message, including any attachments, is for ...{{dropped:16}}
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Help with POSIX
I�m struggling a bit with learning about POSIX objects to do some basic things with objects of this class. Suppose I have the following simple example times - c(03:20, 29:56, 03:30, 21:03, 56:26) aa - strptime(times, %M:%S�) I can do means, and some other basic things, but I cannot correlate the objects with some other variable cor(aa, rnorm(5)) Also, for purposes of a user-interface I have built with shiny, I need for the time to be viewed as simply as minutes:seconds, such as this format(aa, '%M:%S�) But of course after doing this I lose the ability to work with this object as a time variable. Thank you Harold [[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] Plotting using tapply function output
Hi Alexandra, This produces a rather messy plot, but it might get you started: finalData-data.frame(ws=sample(0:100,1300,TRUE), stn=rep(1:13,each=100),hour=rep(1:24,length.out=1300)) statHour = tapply(finalData$ws,list(finalData$stn,finalData$hour),sd) # open a wide device x11(width=13) # leave room for an external legend par(mar=c(5,4,4,6)) matplot(1:24,t(statHour),type=b, main=Wind speed standard deviation by hour of day, xlab=Hour,ylab=Wind speed standard deviation,lty=1) legend(25.2,40,legend=paste(Stn,1:13,sep=),pch=c(0:9,a,b,c), col=1:13,xpd=TRUE) Jim On Tue, Mar 31, 2015 at 10:07 AM, Alexandra Catena amc5...@gmail.com wrote: Hello, I am trying to plot the hourly standard deviation of wind speeds from 13 different measured locations over many years. I imported the data using readLines and into a dataframe called finalData. Using tapply, I determined the standard deviation of the windspeed (ws) for each hour (hour) from every location (stn) using this command line: statHour = tapply(finalData$ws,list(finalData$stn,finalData$hour),sd) I want to plot the standard deviation for each hour of the day, with hours as the x-axis and the standard deviation for the y-axis, and each station as a different color. I've managed to get a boxplot of this, but ideally, I'd like a scatter plot to determine the variations between each instrument throughout the day. The boxplot command is this: boxplot(statHour, names=colnames(statHour),xlab='Hour of the Day',ylab='Standard Deviation of Wind Speed') I also tried to make a dataframe of the tapply output but it ends up using the hours as the column names instead of putting it into the dataframe. Please help!! I have R version 3.1.1 Thanks a lot, Alexandra __ 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] Trying to understand a function passed to lapply
Hi John,
What happens is that you have passed two named arguments to your function
myfun along with the matrix data. Because these arguments have
associated values (delta, SE), these values are substituted into the
expression like this:
x[delta]/x[SE]
which is the return value of myfun. If you just type in:
data[delta]
data[SE]
you will see the values that are successively selected for the calculation
in myfun
Jim
On Tue, Mar 31, 2015 at 12:48 PM, John Sorkin jsor...@grecc.umaryland.edu
wrote:
Colleagues,
I am trying to understand the syntax of a function passed to apply. The
code below generates a matrix, and passes the matrix to a function that is
called by apply. I don't understand the syntax of the function. In some way
the function computes data[,delta]/data[,SE]. I can't understand how
the body of the function, x[c1]/x[c2] refers to the columns data and SE
of the matrix data. Can someone help me understand the syntax?
Thank you,
John
myfun - function(x, c1, c2) x[c1]/x[c2]
apply(data,1,myfun,c1=delta,c2=SE)
CODE:
data-matrix(data=c(-0.70 ,-0.90, -0.50, 20, 20,
-0.30 ,-0.43, -0.17, 43, 43,
-0.50 ,-1.05, 0.05, 16, 18,
0.00 ,-0.21, 0.21, 22, 23,
-1.30 ,-1.48, -1.12, 28, 32,
-0.90 ,-1.01, -0.79, 18, 15,
-0.20 ,-0.47, 0.07, 39, 39,
-0.30 ,-0.83, 0.23, 27, 27),
nrow=8,ncol=5,byrow=TRUE)
dimnames(data) - list(NULL,c(delta,low,high,n1,n2))
data
CI - data[,high]-data[,low]
data - cbind(data,CI)
data
data - cbind(data,SE=data[,CI]/(4*1.96))
data
data - cbind(data,SD=data[,SE]*sqrt(data[,n1]+data[,n2]))
data
myfun - function(x, c1, c2) x[c1]/x[c2]
apply(data,1,myfun,c1=delta,c2=SE)
John David Sorkin M.D., Ph.D.
Professor of Medicine
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology and
Geriatric Medicine
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
Confidentiality Statement:
This email message, including any attachments, is for ...{{dropped:16}}
__
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] Multiple Plots using ggplot
Hi Stephen, Sorry, the data came in bad way. Here is the head of the data. head(data)Date Number.of.Rain.Days Total.rain Start.of.Rain..i. Start.of.Rain..ii. Start.of.Rain..iii. Start.Rain..iv. 1 1952-01-01 86 1139.95292 239 112 112 2 1953-01-01 96977.64698 98 112 112 3 1954-01-01 114 1382.01492 92 120 120 4 1955-01-01 119 1323.086 100 100 125 174 5 1956-01-01 123 1266.44492 92 119 119 6 1957-01-01 124 1235.96492 92 112 112 Frederic Ntirenganya Maseno University, African Maths Initiative, Kenya. Mobile:(+254)718492836 Email: fr...@aims.ac.za https://sites.google.com/a/aims.ac.za/fredo/ On Mon, Mar 30, 2015 at 5:34 PM, stephen sefick ssef...@gmail.com wrote: Hi Frederic, Can you provide a minimal reproducible example including either real data (dput), or simulated data that mimics your situation? This will allow more people to help. Stephen On Mon, Mar 30, 2015 at 8:39 AM, Frederic Ntirenganya ntfr...@gmail.com wrote: Dear All, I want to plot multiple using ggplot function from a data frame of many columns. I want to plot only str1, str2 and str3 and I failed to make it. What I want is to compare str1, str2 and str3 by plotting vertical line. I also need to add points to the plot to be able to separate them. Here is how the data look like and how I tried to make it. Date NumberofRaindays TotalRains str1 str2 str3 1/1/1952 86 1360.5 92 120 112 1/1/1953 96 1100 98 100 110 ... ... df1 -data.frame(data) df1 df2 - melt(df1 , id = 'Date', variable_name = 'start of Rains') df2 ggplot(df2, aes(Date,value)) + geom_line(aes(colour =red),type = h) Kindly any help is welcome. Thanks Regards, Frederic. Frederic Ntirenganya Maseno University, African Maths Initiative, Kenya. Mobile:(+254)718492836 Email: fr...@aims.ac.za https://sites.google.com/a/aims.ac.za/fredo/ [[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. -- Stephen Sefick ** Auburn University Biological Sciences 331 Funchess Hall Auburn, Alabama 36849 ** sas0...@auburn.edu http://www.auburn.edu/~sas0025 ** Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis A big computer, a complex algorithm and a long time does not equal science. -Robert Gentleman [[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] generating phi using function()
Hi,
I am struggling with following function
phi - function(w1, w2, j, k, K){
+ zaehler - (k/K)^(w1-1)*(1-k/K)^(w2-1)
+ nenner - sum( ((1:K)/K)^(w1-1)*(1-(1:K)/K)^(w2-1))
+ return( zaehler/nenner )
+ }
phi(c(1, 1), 44L, 1)
Error in phi(c(1, 1), 44L, 1) : argument k is missing, with no default
Hence, I have changed the function to
phi - function(w, k, K){
+ w1 - w[1]
+ w2 - w[2]
+ zaehler - (k/K)^(w1-1)*(1-k/K)^(w2-1)
+ nenner - sum( ((1:K)/K)^(w1-1)*(1-(1:K)/K)^(w2-1))
+ return( zaehler/nenner )
+ }
Unfortunately, when running the midas regression I get the following error
message
m22.phi- midas_r(rv~mls(rvh,1:max.lag+h1,1,phi), start = list(rvh=c(1,1)))
Error in X[, inds] %*% fun(st) : non-conformable arguments
I guess the problem is w but I do not find a solution how to produce the
formula shown in the attached file where the exponents are w1 and w2,
respectively.
Thanks for your help
From: jlu...@ria.buffalo.edu [mailto:jlu...@ria.buffalo.edu]
Sent: 30 March 2015 16:01
To: T.Riedle
Cc: r-help@r-project.org; R-help
Subject: Re: [R] generating phi using function()
Your function phi has 5 arguments with no defaults. Your call only has 3
arguments. Hence the error message.
phi - function(w1, w2, j, k, K){
+ zaehler - (k/K)^(w1-1)*(1-k/K)^(w2-1)
+ nenner - sum( ((1:K)/K)^(w1-1)*(1-(1:K)/K)^(w2-1))
+ return( zaehler/nenner )
+ }
phi(c(1, 1), 44L, 1)
Error in phi(c(1, 1), 44L, 1) : argument k is missing, with no default
T.Riedle tr...@kent.ac.ukmailto:tr...@kent.ac.uk
Sent by: R-help
r-help-boun...@r-project.orgmailto:r-help-boun...@r-project.org
03/29/2015 08:59 AM
To
r-help@r-project.orgmailto:r-help@r-project.org
r-help@r-project.orgmailto:r-help@r-project.org,
cc
Subject
[R] generating phi using function()
Hi everybody,
I am trying to generate the formula shown in the attachment. My formula so far
looks as follows:
phi - function(w1, w2, j, k, K){
zaehler - (k/K)^(w1-1)*(1-k/K)^(w2-1)
nenner - sum( ((1:K)/K)^(w1-1)*(1-(1:K)/K)^(w2-1))
return( zaehler/nenner )
}
Unfortunately something must be wrong here as I get the following message when
running a midas regression
m22.phi- midas_r(rv~mls(rvh,1:max.lag+h1,1,phi), start = list(rvh=c(1,1)))
Error in phi(c(1, 1), 44L, 1) : argument K is missing, with no default
Called from: .rs.breakOnError(TRUE)
Browse[1] K-125
Browse[1] 125
Could anybody look into my phi formula and tell me what is wrong with it?
Thanks in advance.
__
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Re: [R] Help with POSIX
Hi, On Mar 30, 2015, at 9:15 PM, Doran, Harold hdo...@air.org wrote: I�m struggling a bit with learning about POSIX objects to do some basic things with objects of this class. Suppose I have the following simple example times - c(03:20, 29:56, 03:30, 21:03, 56:26) aa - strptime(times, %M:%S�) I can do means, and some other basic things, but I cannot correlate the objects with some other variable cor(aa, rnorm(5)) You can cast your POSIXlt values to numeric cor(as.numeric(aa), rnorm(5)) Also, for purposes of a user-interface I have built with shiny, I need for the time to be viewed as simply as minutes:seconds, such as this format(aa, '%M:%S�) But of course after doing this I lose the ability to work with this object as a time variable. You may need to keep a copy of your times in a POSIX or numeric format in addition to converting to character. It's hard to tell without more information. Cheers, Ben Thank you Harold [[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] how to deal with changing weighting functions
Hi everybody, Does anybody have an idea how I can generate tau according to the attached formula? The point is that phi changes with k and I thought I could make it by using a for-function in R but I am not sure how to do that. Could anyone help me? Thanks in advance. __ 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] temporal autocorrelation in MCMCglmm
David Villegas Ríos chirleu at gmail.com writes: Hi, For a number of individuals, I have measured several behavioral traits in the wild. Those traits (e.g. home range) can be estimated on different temporal scales, for example daily, weekly or monthly. I want to estimate repeatability of those traits, assuming that the daily/weekly/monthly measurements represent replicates. I have 3 months (90 days) of data for each trait. Two questions: 1) How can assess if there is temporal autocorrelation in my model? I guess that if I consider daily measurements as replicates (90 replicates), I will have some autocorrelation, but if I use just monthly measurements (3 replicates) maybe I avoid it. 2) How can account for temporal autocorrelation in MCMCglmm? Sorry for this pretty basic questions but I haven't found an answer so far. You'll probably be better off asking this question at r-sig-mixed-models (at) r-project.org. As a first pass, you might be able take the residuals from your fit and use acf() to compute the autocorrelation function. Actually, though, you'll probably be better off fitting a 'null' lme() model (fixed=resid~1, random=~1|individual) and then using the ACF() method (not the same thing as acf()) on the resulting model fit. 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] changing column labels for data frames inside a list
On 30 March 2015 at 17:50, Sarah Goslee sarah.gos...@gmail.com wrote:
On Mon, Mar 30, 2015 at 11:43 AM, Sven E. Templer
sven.temp...@gmail.com wrote:
On 30 March 2015 at 17:31, Bert Gunter gunter.ber...@gene.com wrote:
Sarah's statement is correct.
So is yours. They are not contradictory, and I believe Sarah's point
was that the OP needed to learn the appropriate syntax.
That's why I pointed to ?return.
Sarah's statement was not so clear (and might have been misleading) for
me
regarding the R expertise of the OP.
You're right: I totally wasn't clear. I got involved making a working
reproducible example and didn't explain what I was doing. And you are
totally correct, the last expression will be returned. Mea culpa.
No one to blame but the missing reproducible example, I agree!
Sven
Sarah
-- Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom.
Clifford Stoll
On Mon, Mar 30, 2015 at 7:56 AM, Sven E. Templer
sven.temp...@gmail.com
wrote:
On 30 March 2015 at 16:47, Sarah Goslee sarah.gos...@gmail.com
wrote:
colnames(e) - paste0('pop',1:12)
isn't a function and doesn't return anything.
But
function(e){colnames(e) - paste0('pop', 1:2)}
is a function and it returns something (the last evaluated
expression! -
here the paste0 return):
mylist2 - lapply(mylist, function(e){colnames(e) - paste0('pop',
1:2)})
mylist2
[[1]]
[1] pop1 pop2
[[2]]
[1] pop1 pop2
[[3]]
[1] pop1 pop2
from ?return:
If the end of a function is reached without calling return, the value
of
the last evaluated expression is returned.
mylist - list(
+ data.frame(a = runif(10), b = runif(10)),
+ data.frame(c = runif(10), d = runif(10)),
+ data.frame(e = runif(10), f = runif(10)))
mylist2 - lapply(mylist, function(e){colnames(e) - paste0('pop',
1:2);
e})
colnames(mylist2[[1]])
[1] pop1 pop2
Sarah
On Mon, Mar 30, 2015 at 9:54 AM, Vikram Chhatre
crypticline...@gmail.com wrote:
summary(mygenfreqt)
Length Class Mode
dat1.str 59220 -none- numeric
dat2.str 59220 -none- numeric
dat3.str 59220 -none- numeric
head(mylist[[1]])
1 2 3 4 5 6 7 8 910
11
12
L0001.1 0.60 0.500 0.325 0.675 0.600 0.500 0.500 0.375 0.550 0.475
0.350
0.275
L0001.2 0.40 0.500 0.675 0.325 0.400 0.500 0.500 0.625 0.450 0.525
0.650
0.725
I want to change 1:12 to pop1:pop12
mylist- lapply(mylist, function(e) colnames(e) -
paste0('pop',1:12))
What this is doing is replacing the data frames with just names
pop1:pop12. I just want to replace the column labels.
Thanks for any suggestions.
--
Sarah Goslee
http://www.functionaldiversity.org
[[alternative HTML version deleted]]
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