Re: [R] y axis in a stacked bar plot
Hi Michy, I agree with Jeff that you probably don't want a stacked bar chart. What you describe appears to be a display of relative positions of genetic elements in different strains of mice and the place to ask is most likely the Bioconductor mailing list. Jim On Thu, Mar 26, 2015 at 3:51 PM, Jeff Newmiller jdnew...@dcn.davis.ca.us wrote: I don't understand what you expect. Stacked bar charts add the values together. Perhaps that is not really how you want to represent these data? --- 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 25, 2015 9:29:40 PM PDT, Michelle Simon m.si...@har.mrc.ac.uk wrote: Hello, Below is some simple R code I have used to create a stacked bar chart. However the y-axis tick marks do not reflect the data. The minimum and maximum data values I have are 3073351 and 25329814 respectively but the graph has data from 0 and the maximum is different The data file is too large so there is a small sample below. What am I doing wrong? Please help as I cannot be sure the data is represented correctly in the graph (attached). fn -~/Documents/chr7Data.txt x-read.table(fn,check.names = TRUE, header=T) d - data.frame(x) theChart-ggplot(d, aes(x=factor(Congenic),y=Position, fill=Strain, theme_rect=white)) + geom_bar(stat='identity', position = stack ) + scale_fill_manual(values=c(deepskyblue1, red2, green3, steelblue4, lightblue4, gray70)) + ylab(c(Position))+ ggtitle(Strain specific SNP distribution in the congenics) + #coord_cartesian(ylim = c(min(d$Position), max(d$Position))) + #scale_y_continuous(breaks=seq(0, 2, 5000)) + theme_bw()+ theme(plot.title = element_text(size=23), axis.text.x = element_text(size=13, angle=90),#20 before axis.text.y = element_text(size=13), panel.border = element_rect(colour=BLACK,size=0.5), axis.title.x = element_text(size=12), axis.title.y = element_text(size=12,angle = 90), panel.background = element_rect(fill=transparent), legend.text=element_text(size=13), legend.title=element_text(size=12) #plot.background = element_rect(fill = transparent,colour = NA) ) png(~/Documents/sendToRGrp.png,950,750) print(theChart) dev.off() ——— Sample Data — Congenic Position Strain 4201 3073351 hets 4203 3073351 reference 4215 3073351 reference 4333 3073351 cba 4335 3073351 cba 4484 3073351 cba 4485 3073351 reference 4496 3073351 hets 4497 3073351 reference //lots of data// 4201 25329814 reference 4203 25329814 reference 4215 25329814 cba 4333 25329814 balbc 4335 25329814 balbc 4484 25329814 reference 4485 25329814 reference 4496 25329814 balbc 4497 25329814 hets Many thanks, Michy This email may have a PROTECTIVE MARKING, for an explanation please see: http://www.mrc.ac.uk/About/Informationandstandards/Documentmarking/index.htm __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Having trouble with gdata read in
My suggestion is to use XLConnect to read the file: x - C:\\Users\\jh52822\\AppData\\Local\\Temp\\Rtmp6nVgFC\\file385c632aba3.xls require(XLConnect) Loading required package: XLConnect Loading required package: XLConnectJars XLConnect 0.2-10 by Mirai Solutions GmbH [aut], Martin Studer [cre], The Apache Software Foundation [ctb, cph] (Apache POI, Apache Commons Codec), Stephen Colebourne [ctb, cph] (Joda-Time Java library) http://www.mirai-solutions.com , http://miraisolutions.wordpress.com input - f.readXLSheet(x, 1) str(input) 'data.frame': 2266 obs. of 51 variables: $ EIA : num 34 59 87 97 108 118 123 149 150 157 ... $ Entity.Name : chr City of Abbeville City of Abbeville City of Ada Adams Electric Cooperative ... $ State: chr SC LA MN IL ... $ NERC.Region : chr SERC SPP MRO SERC ... $ Filing.Order : num 12 11 1237 392 252 ... $ Q5.MultRegion: chr ... $ Q6.OwnMeters.: chr Yes Yes Yes Yes ... $ Q7.ResMeters : num 3051 4253 857 8154 33670 ... $ Q7.ComMeters : num 531 972 132 155 1719 ... $ Q7.IntMeters : num 0 19 32 NA 626 NA 29 0 2 NA ... $ Q7.TransMeters : num 0 NA NA NA NA NA NA 0 0 NA ... $ Q7.OtherMeters : num 0 NA NA 57 NA NA NA 0 0 NA ... $ Q7...total.meters: num 3582 5244 1021 8366 36015 ... $ Q8.15Min.ResAMI : num 0 NA NA NA NA NA NA NA NA NA ... $ Q8.15Min.ComAMI : num 0 NA NA 155 NA NA NA NA NA NA ... $ Q8.15Min.IndAMI : num 0 NA NA NA NA NA NA NA NA NA ... $ Q8.15Min.TransAMI: num 0 NA NA NA NA NA NA NA NA NA ... $ Q8.15Min.OtherAMI: num 0 NA NA NA NA NA NA NA NA NA ... $ Q8.15Min.TotalAMI: num 0 0 0 155 0 0 0 0 0 0 ... $ Q8.Hourly.ResAMI : num 0 NA NA NA 16100 NA NA NA NA NA ... $ Q8.Hourly.ComAMI : num 0 NA NA NA 1600 NA NA NA NA NA ... Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. On Wed, Mar 25, 2015 at 5:01 PM, Benjamin Baker bba...@reed.edu wrote: Trying to read and clean up the FERC data on Advanced Metering infrastructure. Of course it is in XLS for the first two survey years and then converts to XLSX for the final two. Bad enough that it is all in excel, they had to change the survey design and data format as well. Still, I’m sorting through it. However, when I try and read in the 2008 data, I’m getting this error: ### Wide character in print at /Library/Frameworks/R.framework/Versions/3.1/Resources/library/gdata/perl/ xls2csv.pl line 270. Warning message: In scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : EOF within quoted string ### Here is the code I’m running to get the data: ### install.packages(gdata) library(gdata) fileUrl - http://www.ferc.gov/industries/electric/indus-act/demand-response/2008/survey/ami_survey_responses.xls download.file(fileUrl, destfile=./ami.data/ami-data2008.xls) list.files(ami.data) dateDown.2008 - date() ami.data2008 - read.xls(./ami.data/ami-data2008.xls, sheet=1, header=TRUE) ### Reviewed the data in the XLS file, and both “” and # are present within it. Don’t know how to get the read.xls to ignore them so I can read all the data into my data frame. Tried : ### ami.data2008 - read.xls(./ami.data/ami-data2008.xls, sheet=1, quote=, header=TRUE) ### And it spits out “More columns than column names” output. Been searching this, and I can find some “solutions” for read.table, but nothing specific to read.xls Many thanks, Benjamin Baker — Sent from Mailbox [[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] Converting charter string to vector of numbers (Turgidson)
Hi, See that strsplit() retorns a list object. Then, to transform a character vector (that is an element into the list), to a numèric vector, you should access directly to that element to aply as.numeric. The code is: aux-0.5 1 2 3 4 vectnum-strsplit(aux, ) class(vectnum) ## see that is returning a list numaux-as.numeric(vectnum[[1]])## get the nummerical vector form the vector object that stays into the vectnum list Message: 20 Date: Wed, 25 Mar 2015 10:27:40 -0700 (PDT) From: Turgidson e...@upenn.edu To: r-help@r-project.org Subject: [R] Converting charter string to vector of numbers Message-ID: 1427304460123-4705097.p...@n4.nabble.com Content-Type: text/plain; charset=us-ascii Apologies for bringing up an old topic, but I am not finding working answers. I am also an R newbie, so I'm still on the steep part of the learning curve. I have a character string : 0.5 1 2 3 4 I need to convert it to a vector of numbers. I get as far as strsplit(), which gives me 0.5 1 2 3 4 this is still treated by R as a single item. I need to get it to look like this: 0.5, 1, 2, 3, 4 so that I can use as.numeric() and make it into a vector of numbers... I have been going mad because for text operations, commas and quotes are loaded with meaning, and I keep going around in circles. Any help would be appreciated Thanks Paul -- View this message in context: http://r.789695.n4.nabble.com/Converting-charter-string-to-vector-of-numbers-tp4705097.html Sent from the R help mailing list archive at Nabble.com. Manel Amado i Martí Cap d'Assessoria de Comerç Interior am...@cambrasabadell.org Tel. 93 745 12 63 . Fax 93 745 12 64 Av. Francesc Macià, 35 . 08206 Sabadell Apt. corr. 119 . www.cambrasabadell.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] Why can't I access this type?
On Mar 25, 2015, at 10:14, Henric Winell nilsson.hen...@gmail.com wrote: On 2015-03-25 09:40, Patrick Connolly wrote: On Sun, 22-Mar-2015 at 08:06AM -0800, John Kane wrote: | Well, first off, you have no variable called Name. You have lost | the state names as they are rownames in the matrix state.x77 and | not a variable. | | Try this. It's ugly and I have no idea why I had to do a cbind() You don't have to use cbind | but it seems to work. Personally I find subset easier to read than | the indexing approach. | state - rownames(state.x77) | all.states - as.data.frame(state.x77) | all.states - cbind(state, all.states) ### ? You don't have to use cbind() all.states - within(as.data.frame(state.x77), state - rownames(state.x77)) but I think cbind is simpler to read. | | coldstates - subset(all.states, all.states$Frost 50, | select = c(state,Frost) ) I find the indexing approach coldstates - all.states[all.states$Frost 150, c(state,Frost)] to be the most direct and obvious solution. Tidier, even more so than subset(): require(dplyr) coldstates - all.states %% filter(Frost 150) %% select(state, Frost) Or, easier to see what's happening: coldstates - all.states %% filter(Frost 150) %% select(state, Frost) Well... Opinions may perhaps differ, but apart from '%%' being butt-ugly it's also fairly slow: library(microbenchmark) microbenchmark( + subset(all.states, all.states$Frost 150, select = c(state,Frost)), + all.states[all.states$Frost 150, c(state,Frost)], + all.states %% filter(Frost 150) %% select(state, Frost), + times = 1000L + ) Unit: microseconds expr subset(all.states, all.states$Frost 150, select = c(state, Frost)) all.states[all.states$Frost 150, c(state, Frost)] all.states %% filter(Frost 150) %% select(state, Frost) min lq meanmedianuq max neval cld 139.112 148.673 163.3960 159.1760 170.7895 1763.200 1000 b 104.039 111.973 127.2138 120.4395 128.6640 1381.809 1000 a 1010.076 1033.519 1133.1469 1107.8480 1175.1800 2932.206 1000 c Of course, this doesn't matter for interactive one-off use. But lately I've seen examples of the '%%' operator creeping into functions in packages. However, it would be nice to see a fast pipe operator as part of base R. Henric Winell | | | John Kane | Kingston ON Canada | | | -Original Message- | From: yoursurrogate...@gmail.com | Sent: Sun, 22 Mar 2015 10:39:03 -0400 | To: r-help@r-project.org | Subject: [R] Why can't I access this type? | | Hi, I'm just learning my way around R. I got a bunch of states and would | like to access to get all of the ones where it's cold. But when I do the | following, I will get the following error: | | all.states - as.data.frame(state.x77) | cold.states - all.states[all.states$Frost 150, c(Name, Frost)] | Error in `[.data.frame`(all.states, all.states$Frost 150, c(Name, : |undefined columns selected | | I don't get it. When I look at all.states, this is what I see: | | str(all.states) | 'data.frame': 50 obs. of 8 variables: | $ Population: num 3615 365 2212 2110 21198 ... | $ Income: num 3624 6315 4530 3378 5114 ... | $ Illiteracy: num 2.1 1.5 1.8 1.9 1.1 0.7 1.1 0.9 1.3 2 ... | $ Life Exp : num 69 69.3 70.5 70.7 71.7 ... | $ Murder: num 15.1 11.3 7.8 10.1 10.3 6.8 3.1 6.2 10.7 13.9 ... | $ HS Grad : num 41.3 66.7 58.1 39.9 62.6 63.9 56 54.6 52.6 40.6 ... | $ Frost : num 20 152 15 65 20 166 139 103 11 60 ... | $ Area : num 50708 566432 113417 51945 156361 ... | | What am I messing up? | | [[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. | | | FREE ONLINE PHOTOSHARING - Share your photos online with your friends and family! | Visit http://www.inbox.com/photosharing to find out more! | | __ | 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. I agree with you on the indexing approach. But even after using within, I still get the same error. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE
[R] matrix manipulation question
Hi, I've got a rather large matrix of about 800 rows and 60 columns. Each column is a time-series 800 long. Out of these 60 time series, some have missing values (NA). I want to strip out all columns that have one or more NA values, i.e., only want full time series. This should do the trick: data_no_NA - data[,!apply(is.na(data), 2, any)] I now use data_no_NA as input to a function, which returns output as a matrix of the same size as data_no_NA The trick is that i now need to put these columns back into a new 800 by 60 empty matrix, at their original locations. Any suggestions on how to do that? hopefully without having to use loops. I'm using R/3.0.3 Cheers, Jatin. __ 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
just reverse what you did before. newdata - data newdata[] - NA newdata[,!apply(is.na(data), 2, any)] - myfunction(data_no_NA) On Fri, Mar 27, 2015 at 1:13 AM, Jatin Kala jatin.kala...@gmail.com wrote: Hi, I've got a rather large matrix of about 800 rows and 60 columns. Each column is a time-series 800 long. Out of these 60 time series, some have missing values (NA). I want to strip out all columns that have one or more NA values, i.e., only want full time series. This should do the trick: data_no_NA - data[,!apply(is.na(data), 2, any)] I now use data_no_NA as input to a function, which returns output as a matrix of the same size as data_no_NA The trick is that i now need to put these columns back into a new 800 by 60 empty matrix, at their original locations. Any suggestions on how to do that? hopefully without having to use loops. I'm using R/3.0.3 Cheers, Jatin. __ 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] Vennerable Plots for Publications
Does anyone make Venn diagrams for publication using Vennerable ? I found that the font size is too big when the plot is created at 300 DPI, and there's no option to change it, even when the point size argument to the device is changed. aVenn - Venn(Sets = list(A = 1:5, B = 3:6)) png(forPublication.png, units = in, h = 2.55, w = 2.4, res = 300) # Changing pointsize to a smaller number has no effect on size of the text. plot(aVenn) dev.off() -- Dario Strbenac PhD Student University of Sydney Camperdown NSW 2050 Australia __ 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] Acceso gratuito durante un mes a la plataforma DataCamp...
Muy interesante. Había visto la oferta de 9$, pero NO GRATIS... Como siempre, gracias. Isidro Hidalgo Arellano Observatorio Regional de Empleo Consejería de Empleo y Economía http://www.jccm.es -Mensaje original- De: R-help-es [mailto:r-help-es-boun...@r-project.org] En nombre de Carlos Ortega Enviado el: jueves, 26 de marzo de 2015 18:36 Para: Lista R Asunto: [R-es] Acceso gratuito durante un mes a la plataforma DataCamp... Buenas a todos, La gente de DataCamp se han puesto en contacto con el grupo de Madrid para ofrecernos una oferta para usar gratuitamente su plataforma con diferentes cursos de R durante un mes. Les hemos preguntado si sería posible compartir este ofrecimiento con toda la lista de R-Help-es y no ven mayor problema, les ha parecido muy bien la posibilidad. El link de acceso es el siguiente: https://www.datacamp.com/enroll-meetup ?Cuidado que los cursos son en inglés y para darse de alta, aun siendo gratuitio este primer mes, hay que asociar una tarjeta de crédito o una cuenta Paypal. Pasado este mes el estar dado de alta en la plataforma cuesta ?25$/mes. ?Durante este mes gratuito, uno se puede a puntar a cualquier curso (necesitan de una dedicación de de 5-8 horas aprox.). Los cursos van desde Introducción a R? hasta el uso de la solución RevoScale para BigData de Revolution Analytics. Saludos, Carlos Ortega www.qualityexcellence.es [[alternative HTML version deleted]] ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
[R-es] Acceso gratuito durante un mes a la plataforma DataCamp...
Buenas a todos, La gente de DataCamp se han puesto en contacto con el grupo de Madrid para ofrecernos una oferta para usar gratuitamente su plataforma con diferentes cursos de R durante un mes. Les hemos preguntado si sería posible compartir este ofrecimiento con toda la lista de R-Help-es y no ven mayor problema, les ha parecido muy bien la posibilidad. El link de acceso es el siguiente: https://www.datacamp.com/enroll-meetup Cuidado que los cursos son en inglés y para darse de alta, aun siendo gratuitio este primer mes, hay que asociar una tarjeta de crédito o una cuenta Paypal. Pasado este mes el estar dado de alta en la plataforma cuesta 25$/mes. Durante este mes gratuito, uno se puede a puntar a cualquier curso (necesitan de una dedicación de de 5-8 horas aprox.). Los cursos van desde Introducción a R hasta el uso de la solución RevoScale para BigData de Revolution Analytics. Saludos, Carlos Ortega www.qualityexcellence.es [[alternative HTML version deleted]] ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
Re: [R] list output
I don't think you are correctly keeping track of the structure of your
objects.
(and my email software may mess up the indentation, but I can't prevent it)
I have not tested, but try replacing
out[i]$phi_0 - x$phi_0
out[i]$maxlike - x$maxlike
out[i]$bigp - x$bigp
with
out[[i]] - list(phi_0=x$phi_0,
maxlike=x$maxlike,
bigp=x$bigp
)
(double bracket, as Peter suggested)
Then also replace
aic[i] - -2*out[i]$maxlike + (2*(i+1))
with
aic[i] - -2*out[[i]]$maxlike + (2*(i+1))
To begin understanding this, see this little sequence:
foo - vector('list',3)
foo[[2]] - list(a=1, b=2)
foo[2]$a
NULL
foo[[2]]$a
[1] 1
Also, which() does not return a list object as you are using it, so there
is no reason to unlist() it. Might as well use
which.min(minaic)
while you're at it.
So replace your last three lines with
out[which.min(minaic)]
as the last line of your function, before the closing }
No need to use return()
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 3/26/15, 11:22 AM, Erin Hodgess erinm.hodg...@gmail.com wrote:
Hello!
I am having some trouble with some list output.
Here is my code:
geobunch - function(y) {
out - vector(list,3)
aic - numeric(length=3)
print(str(out))
for(i in 1:3) {
x - geomin(y,i)
print(i)
print(x)
print(str(x))
out[i]$phi_0 - x$phi_0
out[i]$maxlike - x$maxlike
out[i]$bigp - x$bigp
aic[i] - -2*out[i]$maxlike + (2*(i+1))
}
minaic - which(aic==min(aic))
minout - out[unlist(minaic)]
return(minout)
}
And here is the output:
geobunch(ez2a)
List of 3
$ : NULL
$ : NULL
$ : NULL
NULL
[1] 1
$phi_0
[1] 2.856428
$bigp
[1] 0.1584016
$maxlike
[1] -473.0203
List of 3
$ phi_0 : num 2.86
$ bigp : num 0.158
$ maxlike: num -473
NULL
Error in aic[i] - -2 * out[i]$maxlike + (2 * (i + 1)) :
replacement has length zero
In addition: Warning messages:
1: In out[i]$phi_0 - x$phi_0 :
number of items to replace is not a multiple of replacement length
2: In out[i]$maxlike - x$maxlike :
number of items to replace is not a multiple of replacement length
3: In out[i]$bigp - x$bigp :
number of items to replace is not a multiple of replacement length
I know that the problem is in how I am setting up the out variable, but
I'm not sure how to correct it. Also, the bigp variable can take on
different lengths.
This is on R version 3.1.3, on Ubuntu 14.04.
Thank you so much for any help.
Sincerely,
Erin
--
Erin Hodgess
Associate Professor
Department of Mathematical and Statistics
University of Houston - Downtown
mailto: erinm.hodg...@gmail.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.
__
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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-es] Conectividad con Excel
Como te indica Miguel Ángel, no necesitas ODBC para conectar con Excel. Yo también uso el mismo sistema: XLConnect. Tiene una buena vignette: http://cran.r-project.org/web/packages/XLConnect/vignettes/XLConnect.pdf Si quieres profundizar en el tema, Hadley Wickham hace unos días estaba haciendo propaganda por Twitter de haber mejorado mucho un paquete que también conecta con Excel, pero yo no lo he usado. Es el paquete readxl, pero tengo pendiente probarlo. No está en cran, pero lo tienes en GitHub: https://github.com/hadley/readxl Si no quieres complicarte la vida instalando devtools para instalar paquetes desde GitHub, usa XLConnect, que va bien. Suerte. Isidro Hidalgo Arellano Observatorio Regional de Empleo Consejería de Empleo y Economía http://www.jccm.es -Mensaje original- De: R-help-es [mailto:r-help-es-boun...@r-project.org] En nombre de miguel.angel.rodriguez.mui...@sergas.es Enviado el: jueves, 26 de marzo de 2015 15:26 Para: r-help-es@r-project.org Asunto: Re: [R-es] Conectividad con Excel Hola Our Utopy. Te puedo comentar cómo hago yo con XLConnect para acceder a un fichero de Excel (por si es ésto lo que pretendes) Veamos.. require(XLConnect) file.import=/trayecto_del_fichero_Excel/fichero.xlsx file.load=loadWorkbook(file.import, create=FALSE) misdatos=readWorksheet(file.load, sheet=1) Es esto lo que buscas? Un Saludo, Miguel Rodríguez Consellería de Sanidade Xunta de Galicia El 26/03/2015 a las 15:04, Our Utopy escribió: Hola amigos, buenos días Esto sigue avanzando. Estoy ahora tratando de establecer la conectividad de R con Excel y he seguido dos viídeos en YouTube de una profesora llamada Bebilda que se explica muy bien. Con ambos tengo el mismo problema así que planteo uno de ellos, éste es: https://youtu.be/_pSJQO_9I4k?list=PL7DA3FDA21A1A6310 Pero os lo explico brevemente: Cargo el paquete RODBC install.packages(RODBC) Instalo la librería library(RODBC) Todo bien hasta aqui... Se supone, que en esa librería hay unas funciones que me permiten conectar con Excel, esas funciones son: o-odncConnectExcel2007 o-odncConnectExcel DE tal manera que ahora podría abrir una canal canalexcel - odbcConnectExcel2007(lbw.xlsx) Pues esas funciones no están. R me da el error Error: no se pudo encontrar la función odbcConnectExcel2007 lo mismo con la otra función canalexcel - odbcConnectExcel(lbw.xlsx) Y lo curioso es que también me pasa algo similar con la otra posibilidad que es el paquete XLConnect Y es que, además, si miro la documentación de este paquete RODBC no figuran esas funciones. ¿ ? ¿Me podéis nuevamente ayudar? Un saludo 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 ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
Re: [R] list output
Erin,
I think you'll have heard people requesting _reproducible_ examples several
times by now... This one isn't.
Offhand, I'd suspect that the issue has to do with your single-bracket indexing
of the list out. Try out[[i]].
-Peter
On 26 Mar 2015, at 19:22 , Erin Hodgess erinm.hodg...@gmail.com wrote:
Hello!
I am having some trouble with some list output.
Here is my code:
geobunch - function(y) {
out - vector(list,3)
aic - numeric(length=3)
print(str(out))
for(i in 1:3) {
x - geomin(y,i)
print(i)
print(x)
print(str(x))
out[i]$phi_0 - x$phi_0
out[i]$maxlike - x$maxlike
out[i]$bigp - x$bigp
aic[i] - -2*out[i]$maxlike + (2*(i+1))
}
minaic - which(aic==min(aic))
minout - out[unlist(minaic)]
return(minout)
}
And here is the output:
geobunch(ez2a)
List of 3
$ : NULL
$ : NULL
$ : NULL
NULL
[1] 1
$phi_0
[1] 2.856428
$bigp
[1] 0.1584016
$maxlike
[1] -473.0203
List of 3
$ phi_0 : num 2.86
$ bigp : num 0.158
$ maxlike: num -473
NULL
Error in aic[i] - -2 * out[i]$maxlike + (2 * (i + 1)) :
replacement has length zero
In addition: Warning messages:
1: In out[i]$phi_0 - x$phi_0 :
number of items to replace is not a multiple of replacement length
2: In out[i]$maxlike - x$maxlike :
number of items to replace is not a multiple of replacement length
3: In out[i]$bigp - x$bigp :
number of items to replace is not a multiple of replacement length
I know that the problem is in how I am setting up the out variable, but
I'm not sure how to correct it. Also, the bigp variable can take on
different lengths.
This is on R version 3.1.3, on Ubuntu 14.04.
Thank you so much for any help.
Sincerely,
Erin
--
Erin Hodgess
Associate Professor
Department of Mathematical and Statistics
University of Houston - Downtown
mailto: erinm.hodg...@gmail.com
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--
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Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd@cbs.dk Priv: pda...@gmail.com
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[R] list output
Hello!
I am having some trouble with some list output.
Here is my code:
geobunch - function(y) {
out - vector(list,3)
aic - numeric(length=3)
print(str(out))
for(i in 1:3) {
x - geomin(y,i)
print(i)
print(x)
print(str(x))
out[i]$phi_0 - x$phi_0
out[i]$maxlike - x$maxlike
out[i]$bigp - x$bigp
aic[i] - -2*out[i]$maxlike + (2*(i+1))
}
minaic - which(aic==min(aic))
minout - out[unlist(minaic)]
return(minout)
}
And here is the output:
geobunch(ez2a)
List of 3
$ : NULL
$ : NULL
$ : NULL
NULL
[1] 1
$phi_0
[1] 2.856428
$bigp
[1] 0.1584016
$maxlike
[1] -473.0203
List of 3
$ phi_0 : num 2.86
$ bigp : num 0.158
$ maxlike: num -473
NULL
Error in aic[i] - -2 * out[i]$maxlike + (2 * (i + 1)) :
replacement has length zero
In addition: Warning messages:
1: In out[i]$phi_0 - x$phi_0 :
number of items to replace is not a multiple of replacement length
2: In out[i]$maxlike - x$maxlike :
number of items to replace is not a multiple of replacement length
3: In out[i]$bigp - x$bigp :
number of items to replace is not a multiple of replacement length
I know that the problem is in how I am setting up the out variable, but
I'm not sure how to correct it. Also, the bigp variable can take on
different lengths.
This is on R version 3.1.3, on Ubuntu 14.04.
Thank you so much for any help.
Sincerely,
Erin
--
Erin Hodgess
Associate Professor
Department of Mathematical and Statistics
University of Houston - Downtown
mailto: erinm.hodg...@gmail.com
[[alternative HTML version deleted]]
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Re: [R] list output
It's the double bracket issue.
I replaced that, and all is well.
Thanks to Peter and Don.
Sincerely,
Erin
On Thu, Mar 26, 2015 at 3:21 PM, MacQueen, Don macque...@llnl.gov wrote:
I don't think you are correctly keeping track of the structure of your
objects.
(and my email software may mess up the indentation, but I can't prevent it)
I have not tested, but try replacing
out[i]$phi_0 - x$phi_0
out[i]$maxlike - x$maxlike
out[i]$bigp - x$bigp
with
out[[i]] - list(phi_0=x$phi_0,
maxlike=x$maxlike,
bigp=x$bigp
)
(double bracket, as Peter suggested)
Then also replace
aic[i] - -2*out[i]$maxlike + (2*(i+1))
with
aic[i] - -2*out[[i]]$maxlike + (2*(i+1))
To begin understanding this, see this little sequence:
foo - vector('list',3)
foo[[2]] - list(a=1, b=2)
foo[2]$a
NULL
foo[[2]]$a
[1] 1
Also, which() does not return a list object as you are using it, so there
is no reason to unlist() it. Might as well use
which.min(minaic)
while you're at it.
So replace your last three lines with
out[which.min(minaic)]
as the last line of your function, before the closing }
No need to use return()
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 3/26/15, 11:22 AM, Erin Hodgess erinm.hodg...@gmail.com wrote:
Hello!
I am having some trouble with some list output.
Here is my code:
geobunch - function(y) {
out - vector(list,3)
aic - numeric(length=3)
print(str(out))
for(i in 1:3) {
x - geomin(y,i)
print(i)
print(x)
print(str(x))
out[i]$phi_0 - x$phi_0
out[i]$maxlike - x$maxlike
out[i]$bigp - x$bigp
aic[i] - -2*out[i]$maxlike + (2*(i+1))
}
minaic - which(aic==min(aic))
minout - out[unlist(minaic)]
return(minout)
}
And here is the output:
geobunch(ez2a)
List of 3
$ : NULL
$ : NULL
$ : NULL
NULL
[1] 1
$phi_0
[1] 2.856428
$bigp
[1] 0.1584016
$maxlike
[1] -473.0203
List of 3
$ phi_0 : num 2.86
$ bigp : num 0.158
$ maxlike: num -473
NULL
Error in aic[i] - -2 * out[i]$maxlike + (2 * (i + 1)) :
replacement has length zero
In addition: Warning messages:
1: In out[i]$phi_0 - x$phi_0 :
number of items to replace is not a multiple of replacement length
2: In out[i]$maxlike - x$maxlike :
number of items to replace is not a multiple of replacement length
3: In out[i]$bigp - x$bigp :
number of items to replace is not a multiple of replacement length
I know that the problem is in how I am setting up the out variable, but
I'm not sure how to correct it. Also, the bigp variable can take on
different lengths.
This is on R version 3.1.3, on Ubuntu 14.04.
Thank you so much for any help.
Sincerely,
Erin
--
Erin Hodgess
Associate Professor
Department of Mathematical and Statistics
University of Houston - Downtown
mailto: erinm.hodg...@gmail.com
[[alternative HTML version deleted]]
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Erin Hodgess
Associate Professor
Department of Mathematical and Statistics
University of Houston - Downtown
mailto: erinm.hodg...@gmail.com
[[alternative HTML version deleted]]
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Re: [R] basic help with as.Date()
That would be because they are not dates... they don't specify the day.
Either add a day of month to the character data before you convert it, or use
the yearmon class from zoo.
Something like...
as.Date(paste(1,as.character(df$mydate),sep=-), format='%d-%b-%y')
And why in the world are you converting to factor and back? Perhaps because you
are not preventing conversion to factor when you read in your actual data? Do
you know about the stringsAsFactors argument to read.table and friends?
---
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On March 26, 2015 2:11:50 PM PDT, Simon Kiss sjk...@gmail.com wrote:
Hi there: normally I’m quite comfortable with as.Date(). But this data
set is causing problems.
The core of the data frame looks like the sample data frame below, but
my attempt to convert df$mydate to a date object returns only NA. Can
anyone provide a suggestion?
Thank you, Simon Kiss
#sample data frame
df-data.frame(mydate=factor(c('Jan-15', 'Feb-13', 'Mar-11',
'Jul-12')), other=rnorm(4, 3))
#Attempt to convert
as.Date(as.character(df$mydate), format='%b-%y')
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[R] new package: canceR
Dear All, I am pleased to announce the release of new package named canceR http://www.bioconductor.org/packages/devel/bioc/html/canceR.html. The main goal of this package is to facilitate to NOT R user the access of complex cancer genomics data hosted by the MSKCC and available through cgdsr CRAN package. By simple clic, the user can combine Cases (Samples) and Genetic profiles to get rapidly without programming skills all available cancer genomics data (CNA, mRNA, Methylation, Mutation, miRNA, RPPA) in exportable table. Compared to cBioportal http://www.cbioportal.org/, canceR allows user to focus on some Studies and get specific data as Amino Acid changes in Mutation Data for selected gene list. The second part of canceR focus on modeling. I selected some interesting fonctions from others packages (see Vignette) and I adapted them to 1-Associate phenotypes (gene / variable ) to disease 2-Classify genes by phenotype or disease 3-Enrichment gene expression data using GSEA-R (Broad Institute) Other function is in progress to facilitated the visualization of multi Dimensional Genomics Data for multiples cancers using Circos styles. Please fill free for any suggestions. Karim Mezhoud [[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] Acceso gratuito durante un mes a la plataforma DataCamp...
que bien!! justamente hoy he comenzado un curso de los gratis en datacamp. Muchas gracias, le daré buen uso! El día 26 de marzo de 2015, 18:45, Isidro Hidalgo ihida...@jccm.es escribió: Muy interesante. Había visto la oferta de 9$, pero NO GRATIS... Como siempre, gracias. Isidro Hidalgo Arellano Observatorio Regional de Empleo Consejería de Empleo y Economía http://www.jccm.es -Mensaje original- De: R-help-es [mailto:r-help-es-boun...@r-project.org] En nombre de Carlos Ortega Enviado el: jueves, 26 de marzo de 2015 18:36 Para: Lista R Asunto: [R-es] Acceso gratuito durante un mes a la plataforma DataCamp... Buenas a todos, La gente de DataCamp se han puesto en contacto con el grupo de Madrid para ofrecernos una oferta para usar gratuitamente su plataforma con diferentes cursos de R durante un mes. Les hemos preguntado si sería posible compartir este ofrecimiento con toda la lista de R-Help-es y no ven mayor problema, les ha parecido muy bien la posibilidad. El link de acceso es el siguiente: https://www.datacamp.com/enroll-meetup ?Cuidado que los cursos son en inglés y para darse de alta, aun siendo gratuitio este primer mes, hay que asociar una tarjeta de crédito o una cuenta Paypal. Pasado este mes el estar dado de alta en la plataforma cuesta ?25$/mes. ?Durante este mes gratuito, uno se puede a puntar a cualquier curso (necesitan de una dedicación de de 5-8 horas aprox.). Los cursos van desde Introducción a R? hasta el uso de la solución RevoScale para BigData de Revolution Analytics. Saludos, Carlos Ortega www.qualityexcellence.es [[alternative HTML version deleted]] ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
Re: [R] basic help with as.Date()
On Mar 26, 2015, at 4:11 PM, Simon Kiss sjk...@gmail.com wrote:
Hi there: normally I’m quite comfortable with as.Date(). But this data set is
causing problems.
The core of the data frame looks like the sample data frame below, but my
attempt to convert df$mydate to a date object returns only NA. Can anyone
provide a suggestion?
Thank you, Simon Kiss
#sample data frame
df-data.frame(mydate=factor(c('Jan-15', 'Feb-13', 'Mar-11', 'Jul-12')),
other=rnorm(4, 3))
#Attempt to convert
as.Date(as.character(df$mydate), format='%b-%y')
Hi,
R's default date class object requires a full date, with a month, day and year.
You might look at the 'zoo' package on CRAN, which has a yearmon() function.
Regards,
Marc Schwartz
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Re: [R] basic help with as.Date()
On 26/03/2015 21:11, Simon Kiss wrote:
Hi there: normally I’m quite comfortable with as.Date(). But this data set is
causing problems.
The core of the data frame looks like the sample data frame below, but my
attempt to convert df$mydate to a date object returns only NA. Can anyone
provide a suggestion?
Thank you, Simon Kiss
#sample data frame
df-data.frame(mydate=factor(c('Jan-15', 'Feb-13', 'Mar-11', 'Jul-12')),
other=rnorm(4, 3))
#Attempt to convert
as.Date(as.character(df$mydate), format='%b-%y')
You would be on surer ground with something like
as.Date(paste0('01-',as.character(df$mydate)), format='%d-%b-%y')
since it is unclear what dates you expected to get.
--
Brian D. Ripley, rip...@stats.ox.ac.uk
Emeritus Professor of Applied Statistics, University of Oxford
1 South Parks Road, Oxford OX1 3TG, UK
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[R] basic help with as.Date()
Hi there: normally I’m quite comfortable with as.Date(). But this data set is
causing problems.
The core of the data frame looks like the sample data frame below, but my
attempt to convert df$mydate to a date object returns only NA. Can anyone
provide a suggestion?
Thank you, Simon Kiss
#sample data frame
df-data.frame(mydate=factor(c('Jan-15', 'Feb-13', 'Mar-11', 'Jul-12')),
other=rnorm(4, 3))
#Attempt to convert
as.Date(as.character(df$mydate), format='%b-%y')
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[R] Using and abusing %% (was Re: Why can't I access this type?)
On Wed, 25-Mar-2015 at 03:14PM +0100, Henric Winell wrote:
...
| Well... Opinions may perhaps differ, but apart from '%%' being
| butt-ugly it's also fairly slow:
Beauty, it is said, is in the eye of the beholder. I'm impressed by
the way using %% reduces or eliminates complicated nested brackets.
In this tiny example it's not obvious but it's very clear if the
objective is to sort the dataframe by three or four columns and
various lots of aggregation then returning a largish number of
consecutive columns, omitting the rest. It's very easy to see what's
going on without the need for intermediate objects.
|
| .
| Unit: microseconds
|
|expr
| subset(all.states, all.states$Frost 150, select = c(state,
| Frost))
|all.states[all.states$Frost 150,
| c(state, Frost)]
|all.states %% filter(Frost 150) %%
| select(state, Frost)
| min lq meanmedianuq max neval cld
| 139.112 148.673 163.3960 159.1760 170.7895 1763.200 1000 b
| 104.039 111.973 127.2138 120.4395 128.6640 1381.809 1000 a
| 1010.076 1033.519 1133.1469 1107.8480 1175.1800 2932.206 1000 c
It's no surprise that instructing a computer in something closer to
human language is an order of magnitude slower. I'm sure you'd get
something even quicker using machine code. I spend 3 or 4 orders of
magnitude more time writing code than running it. It's much more
important to me to be able to read and modify than it is to have it
run at optimum speed.
|
| Of course, this doesn't matter for interactive one-off use. But
| lately I've seen examples of the '%%' operator creeping into
| functions in packages.
That could indicate that %% is seductively easy to use. It's
probably true that there are places where it should be done the hard
way.
| However, it would be nice to see a fast pipe operator as part of
| base R.
|
|
| Henric Winell
|
--
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
___Patrick Connolly
{~._.~} Great minds discuss ideas
_( Y )_ Average minds discuss events
(:_~*~_:) Small minds discuss people
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Re: [R] Fitting a line on trellis plot
Thanks for the help.
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 Wed, Mar 25, 2015 at 4:48 PM, S Ellison s.elli...@lgcgroup.com wrote:
Call:
lm(formula = curr_data[[tmin_col]] ~ curr_data[[year_col]] -
1 | curr_data[[month_col]])
First, this is not a sensible formula for lm; lm() does not use '|' to
denote grouping. It would be a valid formula for xyplot, in which |
specifies grouping variables. In lm(), '|' is simply being treated as 'or',
which is why one of your coefficients is for ' curr_data[[year_col]] |
curr_data[[month_col]]TRUE'
Second, you should not normally need things like curr_data[[month_col]],
either in lm or xyplot. If curr_data is a data frame, things like
lm(tmin ~ year, data=curr_data)
should work.
Third, 'nested' models in lm use the nesting operator '/', not '|'. So if
you want 12 lines with separate intercept and gradient from an lm, you need
(with month as a factor and the default intercept suppressed)
lm(tmin~month+year/month -1, data=curr_data) #-1 suppresses the intercept
and provides a zero-based intercept for each month
This gives you 12 'month' intercepts and one gradient per month. If you
wanted a common intercept you'd do
lm(tmin~ year/month, data=curr_data)
But beware; the coefficients in both cases cannot be interpreted as a
simple gradient and intercept for each month: if I recall correctly, the
gradients for month2 and on are modelled as an additive increment on the
first month gradient. Use predict() if you want an easy way to predict a
value for a given (possibly fractional) time of year and 'month'.
[Incidentally I don’t immediately see why that is a sensible thing to do -
this fits a monthly summary against a numeric year. But I'm going to assume
you know what you want there.]
Finally, though, this model will not help you much with lattice as there's
no _simple_ way of putting those lines on different panels in a lattice
plot. If you just want a line on each of 12 panels, that's much easier. You
can use the panel() function with panel.lmline to put it there. For
example, if you want to plot a line over the data, use
xyplot(tmin~year|month, curr_data,
panel=function(x, y, ...) {
panel.xyplot(x, y, ...)
panel.lmline(x, y, ...)
}
)
S Ellison
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Re: [R-es] Conectividad con Excel
Hola Our Utopy. Te puedo comentar cómo hago yo con XLConnect para acceder a un fichero de Excel (por si es ésto lo que pretendes) Veamos.. require(XLConnect) file.import=/trayecto_del_fichero_Excel/fichero.xlsx file.load=loadWorkbook(file.import, create=FALSE) misdatos=readWorksheet(file.load, sheet=1) Es esto lo que buscas? Un Saludo, Miguel Rodríguez Consellería de Sanidade Xunta de Galicia El 26/03/2015 a las 15:04, Our Utopy escribió: Hola amigos, buenos días Esto sigue avanzando. Estoy ahora tratando de establecer la conectividad de R con Excel y he seguido dos viídeos en YouTube de una profesora llamada Bebilda que se explica muy bien. Con ambos tengo el mismo problema así que planteo uno de ellos, éste es: https://youtu.be/_pSJQO_9I4k?list=PL7DA3FDA21A1A6310 Pero os lo explico brevemente: Cargo el paquete RODBC install.packages(RODBC) Instalo la librería library(RODBC) Todo bien hasta aqui... Se supone, que en esa librería hay unas funciones que me permiten conectar con Excel, esas funciones son: o-odncConnectExcel2007 o-odncConnectExcel DE tal manera que ahora podría abrir una canal canalexcel - odbcConnectExcel2007(lbw.xlsx) Pues esas funciones no están. R me da el error Error: no se pudo encontrar la función odbcConnectExcel2007 lo mismo con la otra función canalexcel - odbcConnectExcel(lbw.xlsx) Y lo curioso es que también me pasa algo similar con la otra posibilidad que es el paquete XLConnect Y es que, además, si miro la documentación de este paquete RODBC no figuran esas funciones. ¿ ? ¿Me podéis nuevamente ayudar? Un saludo 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
