Re: [R-es] como hacer un simple barchart horizontal ordenado por el valor ...
Eric, En un barplot, el orden que se usa es el de los niveles del factor. Al crear un factor, por defecto este orden es alfabético. Lo que tienes que hacer es reordenar los niveles del factor usando el criterio de la otra variable (podría ser cualquier función aplicada a las observaciones del factor, en tu caso sólo tienes una). Prueba este código: math - read.table(math.txt, header = TRUE, sep=,, stringsAsFactors = TRUE, row.names = 1) math$CNT - reorder(math$CNT, math$mave1, mean) library(lattice) barchart(CNT ~ mave1, data = math[order(math$mave1),], horizontal = TRUE) Por otra parte, ¿qué quieres pintar en la leyenda? Necesitarías alguna otra variable, por ejemplo el continente, el idioma, o algo así... Un saludo, Emilio *Emilio L. Cano* Mobile: +34 665 676 225 skype: emilopezcano twitter: @emilopezcano http://emilio.lcano.com El 19 de mayo de 2015, 7:56, eric ericconchamu...@gmail.com escribió: Estimada comunidad, me resulta muy frustrante no poder hacer un simple barchart como yo quisiera (es la primera vez que intento hacer uno en todo caso :)). Gaste todo el dia buscando y no puedo darle las caracteristicas que necesito: Tengo un simple data.frame que adjunto, con paises y los puntajes que obtuvieron en una prueba y quiero hacer un barchart con los puntajes en orden ascendente, con las barras horizontales y que muestre el nombre de los paises, aunque sea con la fuente pequeña. Uso el siguiente codigo y no resulta: barchart(ave ~ CNT, data=math, horizontal=TRUE) con eso consigo que sea horizontal, pero no se como hacer que las barras se ordenen de acuerdo al valor de ave, lo que sospecho tiene que ver con index.cond, pero no se como usarlo. Tampoco puedo siquiera hacer que aparezca una leyenda con auto.key=TRUE Alguna sugerencia ? Adjunto los datos. Muchas gracias. -- Forest Engineer Master in Environmental and Natural Resource Economics Ph.D. student in Sciences of Natural Resources at La Frontera University Member in AguaDeTemu2030, citizen movement for Temuco with green city standards for living Nota: Las tildes se han omitido para asegurar compatibilidad con algunos lectores de correo. ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-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] Output interpretation: standard error of lm dummy variable
Hi guys I have a statistical question to an analyse I ran in R. It is a dummy variable model with the 5 regions of Denmark as 4 independent dummy variables and price as the dependent variable: price = 10.325 - 0.176*Sjaeland - 0.368 * NJylland - 0.230*MJylland - 0.120* Syddanmark I understand the R^2 = 0.7348 - that it shows the explanatory force of the model (between 0 and 1) My question is simply how to interpret the standard error = 0.7348 on 342199 degrees of freedom? How is it calculated when the model is a dummy variable model. And what does it mean that the F-statistic says that there are 1894 on 4 and 342199 DF (degrees if freedom?) with a p-value 0? I have been searching for hours - and can't quite figure out how R reached the numbers and how to interpret the output of standard error and the p-value of the dummy model. I really hope you can help :) Best Livia __ 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] Output interpretation: standard error of lm dummy variable
Hi Livia, You seem to have mixed up the residual error with the R^2, which is just over 0.02. The bottom line on your summary table says that the obtained F statistic was equal to 1894 (this has been truncated to four significant places). The probability of obtaining that value with your data given the F distribution for 4 numerator and 342199 denominator degrees of freedom is very small, but not less than 0. The notation 2.2e-16 can be roughly translated in English as about as close to zero as this function can calculate. You should note that with that many observations, a significant result is almost guaranteed, but the linear model explains almost none of the variance in prices. Jim On Tue, May 19, 2015 at 6:16 PM, Livia Maria Vestergaard lves...@student.sdu.dk wrote: Hi guys I have a statistical question to an analyse I ran in R. It is a dummy variable model with the 5 regions of Denmark as 4 independent dummy variables and price as the dependent variable: price = 10.325 - 0.176*Sjaeland - 0.368 * NJylland - 0.230*MJylland - 0.120* Syddanmark I understand the R^2 = 0.7348 - that it shows the explanatory force of the model (between 0 and 1) My question is simply how to interpret the standard error = 0.7348 on 342199 degrees of freedom? How is it calculated when the model is a dummy variable model. And what does it mean that the F-statistic says that there are 1894 on 4 and 342199 DF (degrees if freedom?) with a p-value 0? I have been searching for hours - and can't quite figure out how R reached the numbers and how to interpret the output of standard error and the p-value of the dummy model. I really hope you can help :) Best Livia __ 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] Finding Optimum of Stochastic Function
Most of the stochastic optimization methods are directed at multiple optima. You appear to have an imprecisely determined function e.g., time taken for racing driver to get round the track, and indeed this is a different form of stochastic optimization. With Harry Joe of UBC I did quite a bit of work about 20 years ago on response surface minimization, but none of this is (to my knowledge) translated to R. David Wagstaff at Penn State just sent me a msg that he's making some progress translating our Fortran 77 to Fortran 95 (I think to R might actually be easier). I believe Harry had at least a partial C version. reference is Statistics and Computing 13: 277–286, 2003 Numerical optimization and surface estimation with imprecise function evaluations Our approach was to generate some points and model them as a paraboloid and then search near the minimum of that model. All the smarts are in choosing which points to add to or remove from the set of points for modelling the surface. Clearly there are no guarantees, but for some applications we found this worked not too badly. JN On 15-05-19 06:00 AM, r-help-requ...@r-project.org wrote: Message: 11 Date: Mon, 18 May 2015 14:47:49 -0700 From: ivo welch ivo.we...@anderson.ucla.edu To: r-help@r-project.org Subject: [R] Finding Optimum of Stochastic Function Message-ID: CAPr7RtV99V3=3qumkamlridyecor2maq6cqxrv3929madcu...@mail.gmail.com Content-Type: text/plain; charset=UTF-8 Could someone please point me to an optimizer for stochastic functions? (In http://cran.r-project.org/web/views/Optimization.html, I saw methods that use random directions for deterministic functions, which is not the kind of stochastic I need.) For clarification, say I have an outcome function f(x), where x is a vector of, say, 3 choices. f(x) yields a simulated result that depends on random draws. That is, if I run it twice, it will give me different answers. I want to find the value of x that has the highest average f(x). There are apparently well-defined algorithms, such as Robbins-Monro, Kiefer-Wolfowitz, and Spall, although I don't know how they work nor do I need to know much. Presumably, a good algorithm knows not to draw too many points at a given x too early (when far away from the optimum), but to start more scattershot; and not to try to climb too aggressively. Intuitively, I probably want to start from a point, draw in a cloud around this point, and slowly sample-crawl into the direction where values tend to be higher. Ideally, the algorithm would try to solve an updatable least-squares problem to determine its next sample. Pointers appreciated. regards, /iaw Ivo Welch (ivo.we...@gmail.com) [[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] Count unchanged class attribute
John Kane Kingston ON Canada -Original Message- From: gunter.ber...@gene.com Sent: Tue, 19 May 2015 20:04:58 -0700 To: jrkrid...@inbox.com Subject: Re: [R] Count unchanged class attribute Probably or what. This demonstrates a fundamental conundrum: many users or prospective users of R have had little exposure to data structures in their formal education and therefore can be flummoxed by R's fussiness -- as any programming language must necessarily be. Consider: data frames, matrices, lists, classes, objects with attributes (e.g. factors),... == Nonsense, I am sure that weird feeling that my brain was being wrung out like a dishcloth, that I felt for the first 3-5 weeks was due to something I ate and had nothing to due with SPSS or SAS. === Excel, which is basically structureless, of course, exacerbates the problem. Those accustomed to its tolerance (and the confusion that results) expect R to behave the same way. Education is the only recourse, either in formal courses or through R tutorials that strongly emphasize this aspect of interacting with R and especially writing effective code. But that demands effort and, to some extent, aptitude... both of which seem to be in increasingly short supply amidst the worldwide explosion in R's usage. I don't think it's lack of aptitude but I seldom see much in the various tutorials and books that really emphasis data structures or typing so people can spend a lot of time figuring out what a list is. Who, me? I agree that Excel is scary. I believe I mentioned before that I live in fear that some medical spreadsheet will calculate a medical dose on my telephone number rather than my weight. Of course, Excel being Excel, it would probably obligingly translate a character-formatted telephone number into a real number. I must try this the next time I get close to a machine with Excel. === Of course, feel free to disagree... Just my $.02 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 Tue, May 19, 2015 at 7:02 PM, John Kane jrkrid...@inbox.com wrote: Is this a list of data.frames or what? Please have a look at one or both of these for some ideas of how to ask a question and provide information on the problem. The better you can describe what you have and what you need the better people can help. http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example [http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example] and http://adv-r.had.co.nz/Reproducibility.html [http://adv-r.had.co.nz/Reproducibility.html] John Kane Kingston ON Canada -Original Message- From: soe.xi...@gmail.com Sent: Tue, 19 May 2015 23:37:13 +0700 To: r-help@r-project.org Subject: [R] Count unchanged class attribute Maybe someone can help me. Suppose I have data-set like this: Netto Bruto class 1 10 1000 yes 2 100 20 yes Netto Bruto class 1 101 1000 yes 2 100 210 no Netto Bruto class 1 10 10 yes 2 12 28 yes 3 100 20 yes Netto Bruto class 1 120 200 no 2 400 20 yes Netto Bruto class 1 110 12000 yes 2 1100 120 yes 3 120 100 yes 4 1140 125 yes How to calculate the number of classes has changed. The expected result is - class changed 2 - class unchanged 3 Thank you so much. Soe Xiyan [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help [https://stat.ethz.ch/mailman/listinfo/r-help] PLEASE do read the posting guide http://www.R-project.org/posting-guide.html [http://www.R-project.org/posting-guide.html] and provide commented, minimal, self-contained, reproducible code. Can't remember your password? Do you need a strong and secure password? Use Password manager! It stores your passwords protects your account. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help [https://stat.ethz.ch/mailman/listinfo/r-help] PLEASE do read the posting guide http://www.R-project.org/posting-guide.html [http://www.R-project.org/posting-guide.html] and provide commented, minimal, self-contained, reproducible code. Can't remember your password? Do you need a strong and secure password? Use Password manager! It stores your passwords
Re: [R] Count unchanged class attribute
Is this a list of data.frames or what? Please have a look at one or both of these for some ideas of how to ask a question and provide information on the problem. The better you can describe what you have and what you need the better people can help. http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example and http://adv-r.had.co.nz/Reproducibility.html John Kane Kingston ON Canada -Original Message- From: soe.xi...@gmail.com Sent: Tue, 19 May 2015 23:37:13 +0700 To: r-help@r-project.org Subject: [R] Count unchanged class attribute Maybe someone can help me. Suppose I have data-set like this: Netto Bruto class 1 10 1000yes 2 100 20 yes Netto Bruto class 1 101 1000yes 2 100 210 no Netto Bruto class 1 10 10 yes 2 12 28 yes 3 100 20 yes Netto Bruto class 1 120 200 no 2 400 20 yes Netto Bruto class 1 110 12000 yes 2 1100120 yes 3 120 100 yes 4 1140125 yes How to calculate the number of classes has changed. The expected result is - class changed2 - class unchanged 3 Thank you so much. Soe Xiyan [[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. Can't remember your password? Do you need a strong and secure password? Use Password manager! It stores your passwords protects your account. __ 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] Count unchanged class attribute
On 20/05/15 04:37, Soe Xiyan wrote: Maybe someone can help me. Suppose I have data-set like this: Netto Bruto class 1 10 1000yes 2 100 20 yes Netto Bruto class 1 101 1000yes 2 100 210 no Netto Bruto class 1 10 10 yes 2 12 28 yes 3 100 20 yes Netto Bruto class 1 120 200 no 2 400 20 yes Netto Bruto class 1 110 12000 yes 2 1100120 yes 3 120 100 yes 4 1140125 yes How to calculate the number of classes has changed. The expected result is - class changed2 - class unchanged 3 The actual structure of your data set is unclear. You appear to have your data stored in a number of separate data frames, but this is not made explicit. The details must be specified in order for anyone to be able to help you. Moreover it is not at all clear what you want to achieve. Why is class changed equal to 2 and class unchanged equal to 3? I count 2 no-s and 11 yes-s (not 3 yes-s). There are many clever people who subscribe to the R-help list, but few if any of them are mind-readers. cheers, 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] Count unchanged class attribute
Probably or what. This demonstrates a fundamental conundrum: many users or prospective users of R have had little exposure to data structures in their formal education and therefore can be flummoxed by R's fussiness -- as any programming language must necessarily be. Consider: data frames, matrices, lists, classes, objects with attributes (e.g. factors),... Excel, which is basically structureless, of course, exacerbates the problem. Those accustomed to its tolerance (and the confusion that results) expect R to behave the same way. Education is the only recourse, either in formal courses or through R tutorials that strongly emphasize this aspect of interacting with R and especially writing effective code. But that demands effort and, to some extent, aptitude... both of which seem to be in increasingly short supply amidst the worldwide explosion in R's usage. Of course, feel free to disagree... Just my $.02 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 Tue, May 19, 2015 at 7:02 PM, John Kane jrkrid...@inbox.com wrote: Is this a list of data.frames or what? Please have a look at one or both of these for some ideas of how to ask a question and provide information on the problem. The better you can describe what you have and what you need the better people can help. http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example and http://adv-r.had.co.nz/Reproducibility.html John Kane Kingston ON Canada -Original Message- From: soe.xi...@gmail.com Sent: Tue, 19 May 2015 23:37:13 +0700 To: r-help@r-project.org Subject: [R] Count unchanged class attribute Maybe someone can help me. Suppose I have data-set like this: Netto Bruto class 1 10 1000yes 2 100 20 yes Netto Bruto class 1 101 1000yes 2 100 210 no Netto Bruto class 1 10 10 yes 2 12 28 yes 3 100 20 yes Netto Bruto class 1 120 200 no 2 400 20 yes Netto Bruto class 1 110 12000 yes 2 1100120 yes 3 120 100 yes 4 1140125 yes How to calculate the number of classes has changed. The expected result is - class changed2 - class unchanged 3 Thank you so much. Soe Xiyan [[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. Can't remember your password? Do you need a strong and secure password? Use Password manager! It stores your passwords protects your account. __ 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] Error in dimnames(phi) - list(rn, dims) : length of 'dimnames' [2] not equal to array extent
Dear Oleg, Thank you so much for your advice! But I tried to update the packages and check whether the problem is fixed. But I end up with the same error. Error in dimnames(phi) - list(rn, dims) : length of 'dimnames' [2] not equal to array extent It seems to me that, what I got to update from R-forge is version 0.60 not version 0.61. Please see below the script I got while updating the package: trying URL 'http://R-Forge.R-project.org/bin/windows/contrib/3.1/ca_0.60.zip ' Content type 'application/zip' length 98081 bytes (95 KB) opened URL downloaded 95 KB Do you think version 0.60 is not fixing the problem? Kindly, Yonas On Mon, May 18, 2015 at 6:52 PM, Oleg Nenadić onen...@uni-goettingen.de wrote: David, Thanks for forwarding this to me. Yonas, Please try updating your ca package to version 0.61 which fixes this issue. This is not the latest official CRAN version, so get it from R-forge via update.packages(ca, repos = http://r-forge.r-project.org;) All the best, Oleg. On 18/05/2015 16:27, David L Carlson wrote: I think this is a bug in the current version of ca() in package ca. I am copying the package maintainer with this example: # Reproducible example from manual page for ca(): library(ca) data(author) author.ca - ca(author) # No problem author.ca - ca(author, nd=3) Error in dimnames(phi) - list(rn, dims) : length of 'dimnames' [2] not equal to array extent library(MASS) author.ca - corresp(author, nf=3) # No problem So the MASS version of correspondence analysis, corresp(), is able to extract three dimensions (actually up to 11) from author. I am certain I have used ca() in the past and extracted more than two dimensions from similar tables. - 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 Jim Lemon Sent: Sunday, May 17, 2015 7:31 PM Cc: r-help mailing list Subject: Re: [R] Error in dimnames(phi) - list(rn, dims) : length of 'dimnames' [2] not equal to array extent Hi Yonas, If this is the ca function from the package of the same name, it looks to me as though your data set is only two dimensions and you are requesting 3 dimensions in the output. Have you tried calling ca with the default nd=NA? Jim On Mon, May 18, 2015 at 3:22 AM, Jeff Newmiller jdnew...@dcn.davis.ca.us wrote: You desperately need to study [1] and the Posting Guide mentioned at the bottom of this and every other message on this list. [1] http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example --- 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 May 17, 2015 7:31:08 AM PDT, Yonas Yohannes yo...@wku.edu.et wrote: Dears, I have presence and absence data set (8 rows and 33 columns) and when I want to get out of the default two dimensions command using summary(ca(mydata, nd=3)) the following error message displyed: Error in dimnames(phi) - list(rn, dims) : length of 'dimnames' [2] not equal to array extent Please help! So many thanks in advance! Kindly, * mail%3ayo...@wku.edu.et* [[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-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
[R] Count number in r
Dear R users,
Could someone help me on this? I have this kind of data set:
structure(list(Year = c(1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L), Month = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L), Rain = c(58.9, 74.6, 17.7, 7.8, 1.2, 1, 5.3, 0.7,
1.2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 10.4, 17.5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)), .Names = c(Year,
Month, Rain), class = data.frame, row.names = c(NA, -44L
))
I want to count data in Rain that is greater and equal to 0.1 mm
according to their Month and Year. I have used this code, but it seems
so wrong.
raindat - read.csv('my data set',header=TRUE)
yearcorr-min(raindat$Year)-1
years-unique(raindat$Year)
rainmonth-as.data.frame(matrix(0,nrow=2,ncol=12))
for(year in years) {
for(month in 1:12) {
if(any(raindat$Year==yearraindat$Month==month))
rainmonth[year-yearcorr,month]-
length((which(raindat$Rain =
0.1))[raindat$Year==yearraindat$Month==month])
}
}
rownames(rainmonth)-years
names(rainmonth)-month.abb
rainmonth
Thank you so much.
Arshad
[[alternative HTML version deleted]]
__
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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] Count number in r
Dear Arshad,
Here is a possibility using tapply():
with(d, tapply(Rain, list(Month, Year), function(x) sum(x .1)))
##1971
#1 12
#20
where d is your data.frame(). See also ?aggregate and ?ave.
Best,
Jorge.-
On Tue, May 19, 2015 at 8:10 PM, Hafizuddin Arshad
hafizuddinarsha...@gmail.com wrote:
Dear R users,
Could someone help me on this? I have this kind of data set:
structure(list(Year = c(1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L), Month = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L), Rain = c(58.9, 74.6, 17.7, 7.8, 1.2, 1, 5.3, 0.7,
1.2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 10.4, 17.5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)), .Names = c(Year,
Month, Rain), class = data.frame, row.names = c(NA, -44L
))
I want to count data in Rain that is greater and equal to 0.1 mm
according to their Month and Year. I have used this code, but it seems
so wrong.
raindat - read.csv('my data set',header=TRUE)
yearcorr-min(raindat$Year)-1
years-unique(raindat$Year)
rainmonth-as.data.frame(matrix(0,nrow=2,ncol=12))
for(year in years) {
for(month in 1:12) {
if(any(raindat$Year==yearraindat$Month==month))
rainmonth[year-yearcorr,month]-
length((which(raindat$Rain =
0.1))[raindat$Year==yearraindat$Month==month])
}
}
rownames(rainmonth)-years
names(rainmonth)-month.abb
rainmonth
Thank you so much.
Arshad
[[alternative HTML version deleted]]
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Re: [R] Error in dimnames(phi) - list(rn, dims) : length of 'dimnames' [2] not equal to array extent
Dear Yonas, It usually takes some time until a Windows binary version is available on R-Forge. It is version 0.61 you want here. So, you can either install the source version using install.packages(ca, repos=http://R-Forge.R-project.org;, type=source) or manually download and install the binary package from https://r-forge.r-project.org/R/?group_id=1859 Kind regards, Oleg. On 19.05.2015 08:23, Yonas Yohannes wrote: Dear Oleg, Thank you so much for your advice! But I tried to update the packages and check whether the problem is fixed. But I end up with the same error. Error in dimnames(phi) - list(rn, dims) : length of 'dimnames' [2] not equal to array extent It seems to me that, what I got to update from R-forge is version 0.60 not version 0.61. Please see below the script I got while updating the package: trying URL 'http://R-Forge.R-project.org/bin/windows/contrib/3.1/ca_0.60.zip' Content type 'application/zip' length 98081 bytes (95 KB) opened URL downloaded 95 KB Do you think version 0.60 is not fixing the problem? Kindly, Yonas On Mon, May 18, 2015 at 6:52 PM, Oleg Nenadić onen...@uni-goettingen.de mailto:onen...@uni-goettingen.de wrote: David, Thanks for forwarding this to me. Yonas, Please try updating your ca package to version 0.61 which fixes this issue. This is not the latest official CRAN version, so get it from R-forge via update.packages(ca, repos = http://r-forge.r-project.org;) All the best, Oleg. On 18/05/2015 16:27, David L Carlson wrote: I think this is a bug in the current version of ca() in package ca. I am copying the package maintainer with this example: # Reproducible example from manual page for ca(): library(ca) data(author) author.ca http://author.ca - ca(author) # No problem author.ca http://author.ca - ca(author, nd=3) Error in dimnames(phi) - list(rn, dims) : length of 'dimnames' [2] not equal to array extent library(MASS) author.ca http://author.ca - corresp(author, nf=3) # No problem So the MASS version of correspondence analysis, corresp(), is able to extract three dimensions (actually up to 11) from author. I am certain I have used ca() in the past and extracted more than two dimensions from similar tables. - 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 mailto:r-help-boun...@r-project.org] On Behalf Of Jim Lemon Sent: Sunday, May 17, 2015 7:31 PM Cc: r-help mailing list Subject: Re: [R] Error in dimnames(phi) - list(rn, dims) : length of 'dimnames' [2] not equal to array extent Hi Yonas, If this is the ca function from the package of the same name, it looks to me as though your data set is only two dimensions and you are requesting 3 dimensions in the output. Have you tried calling ca with the default nd=NA? Jim On Mon, May 18, 2015 at 3:22 AM, Jeff Newmiller jdnew...@dcn.davis.ca.us mailto:jdnew...@dcn.davis.ca.us wrote: You desperately need to study [1] and the Posting Guide mentioned at the bottom of this and every other message on this list. [1] http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example --- Jeff Newmiller The . . Go Live... DCN:jdnew...@dcn.davis.ca.us mailto:jdnew...@dcn.davis.ca.us Basics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/Batteries O.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. On May 17, 2015 7:31:08 AM PDT, Yonas Yohannes yo...@wku.edu.et mailto:yo...@wku.edu.et wrote: Dears, I have presence and absence data set (8 rows and 33 columns) and when I want to get out of the default two dimensions command using summary(ca(mydata, nd=3)) the following error message displyed: Error in dimnames(phi) - list(rn, dims) : length of 'dimnames' [2] not equal to array extent Please help! So many thanks in advance! Kindly, * mail%3ayo...@wku.edu.et mailto:mail%253ayo...@wku.edu.et* [[alternative HTML version deleted]] __ R-help@r-project.org mailto:R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
[R-es] Formador en R en México DF....
Hola, Eduard (le conocemos del grupo de Periodismo de Datos) nos ha pedido ayuda para localizar a un formador en R en México DF. Si alguien de la lista está interesado, por favor que se ponga en contacto con él o con Tamar (ambos en copia) para concretar detalles de la colaboración. G racias, 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] Count number in r
Here's an approach using xtabs() if you want the output as a table:
flag - as.integer(d$Rain=.1)
xtabs(flag~Year+Month, d)
Month
Year1 2
1971 12 0
-
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 Jorge I Velez
Sent: Tuesday, May 19, 2015 8:21 AM
To: Hafizuddin Arshad
Cc: R Help
Subject: Re: [R] Count number in r
Dear Arshad,
Here is a possibility using tapply():
with(d, tapply(Rain, list(Month, Year), function(x) sum(x .1)))
##1971
#1 12
#20
where d is your data.frame(). See also ?aggregate and ?ave.
Best,
Jorge.-
On Tue, May 19, 2015 at 8:10 PM, Hafizuddin Arshad
hafizuddinarsha...@gmail.com wrote:
Dear R users,
Could someone help me on this? I have this kind of data set:
structure(list(Year = c(1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L), Month = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L), Rain = c(58.9, 74.6, 17.7, 7.8, 1.2, 1, 5.3, 0.7,
1.2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 10.4, 17.5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)), .Names = c(Year,
Month, Rain), class = data.frame, row.names = c(NA, -44L
))
I want to count data in Rain that is greater and equal to 0.1 mm
according to their Month and Year. I have used this code, but it seems
so wrong.
raindat - read.csv('my data set',header=TRUE)
yearcorr-min(raindat$Year)-1
years-unique(raindat$Year)
rainmonth-as.data.frame(matrix(0,nrow=2,ncol=12))
for(year in years) {
for(month in 1:12) {
if(any(raindat$Year==yearraindat$Month==month))
rainmonth[year-yearcorr,month]-
length((which(raindat$Rain =
0.1))[raindat$Year==yearraindat$Month==month])
}
}
rownames(rainmonth)-years
names(rainmonth)-month.abb
rainmonth
Thank you so much.
Arshad
[[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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[R] Count unchanged class attribute
Maybe someone can help me. Suppose I have data-set like this: Netto Bruto class 1 10 1000yes 2 100 20 yes Netto Bruto class 1 101 1000yes 2 100 210 no Netto Bruto class 1 10 10 yes 2 12 28 yes 3 100 20 yes Netto Bruto class 1 120 200 no 2 400 20 yes Netto Bruto class 1 110 12000 yes 2 1100120 yes 3 120 100 yes 4 1140125 yes How to calculate the number of classes has changed. The expected result is - class changed2 - class unchanged 3 Thank you so much. Soe Xiyan [[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] RJDBC and Unicode characters on Windows
Hello, I am using the RJDBC library to connect to a SQL database and pull out text that includes Hebrew characters. If I do this on a Mac, the data frames I construct display the characters properly. If I do it on a Windows PC, then the characters are converted to strings like this: U+05DEU+05E2U+05D5U+05EA U+05D7U+05D8U+05D9U+05DD I have tried UTF-8, UCS-2LE, UTF-8-BOM, and UTF-16LE encodings with no luck. Any suggestions on getting the data to display properly? Thanks, Josh [[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] Count number in r
And another approach just for the heck of it.
library(plyr)
# where dat1 is your data
dd1 - subset(dat1, Rain = .01)
dd1$Year - as.factor(dd1$Year)
dd1$Month - as.factor(dd1$Month)
count (dd1, .(Year, Month))
John Kane
Kingston ON Canada
-Original Message-
From: hafizuddinarsha...@gmail.com
Sent: Tue, 19 May 2015 03:10:32 -0700
To: r-help@r-project.org
Subject: [R] Count number in r
Dear R users,
Could someone help me on this? I have this kind of data set:
structure(list(Year = c(1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L), Month = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L), Rain = c(58.9, 74.6, 17.7, 7.8, 1.2, 1, 5.3, 0.7,
1.2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 10.4, 17.5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)), .Names =
c(Year,
Month, Rain), class = data.frame, row.names = c(NA, -44L
))
I want to count data in Rain that is greater and equal to 0.1 mm
according to their Month and Year. I have used this code, but it
seems
so wrong.
raindat - read.csv('my data set',header=TRUE)
yearcorr-min(raindat$Year)-1
years-unique(raindat$Year)
rainmonth-as.data.frame(matrix(0,nrow=2,ncol=12))
for(year in years) {
for(month in 1:12) {
if(any(raindat$Year==yearraindat$Month==month))
rainmonth[year-yearcorr,month]-
length((which(raindat$Rain =
0.1))[raindat$Year==yearraindat$Month==month])
}
}
rownames(rainmonth)-years
names(rainmonth)-month.abb
rainmonth
Thank you so much.
Arshad
[[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.
Can't remember your password? Do you need a strong and secure password?
Use Password manager! It stores your passwords protects your account.
__
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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] Count number in r
Dear all,
I am kindly requesting for help on how I can count pixels with value less
and equal to -0.08 for a raster stack.
Thanks for your help
John
On Tue, May 19, 2015 at 5:57 PM, John Kane jrkrid...@inbox.com wrote:
And another approach just for the heck of it.
library(plyr)
# where dat1 is your data
dd1 - subset(dat1, Rain = .01)
dd1$Year - as.factor(dd1$Year)
dd1$Month - as.factor(dd1$Month)
count (dd1, .(Year, Month))
John Kane
Kingston ON Canada
-Original Message-
From: hafizuddinarsha...@gmail.com
Sent: Tue, 19 May 2015 03:10:32 -0700
To: r-help@r-project.org
Subject: [R] Count number in r
Dear R users,
Could someone help me on this? I have this kind of data set:
structure(list(Year = c(1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L), Month = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L), Rain = c(58.9, 74.6, 17.7, 7.8, 1.2, 1, 5.3, 0.7,
1.2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 10.4, 17.5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)), .Names =
c(Year,
Month, Rain), class = data.frame, row.names = c(NA, -44L
))
I want to count data in Rain that is greater and equal to 0.1 mm
according to their Month and Year. I have used this code, but it
seems
so wrong.
raindat - read.csv('my data set',header=TRUE)
yearcorr-min(raindat$Year)-1
years-unique(raindat$Year)
rainmonth-as.data.frame(matrix(0,nrow=2,ncol=12))
for(year in years) {
for(month in 1:12) {
if(any(raindat$Year==yearraindat$Month==month))
rainmonth[year-yearcorr,month]-
length((which(raindat$Rain =
0.1))[raindat$Year==yearraindat$Month==month])
}
}
rownames(rainmonth)-years
names(rainmonth)-month.abb
rainmonth
Thank you so much.
Arshad
[[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.
Can't remember your password? Do you need a strong and secure password?
Use Password manager! It stores your passwords protects your account.
__
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.
--
John Wasige
Birds born in a Cage think Flying is an illness.
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Urgent :) Output interpretation: standard error of lm dummy variable
Hi :) I have an an examination tomorrow, and don't quite understand how R calculated the values. Yes. You are right R^2=0.02166 :) As mentioned It is a dummy variable model with the 5 regions of Denmark as 4 independent dummy variables and price as the dependent variable. price = 10.325 - 0.176*Sjaeland - 0.368 * NJylland - 0.230*MJylland - 0.120* Syddanmark I will probably be asked how to interpret the standard error = 0.7348 on 342199 degrees of freedom (= 342 204 observations - 5 categories); about the 5 standard errors for the beta values, the F-statistic = 1894 on 4 categories and the p-value ≈ 0. But I don't quite understand how R reached the outputs and what parameter are F-distributed, what the standard errors says something about and the standard errors and the F-statistic = 1894 when it is a dummy variable model. Hopefully the answer is out there somewhere and you can help :) Best Livia Fra: Livia Maria Vestergaard Sendt: 19. maj 2015 10:16 Til: r-help@r-project.org Emne: [R] Output interpretation: standard error of lm dummy variable Hi guys I have a statistical question to an analyse I ran in R. It is a dummy variable model with the 5 regions of Denmark as 4 independent dummy variables and price as the dependent variable: price = 10.325 - 0.176*Sjaeland - 0.368 * NJylland - 0.230*MJylland - 0.120* Syddanmark I understand the R^2 = 0.7348 - that it shows the explanatory force of the model (between 0 and 1) My question is simply how to interpret the standard error = 0.7348 on 342199 degrees of freedom? How is it calculated when the model is a dummy variable model. And what does it mean that the F-statistic says that there are 1894 on 4 and 342199 DF (degrees if freedom?) with a p-value 0? I have been searching for hours - and can't quite figure out how R reached the numbers and how to interpret the output of standard error and the p-value of the dummy model. I really hope you can help :) Best Livia __ 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] Fixed effects regression and robust regression
Hello: I am using R 3.0.2. I have panel data on countries' renewable energy net generation (and installed capacity) over time. I am regressing these dependent variables on various socioeconomic variables, as well as binary policy variables. I have have done basic OLS, but I wanted to explore both fixed effects models, as there are likely significant country effects (using plm) and robust regression (using rlm), as Q-Q plots indicate that there are some strong outliers. This might be a question of apples and oranges, but how do I compare the goodness of fit of the fixed effects models with the robust regression models? One can use F-tests to compare OLS and the fixed effects, and since the OLS and robust regressions have the same number of DFs, looking at the residual standard error is insightful. Any help would be appreciated. Cheers, Michael [[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] Fixed effects regression and robust regression
On Tue, May 19, 2015 at 1:23 PM, michael westphal via R-help r-help@r-project.org wrote: You can't compare them (statistically -- you can of course draw pictures). Note, from ?rlm: Note that the df.residual component is deliberately set to NA to avoid inappropriate estimation of the residual scale from the residual mean square by lm methods. Further questions should probably go to a statistics list like stats.stackexchange.com, as statistical questions are generally offtopic here. 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 __ 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] Count number in r
If nothing suggested in this thread help I'd suggest asking in R-sig-Geo where
they will be more familiar with the issues.
Please do not post in HTML. It can serious mangle code to the point it is
indecipherable.
John Kane
Kingston ON Canada
-Original Message-
From: johnwas...@gmail.com
Sent: Tue, 19 May 2015 18:26:35 +0200
To: jrkrid...@inbox.com
Subject: Re: [R] Count number in r
Dear all,
I am kindly requesting for help on how I can count pixels with value less and
equal to -0.08 for a raster stack.
Thanks for your help
John
On Tue, May 19, 2015 at 5:57 PM, John Kane jrkrid...@inbox.com wrote:
And another approach just for the heck of it.
library(plyr)
# where dat1 is your data
dd1 - subset(dat1, Rain = .01)
dd1$Year - as.factor(dd1$Year)
dd1$Month - as.factor(dd1$Month)
count (dd1, .(Year, Month))
John Kane
Kingston ON Canada
-Original Message-
From: hafizuddinarsha...@gmail.com
Sent: Tue, 19 May 2015 03:10:32 -0700
To: r-help@r-project.org
Subject: [R] Count number in r
Dear R users,
Could someone help me on this? I have this kind of data set:
structure(list(Year = c(1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
1971L, 1971L), Month = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L), Rain = c(58.9, 74.6, 17.7, 7.8, 1.2, 1, 5.3, 0.7,
1.2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 10.4, 17.5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)), .Names =
c(Year,
Month, Rain), class = data.frame, row.names = c(NA, -44L
))
I want to count data in Rain that is greater and equal to 0.1 mm
according to their Month and Year. I have used this code, but it
seems
so wrong.
raindat - read.csv('my data set',header=TRUE)
yearcorr-min(raindat$Year)-1
years-unique(raindat$Year)
rainmonth-as.data.frame(matrix(0,nrow=2,ncol=12))
for(year in years) {
for(month in 1:12) {
if(any(raindat$Year==yearraindat$Month==month))
rainmonth[year-yearcorr,month]-
length((which(raindat$Rain =
0.1))[raindat$Year==yearraindat$Month==month])
}
}
rownames(rainmonth)-years
names(rainmonth)-month.abb
rainmonth
Thank you so much.
Arshad
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
[https://stat.ethz.ch/mailman/listinfo/r-help]
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
[http://www.R-project.org/posting-guide.html]
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Re: [R] Urgent :) Output interpretation: standard error of lm dummy variable
Technically it is not on topic to discuss the statistics behind R calculations here, and certainly not our job to do so within the context of your educational institution's schedule. However, you have the power to read the source code of any function in R or contributed packages, so go for it. Just enter the name of most functions without parentheses at the R command line. Page 43 of [1] should help for more deeply hidden code. [1] http://cran.r-project.org/doc/Rnews/Rnews_2006-4.pdf --- 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 May 19, 2015 1:37:13 PM PDT, Livia Maria Vestergaard lves...@student.sdu.dk wrote: Hi :) I have an an examination tomorrow, and don't quite understand how R calculated the values. Yes. You are right R^2=0.02166 :) As mentioned It is a dummy variable model with the 5 regions of Denmark as 4 independent dummy variables and price as the dependent variable. price = 10.325 - 0.176*Sjaeland - 0.368 * NJylland - 0.230*MJylland - 0.120* Syddanmark I will probably be asked how to interpret the standard error = 0.7348 on 342199 degrees of freedom (= 342 204 observations - 5 categories); about the 5 standard errors for the beta values, the F-statistic = 1894 on 4 categories and the p-value ≈ 0. But I don't quite understand how R reached the outputs and what parameter are F-distributed, what the standard errors says something about and the standard errors and the F-statistic = 1894 when it is a dummy variable model. Hopefully the answer is out there somewhere and you can help :) Best Livia Fra: Livia Maria Vestergaard Sendt: 19. maj 2015 10:16 Til: r-help@r-project.org Emne: [R] Output interpretation: standard error of lm dummy variable Hi guys I have a statistical question to an analyse I ran in R. It is a dummy variable model with the 5 regions of Denmark as 4 independent dummy variables and price as the dependent variable: price = 10.325 - 0.176*Sjaeland - 0.368 * NJylland - 0.230*MJylland - 0.120* Syddanmark I understand the R^2 = 0.7348 - that it shows the explanatory force of the model (between 0 and 1) My question is simply how to interpret the standard error = 0.7348 on 342199 degrees of freedom? How is it calculated when the model is a dummy variable model. And what does it mean that the F-statistic says that there are 1894 on 4 and 342199 DF (degrees if freedom?) with a p-value 0? I have been searching for hours - and can't quite figure out how R reached the numbers and how to interpret the output of standard error and the p-value of the dummy model. I really hope you can help :) Best Livia __ 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.
