[R] is there colNA for a image? (i.e. How to set the color of NAs or background in an image)
Hi everyone, I am using a layout of two images, one of which is an image of a raster. In this one image I am using a color palette to show the gradient of temperature in the world.By default the color of the continents (which I set to NA in the raster) appears as white. I would like the NA values or background of just that one image to be of a given color. is this possible? I know I can use bg in par but that will change the background color of the entire window. I found I can use plot instead of image, which has the option of setting colNA to any color. My problem is that that the plot does not work well in the layout with the other image. Any help very much appreciated. Again, my question is how to set the background color of an image in a layout?. Thanks, Camilo Camilo Mora, Ph.D. Department of Geography, University of Hawaii __ R-help@r-project.org mailing list 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] How to concatenate a several rows according with a column ?
HI, Try this: #Reformatted the ddply results to match your desired output. dat1-read.table(text= a b c d E001234 TSA IP234 like_domain E001234 TSB IP234 like_domain E001234 TSC IP234 like_domain E001234 TSD IP234 like_domain E001235 TSA IP235 two_domain E001235 TSD IP235 two_domain E001235 TSS IP235 two_domain E001236 TSP IP236 like_domain E001236 TST IP236 like_domain E001237 TSV IP237 simple_domain ,sep=,header=TRUE,stringsAsFactors=FALSE) dat2-ddply(dat1,.(a,c,d), paste,sep=,) dat2-dat2[,c(1,5,2:3)] colnames(dat2)-colnames(dat1) dat3-data.frame(sapply(dat2,function(x) gsub(c|\\(|\\\|\\),,x))) dat3 # a b c d #1 E001234 TSA, TSB, TSC, TSD IP234 like_domain #2 E001235 TSA, TSD, TSS IP235 two_domain #3 E001236 TSP, TST IP236 like_domain #4 E001237 TSV IP237 simple_domain A.K. - Original Message - From: tgodoy tingola...@hotmail.com To: r-help@r-project.org Cc: Sent: Friday, August 3, 2012 1:00 PM Subject: [R] How to concatenate a several rows according with a column ? Hi, I'm a new user or R and I try to concatenate a several rows according with the value in a column. this is my data.frame and I want to concatenate my data.frame according with the column b and make a new data.frame with the information in the others columns. table1 a b c d 1 E001234 TSA IP234 like_domain 2 E001234 TSB IP234 like_domain 3 E001234 TSC IP234 like_domain 4 E001234 TSD IP234 like_domain 5 E001235 TSA IP235 two_domain 6 E001235 TSD IP235 two_domain 7 E001235 TSS IP235 two_domain 8 E001236 TSP IP236 like_domain 9 E001236 TST IP236 like_domain 10 E001237 TSV IP237 simple_domain I want my table in this way table_new a b c d 1 E001234 TSA, TSB, TSC, TSD IP234 like_domain 2 E001235 TSA, TSD, TSS IP235 two_domain 3 E001236 TSP, TSP, TST IP236 like_domain 4 E001237 TSV IP237 simple_domain How can I do this in R? Thanks -- View this message in context: http://r.789695.n4.nabble.com/How-to-concatenate-a-several-rows-according-with-a-column-tp4639072.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list 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 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] How to concatenate a several rows according with a column ?
On Aug 3, 2012, at 10:00 AM, tgodoy wrote: Hi, I'm a new user or R and I try to concatenate a several rows according with the value in a column. this is my data.frame and I want to concatenate my data.frame according with the column b and make a new data.frame with the information in the others columns. table1 - -read.table(text= ab c d 1 E001234 TSAIP234 like_domain 2 E001234 TSBIP234 like_domain 3 E001234 TSCIP234 like_domain 4 E001234 TSDIP234 like_domain 5 E001235 TSAIP235 two_domain 6 E001235 TSDIP235 two_domain 7 E001235 TSSIP235 two_domain 8 E001236 TSPIP236 like_domain 9 E001236 TSTIP236 like_domain 10E001237 TSVIP237 simple_domain,header=TRUE,stringsAsFactors=FALSE) aggrdat - with(table1, aggregate(b, list(a,c,d), FUN=paste, sep=,) ) names(aggrdat) - names(table1)[c(2:4,1)] aggrdat b c d a 1 E001234 IP234 like_domain TSA, TSB, TSC, TSD 2 E001236 IP236 like_domain TSP, TST 3 E001237 IP237 simple_domainTSV 4 E001235 IP235two_domain TSA, TSD, TSS Swapping the column position is left as an exercise. -- David. I want my table in this way table_new ab c d 1 E001234 TSA, TSB, TSC, TSDIP234 like_domain 2 E001235TSA, TSD, TSSIP235 two_domain 3 E001236 TSP, TSP, TSTIP236 like_domain 4 E001237 TSVIP237 simple_domain How can I do this in R? Thanks -- View this message in context: http://r.789695.n4.nabble.com/How-to-concatenate-a-several-rows-according-with-a-column-tp4639072.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list 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. David Winsemius, MD Alameda, CA, USA __ R-help@r-project.org mailing list 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] How to concatenate a several rows according with a column ?
Hello, Inline. Em 04-08-2012 09:34, David Winsemius escreveu: On Aug 3, 2012, at 10:00 AM, tgodoy wrote: Hi, I'm a new user or R and I try to concatenate a several rows according with the value in a column. this is my data.frame and I want to concatenate my data.frame according with the column b and make a new data.frame with the information in the others columns. table1 - -read.table(text= ab c d 1 E001234 TSAIP234 like_domain 2 E001234 TSBIP234 like_domain 3 E001234 TSCIP234 like_domain 4 E001234 TSDIP234 like_domain 5 E001235 TSAIP235 two_domain 6 E001235 TSDIP235 two_domain 7 E001235 TSSIP235 two_domain 8 E001236 TSPIP236 like_domain 9 E001236 TSTIP236 like_domain 10E001237 TSVIP237 simple_domain,header=TRUE,stringsAsFactors=FALSE) aggrdat - with(table1, aggregate(b, list(a,c,d), FUN=paste, sep=,) ) names(aggrdat) - names(table1)[c(2:4,1)] It's a column order issue, not a names one. Using Jean's form of aggregate, aggrdat - aggregate(b ~ a + c + d, data = table1, paste, sep=,) (table2 - aggrdat[, c(1, 4, 2, 3)]) Hope this helps, Rui Barradas aggrdat b c d a 1 E001234 IP234 like_domain TSA, TSB, TSC, TSD 2 E001236 IP236 like_domain TSP, TST 3 E001237 IP237 simple_domainTSV 4 E001235 IP235two_domain TSA, TSD, TSS Swapping the column position is left as an exercise. __ R-help@r-project.org mailing list 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] Writing Faroes lettsers in text and plot ?
Depends on your OS probably. I would just add a Faroese keyboard to my Ubuntu setup and switch between my normal keyboard and Faoese as required. John Kane Kingston ON Canada -Original Message- From: klausflemlo...@mail.tele.dk Sent: Fri, 3 Aug 2012 10:08:48 -0700 (PDT) To: r-help@r-project.org Subject: [R] Writing Faroes lettsers in text and plot ? I need to write Faroese letters in plain tekst and in plots. How can I do this ? http://en.wikipedia.org/wiki/Faroese_language Faroese_language -- View this message in context: http://r.789695.n4.nabble.com/Writing-Faroes-lettsers-in-text-and-plot-tp4639074.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list 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 3D EARTH SCREENSAVER - Watch the Earth right on your desktop! __ R-help@r-project.org mailing list 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] metafor- interpretation of moderators test for raw proportions
At 17:03 03/08/2012, cpanderson wrote: Wolfgang, Thanks for your quick response. You are correct- indeed I had inadvertently left out that i had set intercept = FALSE in rma. Dear Christopher, It does help if you paste in the exact code you used, see also below Following your suggestions, I get the following results: Test of Moderators (coefficient(s) 2,3): QM(df = 2) = 0.2207, p-val = 0.8955 Model Results: estimate se zvalpvalci.lb ci.ub intrcpt0.6498 0.0492 13.2160 .0001 0.5534 0.7462 *** Complex0.0457 0.1007 0.4538 0.6500 -0.1517 0.2430 Dome 0.0244 0.1135 0.2148 0.8299 -0.1980 0.2468 This may be an even dumber question than my first one, but if you have time I'd appreciate knowing how this works. Originally I had tried the syntax mods = ~ Technique. I ended up getting the following error message: Error in qr.solve(wX, diag(k)) : singular matrix 'a' in solve if I supply the argument that mods =~ factor(Technique), I don't get the singular matrix 'a' message. But I don't understand why that should make a difference, because dat$Technique is already an object of class factor: Pasting in the exact code would help here and maybe also giving the output of traceback(). You might also consider pasting in the result of summary(dat) unless it is really huge. class(dat$Technique) [1] factor Thanks again. Christopher -- View this message in context: http://r.789695.n4.nabble.com/metafor-interpretation-of-moderators-test-for-raw-proportions-tp4638972p4639057.html Sent from the R help mailing list archive at Nabble.com. Michael Dewey i...@aghmed.fsnet.co.uk http://www.aghmed.fsnet.co.uk/home.html __ R-help@r-project.org mailing list 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] Partial Likelihood
Hello everyone, i would like to ask if everyone knows how to perfom a glm partial likelihood estimation in a time series whrere dependence exists. lets say that i want to perform a logistic regression for binary data (0, 1) with binary responses which a re the previous days. for example: logistic-glm(dat$Day~dat$Day1+dat$Day2, family=binomial(link=logit)) where dat$Day (0 or 1) is the current day and dat$Day1 is one day before (0 or 1). is it possible that R performs partial likelihood estimation automatically? thank you in advance Konstantinos Mammas -- View this message in context: http://r.789695.n4.nabble.com/Partial-Likelihood-tp4639159.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list 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] Error message 'x' must be numeric
http://r.789695.n4.nabble.com/file/n4639158/elgas2.csv elgas2.csv http://r.789695.n4.nabble.com/file/n4639158/histogramselgas.R histogramselgas.R Hello, I want to plot 9 histograms, and I prepare the data for this operation with the usual command, sapply(x, is.numeric), then I get a confirmation that all data are numeric. Strangely, when I want to plot the histograms I get the error message: 'x' must be numeric. My 'x' is already numeric and I have no idea what's wrong. Could somebody tell me where the error is? I attach my .csv data and my R code. Thank you. -- View this message in context: http://r.789695.n4.nabble.com/Error-message-x-must-be-numeric-tp4639158.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list 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] Questionnaire Analysis virtually without continuous Variables
Hello! I am doing an analysis on a questionnaire of hunters taken in 4 different districts of some mysterious foreign country. The aim of the study was to gather info on the factors that determine the hunting success of a peculiarly beautiful bird in that area. All variables are factors, i.e. they are variables such as Use of Guns - yes / no, Use of Dogs - yes / no and the likes. The response is upposed to be number of Birds caught, which was designed to be the only continuous variable. However, in reality the number of caught birds is between 0 and 1, sometimes hunters answered with 2. Unfortunately, it is not the questioner who is burdened with the analysis, but me. I am struggling to find an appropriate approach to the analysis. I don't really consider this as count data, since it would be very vulnerable to overinflation (and a steep decline for counts above 0). I can't really suggest binomial models either, since the lack of explanatory, continuous data renders such an approach quite vague. I also struggle with the random design of the survey (households nested within villages nested within districts). Adding to that, hunters don't even target the bird as their prime objective. The bird is essentially a by-catch, most often used for instant consumption on the hunting trip. I therefore doubt that any analysis makes more than a little sense, but I will not yet succumb to failure. Any ideas? Thanks in advance! PS: I just realized that this is not a question related to R but to statistics in general. Apologies for that! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to assing unique ID in a table and do regression
Hi R- User I have very big data set (5000 rows). I wanted to make classes based on a column of that table (that column has the data which is continuous .) After converting into different class, this class would be Unique ID. I want to run regression for each ID. For example I have a data set dput(dat) structure(list(ID = c(0.1, 0.8, 0.1, 1.5, 1.1, 0.9, 1.8, 2.5, 2, 2.5, 2.8, 3, 3.1, 3.2, 3.9, 1, 4, 4.7, 4.3, 4.9, 2.1, 2.4), S = c(4L, 7L, 9L, 10L, 10L, 8L, 8L, 8L, 17L, 18L, 13L, 13L, 11L, 1L, 10L, 20L, 22L, 20L, 18L, 16L, 7L, 20L), en2 = c(-2.5767, -2.5767, -2.5767, -2.5767, -2.5767, -2.5767, -2.5767, -2.5347, -2.5347, -2.5347, -2.5347, -2.5347, -2.5347, -2.4939, -2.4939, -2.4939, -2.4939, -2.4939, -2.4939, -2.4939, -2.4543, -2.4543 ), en3 = c(-1.1785, -0.6596, -0.6145, -0.6437, -0.6593, -0.7811, -1.1785, -1.1785, -1.1785, -0.6596, -0.6145, -0.6437, -0.6593, -1.1785, -0.1342, -0.2085, -0.4428, -0.5125, -0.8075, -1.1785, -1.1785, -0.1342), en4 = c(-1.4445, -1.3645, -1.1634, -0.7735, -0.6931, -1.1105, -1.4127, -1.5278, -1.4445, -1.3645, -1.1634, -0.7735, -0.6931, -1.0477, -0.8655, -0.1759, 0.1203, -0.2962, -0.4473, -1.0436, -0.9705, -0.8953), en5 = c(-0.4783, -0.3296, -0.2026, -0.3579, -0.5154, -0.5726, -0.6415, -0.3996, -0.4529, -0.5762, -0.561, -0.6891, -0.7408, -0.6287, -0.4337, -0.4586, -0.5249, -0.6086, -0.7076, -0.7114, -0.4952, 0.1091)), .Names = c(ID, S, en2, en3, en4, en5), class = data.frame, row.names = c(NA, -22L)) Here ID has continuous value, I want to make groups with value 0-1, 1-2, 2-3, 3-4 from the column ID. and then. I wanted to run regression with S (dependent variable) and en2 (independent variable); again regression of S and en3 , and so on. After that, I wanted to have a table with r2 and p value. would you help me how I can do it? I was trying it manually - but it took so much time. therefore I thought to write you for your help. Thanks for your help. Kristi [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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] Writing Faroes lettsers in text and plot ?
Klaus: John Kane is right to point out that the procedure will depend on your OS. It will also depend on what interface you use for entering text into R. All the letters you need for Faroese are present in the basic iso-8859-1 Western European character set (also known as ISO-Latin1) which should be available on just about any current operating system. It is how you access the non-US/English alphabetical characters which will depend on OS and interface. In some cases (especially in Windows) there will be a drop-down menu of available characters in which you point-and-click to enter the character you want. Or your OS may allow you to set up your keyboard configuration so that certain key-combinations give special characters. Or you may use an editor program which itself can have a special key setup. The mail program in which I am typing this reply uses iso-8859-1 and has its own keyboard setup involving a system of digraphs. Each digraph is initiated by pressing Control-D ([^D]), followed by a sequence of two letters, so that [^D]AE gives Æ, for instance. Thus, in addition to the usual A-Z, a-z: [^D]A': Á[^D]D-: Ð[^D]I': Í[^D]O': Ó [^D]U': Ú[^D]Y': Ý[^D]AE: Æ[^D]O/: Ø with the corresponding digraphs for lower-case letters, such as [^D]ae: æ etc. I believe that may be enough for Faroese; however, if needed other Nordic letters are available in iso-8859-1, such as þ ([^D]th), Þ ([^D]TH), Ö ([^D]O], ö ([^D]o0, etc. When using R (in Linux), as program editor I usually use vim, which operates a similar digraph system (but with Control-K ([^K]) instead of [^D]). One way of entering such characters directly into R is to first enter them into a vim console window, and then (with mouse) copy and paste into the R command line. However, the R command-line window can have its own configuration as well. The choice is up to you, in the context of your OS and resources. It will probably be possible to give you direct and specific advice, if you let us know what your system set-up is, and how you access R. Hoping this helps, Ted. On 04-Aug-2012 13:25:07 John Kane wrote: Depends on your OS probably. I would just add a Faroese keyboard to my Ubuntu setup and switch between my normal keyboard and Faoese as required. John Kane Kingston ON Canada -Original Message- From: klausflemlo...@mail.tele.dk Sent: Fri, 3 Aug 2012 10:08:48 -0700 (PDT) To: r-help@r-project.org Subject: [R] Writing Faroes lettsers in text and plot ? I need to write Faroese letters in plain tekst and in plots. How can I do this ? http://en.wikipedia.org/wiki/Faroese_language Faroese_language -- View this message in context: http://r.789695.n4.nabble.com/Writing-Faroes-lettsers-in-text-and-plot-tp4639 074.html Sent from the R help mailing list archive at Nabble.com. - E-Mail: (Ted Harding) ted.hard...@wlandres.net Date: 04-Aug-2012 Time: 15:41:39 This message was sent by XFMail __ R-help@r-project.org mailing list 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 message 'x' must be numeric
On 04.08.2012 12:18, Ivette wrote: http://r.789695.n4.nabble.com/file/n4639158/elgas2.csv elgas2.csv http://r.789695.n4.nabble.com/file/n4639158/histogramselgas.R histogramselgas.R Hello, I want to plot 9 histograms, and I prepare the data for this operation with the usual command, sapply(x, is.numeric), then I get a confirmation that all data are numeric. Strangely, when I want to plot the histograms I get the error message: 'x' must be numeric. My 'x' is already numeric and I have no idea what's wrong. Could somebody tell me where the error is? I attach my .csv data and my R code. You diverse x are data.frames rather than numeric vectors, just the columns of your data.frames are numeric. Hence use, e.g., x1[,1] do access the first column of x1 etc. Uwe Ligges Thank you. -- View this message in context: http://r.789695.n4.nabble.com/Error-message-x-must-be-numeric-tp4639158.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list 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 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] how to assing unique ID in a table and do regression
Hello, Try the following. id.groups - with(dat, cut(ID, breaks=0:ceiling(max(ID sp - split(dat, id.groups) regressors - grep(en, names(dat)) models - lapply(sp, function(.df) lapply(regressors, function(x) lm(.df[[S]] ~ .df[[x]]))) mod.summ - lapply(models, function(x) lapply(x, summary)) # First R2 mod.r2 - lapply(mod.summ, function(x) lapply(x, `[[`, r.squared)) mod.r2 # Now p-values mod.coef - lapply(mod.summ, function(x) lapply(x, coef)) mod.pvalue - lapply(mod.coef, function(x) lapply(x, `[`, , 4)) # p-values in matrix form, columns are 'en2', en3', etc #lapply(mod.pvalue, function(x) do.call(cbind, x)) Hope this helps, Rui Barradas Em 04-08-2012 15:22, Kristi Glover escreveu: Hi R- User I have very big data set (5000 rows). I wanted to make classes based on a column of that table (that column has the data which is continuous .) After converting into different class, this class would be Unique ID. I want to run regression for each ID. For example I have a data set dput(dat) structure(list(ID = c(0.1, 0.8, 0.1, 1.5, 1.1, 0.9, 1.8, 2.5, 2, 2.5, 2.8, 3, 3.1, 3.2, 3.9, 1, 4, 4.7, 4.3, 4.9, 2.1, 2.4), S = c(4L, 7L, 9L, 10L, 10L, 8L, 8L, 8L, 17L, 18L, 13L, 13L, 11L, 1L, 10L, 20L, 22L, 20L, 18L, 16L, 7L, 20L), en2 = c(-2.5767, -2.5767, -2.5767, -2.5767, -2.5767, -2.5767, -2.5767, -2.5347, -2.5347, -2.5347, -2.5347, -2.5347, -2.5347, -2.4939, -2.4939, -2.4939, -2.4939, -2.4939, -2.4939, -2.4939, -2.4543, -2.4543 ), en3 = c(-1.1785, -0.6596, -0.6145, -0.6437, -0.6593, -0.7811, -1.1785, -1.1785, -1.1785, -0.6596, -0.6145, -0.6437, -0.6593, -1.1785, -0.1342, -0.2085, -0.4428, -0.5125, -0.8075, -1.1785, -1.1785, -0.1342), en4 = c(-1.4445, -1.3645, -1.1634, -0.7735, -0.6931, -1.1105, -1.4127, -1.5278, -1.4445, -1.3645, -1.1634, -0.7735, -0.6931, -1.0477, -0.8655, -0.1759, 0.1203, -0.2962, -0.4473, -1.0436, -0.9705, -0.8953), en5 = c(-0.4783, -0.3296, -0.2026, -0.3579, -0.5154, -0.5726, -0.6415, -0.3996, -0.4529, -0.5762, -0.561, -0.6891, -0.7408, -0.6287, -0.4337, -0.4586, -0.5249, -0.6086, -0.7076, -0.7114, -0.4952, 0.1091)), .Names = c(ID, S, en2, en3, en4, en5), class = data.frame, row.names = c(NA, -22L)) Here ID has continuous value, I want to make groups with value 0-1, 1-2, 2-3, 3-4 from the column ID. and then. I wanted to run regression with S (dependent variable) and en2 (independent variable); again regression of S and en3 , and so on. After that, I wanted to have a table with r2 and p value. would you help me how I can do it? I was trying it manually - but it took so much time. therefore I thought to write you for your help. Thanks for your help. Kristi [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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 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] Questionnaire Analysis virtually without continuous Variables
On Sat, Aug 4, 2012 at 9:12 AM, Sacha Viquerat dawa.ya.m...@googlemail.com wrote: Hello! I am doing an analysis on a questionnaire of hunters taken in 4 different districts of some mysterious foreign country. The aim of the study was to gather info on the factors that determine the hunting success of a peculiarly beautiful bird in that area. All variables are factors, i.e. they are variables such as Use of Guns - yes / no, Use of Dogs - yes / no and the likes. The response is upposed to be number of Birds caught, which was designed to be the only continuous variable. However, in reality the number of caught birds is between 0 and 1, sometimes hunters answered with 2. Unfortunately, it is not the questioner who is burdened with the analysis, but me. I am struggling to find an appropriate approach to the analysis. I don't really consider this as count data, since it would be very vulnerable to overinflation (and a steep decline for counts above 0). I can't really suggest binomial models either, since the lack of explanatory, continuous data renders such an approach quite vague. I also struggle with the random design of the survey (households nested within villages nested within districts). Adding to that, hunters don't even target the bird as their prime objective. The bird is essentially a by-catch, most often used for instant consumption on the hunting trip. I therefore doubt that any analysis makes more than a little sense, but I will not yet succumb to failure. Any ideas? Thanks in advance! Hi Sacha, This sounds a good deal like homework to me (some mysterious foreign country) and this list has a no homework policy so unfortunately, I don't think you'll be able to get much help here. Best of luck with your analysis however! Michael PS: I just realized that this is not a question related to R but to statistics in general. Apologies for that! __ R-help@r-project.org mailing list 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 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] metafor- interpretation of moderators test for raw proportions
Just to follow up on what Michael wrote: I cannot reproduce that error. For example, this all works as intended: data(dat.bcg) dat - escalc(measure=RR, ai=tpos, bi=tneg, ci=cpos, di=cneg, data=dat.bcg, append=TRUE) rma(yi, vi, mods = ~ alloc, data=dat) ### 'alloc' automatically converted to a factor dat$alloc - factor(dat$alloc) ### explicitly make 'alloc' a factor rma(yi, vi, mods = ~ alloc, data=dat) ### works as before rma(yi, vi, mods = ~ factor(alloc), data=dat) ### factor() not necessary, but works If you can provide a reproducible example, I'll be glad to look into the issue. Aside from that, the error you got occurs when the design matrix is not of full rank. For example: X - model.matrix(~ factor(alloc) - 1, data=dat) rma(yi, vi, mods = X, data=dat) will fail, because the model now has an intercept plus the 3 dummy variables for the 3 levels (setting intercept=FALSE will make this work). It seems to me that this is what happened (since in your previous post, you showed that you coded the three levels of your factor manually). But this is something different than what you describe below, so I don't know for sure. Best, Wolfgang From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of cpanderson [christopher.p.ander...@healthpartners.com] Sent: Friday, August 03, 2012 6:03 PM To: r-help@r-project.org Subject: Re: [R] metafor- interpretation of moderators test for raw proportions Wolfgang, Thanks for your quick response. You are correct- indeed I had inadvertently left out that i had set intercept = FALSE in rma. Following your suggestions, I get the following results: Test of Moderators (coefficient(s) 2,3): QM(df = 2) = 0.2207, p-val = 0.8955 Model Results: estimate se zvalpvalci.lb ci.ub intrcpt0.6498 0.0492 13.2160 .0001 0.5534 0.7462 *** Complex0.0457 0.1007 0.4538 0.6500 -0.1517 0.2430 Dome 0.0244 0.1135 0.2148 0.8299 -0.1980 0.2468 This may be an even dumber question than my first one, but if you have time I'd appreciate knowing how this works. Originally I had tried the syntax mods = ~ Technique. I ended up getting the following error message: Error in qr.solve(wX, diag(k)) : singular matrix 'a' in solve if I supply the argument that mods =~ factor(Technique), I don't get the singular matrix 'a' message. But I don't understand why that should make a difference, because dat$Technique is already an object of class factor: class(dat$Technique) [1] factor Thanks again. Christopher __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Partial Likelihood
Sounds like generalized linear mixed modeling (glmm) to me. Try posting to the r-sig-mixed-models list rather than here to increase the likelihood of a useful response. -- Bert On Sat, Aug 4, 2012 at 3:55 AM, doctoratza mamma...@live.com wrote: Hello everyone, i would like to ask if everyone knows how to perfom a glm partial likelihood estimation in a time series whrere dependence exists. lets say that i want to perform a logistic regression for binary data (0, 1) with binary responses which a re the previous days. for example: logistic-glm(dat$Day~dat$Day1+dat$Day2, family=binomial(link=logit)) where dat$Day (0 or 1) is the current day and dat$Day1 is one day before (0 or 1). is it possible that R performs partial likelihood estimation automatically? thank you in advance Konstantinos Mammas -- View this message in context: http://r.789695.n4.nabble.com/Partial-Likelihood-tp4639159.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list 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. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ R-help@r-project.org mailing list 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] DAtes
Hi all, I´m trying to convert as a data frame (with format date) this copied excel column of dates (exposed below), I have tried to save them in a txt file tfr-read.table(tfra.txt) tfr-data.frame(tfr) I have tried several things, as date, so on, but always error. And it makes Error en as.Date.default(tfr, %m/%d/%y) : do not know how to convert 'a' to class Date Can anyone give me a clue or a gide to achieve this final result. 14/12/2000 15/12/2000 18/12/2000 19/12/2000 20/12/2000 21/12/2000 22/12/2000 25/12/2000 26/12/2000 27/12/2000 28/12/2000 29/12/2000 01/01/2001 02/01/2001 03/01/2001 04/01/2001 05/01/2001 08/01/2001 09/01/2001 10/01/2001 11/01/2001 12/01/2001 15/01/2001 16/01/2001 17/01/2001 18/01/2001 19/01/2001 22/01/2001 23/01/2001 24/01/2001 25/01/2001 26/01/2001 29/01/2001 30/01/2001 31/01/2001 01/02/2001 02/02/2001 05/02/2001 06/02/2001 07/02/2001 08/02/2001 09/02/2001 12/02/2001 13/02/2001 14/02/2001 15/02/2001 16/02/2001 19/02/2001 20/02/2001 21/02/2001 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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] DAtes
Hello, Works with me: x - scan(what = character, text = 14/12/2000 15/12/2000 18/12/2000 19/12/2000 20/12/2000 21/12/2000 22/12/2000 25/12/2000 26/12/2000 27/12/2000 28/12/2000 29/12/2000 01/01/2001 02/01/2001 03/01/2001 04/01/2001 05/01/2001 08/01/2001 09/01/2001 10/01/2001 11/01/2001 12/01/2001 15/01/2001 16/01/2001 17/01/2001 18/01/2001 19/01/2001 22/01/2001 23/01/2001 24/01/2001 25/01/2001 26/01/2001 29/01/2001 30/01/2001 31/01/2001 01/02/2001 02/02/2001 05/02/2001 06/02/2001 07/02/2001 08/02/2001 09/02/2001 12/02/2001 13/02/2001 14/02/2001 15/02/2001 16/02/2001 19/02/2001 20/02/2001 21/02/2001) as.Date(x, format = %d/%m/%Y) You've posted the error message but not the command. If it is what it seems to be, by looking at the error message, then you've passed the wrong format argument. There is NO month 14 (first date of all!) Hope this helps, Rui Barradas Em 04-08-2012 17:09, Trying To learn again escreveu: Hi all, I´m trying to convert as a data frame (with format date) this copied excel column of dates (exposed below), I have tried to save them in a txt file tfr-read.table(tfra.txt) tfr-data.frame(tfr) I have tried several things, as date, so on, but always error. And it makes Error en as.Date.default(tfr, %m/%d/%y) : do not know how to convert 'a' to class Date Can anyone give me a clue or a gide to achieve this final result. 14/12/2000 15/12/2000 18/12/2000 19/12/2000 20/12/2000 21/12/2000 22/12/2000 25/12/2000 26/12/2000 27/12/2000 28/12/2000 29/12/2000 01/01/2001 02/01/2001 03/01/2001 04/01/2001 05/01/2001 08/01/2001 09/01/2001 10/01/2001 11/01/2001 12/01/2001 15/01/2001 16/01/2001 17/01/2001 18/01/2001 19/01/2001 22/01/2001 23/01/2001 24/01/2001 25/01/2001 26/01/2001 29/01/2001 30/01/2001 31/01/2001 01/02/2001 02/02/2001 05/02/2001 06/02/2001 07/02/2001 08/02/2001 09/02/2001 12/02/2001 13/02/2001 14/02/2001 15/02/2001 16/02/2001 19/02/2001 20/02/2001 21/02/2001 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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 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] Partial Likelihood
In addition to Bert's suggestion of r sig mixed models (which I second), I would encourage you to create a more detailed example and explanation of what you hope to accomplish. Sounds a bit like an auto regressive structure, but more details would be good. Cheers, Josh On Aug 4, 2012, at 9:34, Bert Gunter gunter.ber...@gene.com wrote: Sounds like generalized linear mixed modeling (glmm) to me. Try posting to the r-sig-mixed-models list rather than here to increase the likelihood of a useful response. -- Bert On Sat, Aug 4, 2012 at 3:55 AM, doctoratza mamma...@live.com wrote: Hello everyone, i would like to ask if everyone knows how to perfom a glm partial likelihood estimation in a time series whrere dependence exists. lets say that i want to perform a logistic regression for binary data (0, 1) with binary responses which a re the previous days. for example: logistic-glm(dat$Day~dat$Day1+dat$Day2, family=binomial(link=logit)) where dat$Day (0 or 1) is the current day and dat$Day1 is one day before (0 or 1). is it possible that R performs partial likelihood estimation automatically? thank you in advance Konstantinos Mammas -- View this message in context: http://r.789695.n4.nabble.com/Partial-Likelihood-tp4639159.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list 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. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ R-help@r-project.org mailing list 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 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] Questionnaire Analysis virtually without continuous Variables
Hi Sacha, You're right that this is not an R related question really (would be better somewhere like crossvalidated.com). If basically everyone catches 0/1 birds, then I would consider dichotomizing: Y - as.integer(caught = 1) then check cross tabs to make sure there are no zero cells between predictors and outcome: xtabs(~Y + dogs + guns, data=yourdata) then use the glmer() function to model the nested random effects. m - glmer(Y ~ dog + gun + (1 | household) + (1 | village) + (1 | district), data = yourdata, family=binomial) summary(m) Cheers, Josh On Aug 4, 2012, at 7:12, Sacha Viquerat dawa.ya.m...@googlemail.com wrote: Hello! I am doing an analysis on a questionnaire of hunters taken in 4 different districts of some mysterious foreign country. The aim of the study was to gather info on the factors that determine the hunting success of a peculiarly beautiful bird in that area. All variables are factors, i.e. they are variables such as Use of Guns - yes / no, Use of Dogs - yes / no and the likes. The response is upposed to be number of Birds caught, which was designed to be the only continuous variable. However, in reality the number of caught birds is between 0 and 1, sometimes hunters answered with 2. Unfortunately, it is not the questioner who is burdened with the analysis, but me. I am struggling to find an appropriate approach to the analysis. I don't really consider this as count data, since it would be very vulnerable to overinflation (and a steep decline for counts above 0). I can't really suggest binomial models either, since the lack of explanatory, continuous data renders such an! approach quite vague. I also struggle with the random design of the survey (households nested within villages nested within districts). Adding to that, hunters don't even target the bird as their prime objective. The bird is essentially a by-catch, most often used for instant consumption on the hunting trip. I therefore doubt that any analysis makes more than a little sense, but I will not yet succumb to failure. Any ideas? Thanks in advance! PS: I just realized that this is not a question related to R but to statistics in general. Apologies for that! __ R-help@r-project.org mailing list 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 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] DAtes
Hi, Try this: tfr-read.table(text= 14/12/2000 15/12/2000 18/12/2000 19/12/2000 20/12/2000 21/12/2000 22/12/2000 25/12/2000 26/12/2000 27/12/2000 28/12/2000 29/12/2000 01/01/2001 02/01/2001 03/01/2001 04/01/2001 05/01/2001 08/01/2001 09/01/2001 10/01/2001 11/01/2001 12/01/2001 15/01/2001 16/01/2001 17/01/2001 18/01/2001 19/01/2001 22/01/2001 23/01/2001 24/01/2001 25/01/2001 26/01/2001 29/01/2001 30/01/2001 31/01/2001 01/02/2001 02/02/2001 05/02/2001 06/02/2001 07/02/2001 08/02/2001 09/02/2001 12/02/2001 13/02/2001 14/02/2001 15/02/2001 16/02/2001 19/02/2001 20/02/2001 21/02/2001 ,sep=,header=FALSE,fill=TRUE,stringsAsFactors=FALSE) as.Date(tfr$V1,format=%d/%m/%Y) [1] 2000-12-14 2000-12-22 2001-01-01 2001-01-09 2001-01-17 [6] 2001-01-25 2001-02-02 2001-02-12 2001-02-20 A.K. - Original Message - From: Trying To learn again tryingtolearnag...@gmail.com To: r-help@r-project.org Cc: Sent: Saturday, August 4, 2012 12:09 PM Subject: [R] DAtes Hi all, I´m trying to convert as a data frame (with format date) this copied excel column of dates (exposed below), I have tried to save them in a txt file tfr-read.table(tfra.txt) tfr-data.frame(tfr) I have tried several things, as date, so on, but always error. And it makes Error en as.Date.default(tfr, %m/%d/%y) : do not know how to convert 'a' to class Date Can anyone give me a clue or a gide to achieve this final result. 14/12/2000 15/12/2000 18/12/2000 19/12/2000 20/12/2000 21/12/2000 22/12/2000 25/12/2000 26/12/2000 27/12/2000 28/12/2000 29/12/2000 01/01/2001 02/01/2001 03/01/2001 04/01/2001 05/01/2001 08/01/2001 09/01/2001 10/01/2001 11/01/2001 12/01/2001 15/01/2001 16/01/2001 17/01/2001 18/01/2001 19/01/2001 22/01/2001 23/01/2001 24/01/2001 25/01/2001 26/01/2001 29/01/2001 30/01/2001 31/01/2001 01/02/2001 02/02/2001 05/02/2001 06/02/2001 07/02/2001 08/02/2001 09/02/2001 12/02/2001 13/02/2001 14/02/2001 15/02/2001 16/02/2001 19/02/2001 20/02/2001 21/02/2001 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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 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] DAtes
HI, I guess this will be more appropriate: tfr-read.table(text= 14/12/2000 15/12/2000 18/12/2000 19/12/2000 20/12/2000 21/12/2000 22/12/2000 25/12/2000 26/12/2000 27/12/2000 28/12/2000 29/12/2000 01/01/2001 02/01/2001 03/01/2001 04/01/2001 05/01/2001 08/01/2001 09/01/2001 10/01/2001 11/01/2001 12/01/2001 15/01/2001 16/01/2001 17/01/2001 18/01/2001 19/01/2001 22/01/2001 23/01/2001 24/01/2001 25/01/2001 26/01/2001 29/01/2001 30/01/2001 31/01/2001 01/02/2001 02/02/2001 05/02/2001 06/02/2001 07/02/2001 08/02/2001 09/02/2001 12/02/2001 13/02/2001 14/02/2001 15/02/2001 16/02/2001 19/02/2001 20/02/2001 21/02/2001 ,sep=,header=FALSE,fill=TRUE,stringsAsFactors=FALSE) tfr1-data.frame(dates=c(tfr[,1],tfr[,2],tfr[,3],tfr[,4],tfr[,5],tfr[,6])) tfr2-tfr1[-c(27,36,45),] head(data.frame(dates=as.Date(tfr2,format=%d/%m/%Y))) dates 1 2000-12-14 2 2000-12-22 3 2001-01-01 4 2001-01-09 5 2001-01-17 6 2001-01-25 # or you can do this: tfr1-as.vector(c(tfr[,1],tfr[,2],tfr[,3],tfr[,4],tfr[,5],tfr[,6])) tfr2-as.data.frame(as.Date(tfr1,format=%d/%m/%Y)) colnames(tfr2)-dates head(tfr2) dates 1 2000-12-14 2 2000-12-22 3 2001-01-01 4 2001-01-09 5 2001-01-17 6 2001-01-25 A.K. - Original Message - From: Trying To learn again tryingtolearnag...@gmail.com To: r-help@r-project.org Cc: Sent: Saturday, August 4, 2012 12:09 PM Subject: [R] DAtes Hi all, I´m trying to convert as a data frame (with format date) this copied excel column of dates (exposed below), I have tried to save them in a txt file tfr-read.table(tfra.txt) tfr-data.frame(tfr) I have tried several things, as date, so on, but always error. And it makes Error en as.Date.default(tfr, %m/%d/%y) : do not know how to convert 'a' to class Date Can anyone give me a clue or a gide to achieve this final result. 14/12/2000 15/12/2000 18/12/2000 19/12/2000 20/12/2000 21/12/2000 22/12/2000 25/12/2000 26/12/2000 27/12/2000 28/12/2000 29/12/2000 01/01/2001 02/01/2001 03/01/2001 04/01/2001 05/01/2001 08/01/2001 09/01/2001 10/01/2001 11/01/2001 12/01/2001 15/01/2001 16/01/2001 17/01/2001 18/01/2001 19/01/2001 22/01/2001 23/01/2001 24/01/2001 25/01/2001 26/01/2001 29/01/2001 30/01/2001 31/01/2001 01/02/2001 02/02/2001 05/02/2001 06/02/2001 07/02/2001 08/02/2001 09/02/2001 12/02/2001 13/02/2001 14/02/2001 15/02/2001 16/02/2001 19/02/2001 20/02/2001 21/02/2001 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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 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] AFT model time-dependent with weibull distribution
Dear R-community, I have tried to estimate an EXPONENTIEL accelerated failure time(AFT) power rule model with time-independent . For that purpose, I have used the eha package. Please, consider this example: vi Ti 1 265.79 2 26 1579.52 3 26 2323.70 4 28 68.85 5 28 426.07 6 28 110.29 7 28 108.29 8 28 1067.60 9 30 17.05 10 30 22.66 11 30 21.02 12 30 175.88 13 30 139.07 14 30 144.12 15 30 20.46 16 30 43.40 17 30 194.90 18 30 47.30 19 307.74 20 320.40 21 32 82.85 22 329.88 23 32 89.29 24 32 215.10 25 321.75 26 320.79 27 32 15.93 28 323.91 29 320.27 30 320.69 31 32 100.58 32 32 27.80 33 32 13.95 34 32 53.24 35 340.96 36 344.15 37 340.19 38 340.78 39 348.01 40 34 31.75 41 347.35 42 346.50 43 348.27 44 34 33.91 45 34 32.52 46 343.16 47 344.85 48 342.78 49 344.67 50 341.31 51 34 12.06 52 34 36.71 53 34 72.89 54 361.97 55 360.59 56 362.58 57 361.69 58 362.71 59 36 25.50 60 360.35 61 360.99 62 363.99 63 363.67 64 362.07 65 360.96 66 365.35 67 362.90 68 36 13.77 69 380.47 70 380.73 71 381.40 72 380.74 73 380.39 74 381.13 75 380.09 76 382.38 aftexp-aftreg(Surv(time,status) ~ vi, data=data.frame(dataexp), dist=exponentiel) Error in Surv(time, status) : Time variable is not numeric aftexp-aftreg(Surv(Ti,status) ~ vi, data=data.frame(dataexp), dist=exponentiel) Error in Surv(Ti, status) : object 'status' not found status- rep(1, 76) status [1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 [39] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 cbind(dataexp, status) vi Ti status 1 265.79 1 2 26 1579.52 1 3 26 2323.70 1 4 28 68.85 1 5 28 426.07 1 6 28 110.29 1 7 28 108.29 1 8 28 1067.60 1 9 30 17.05 1 10 30 22.66 1 11 30 21.02 1 12 30 175.88 1 13 30 139.07 1 14 30 144.12 1 15 30 20.46 1 16 30 43.40 1 17 30 194.90 1 18 30 47.30 1 19 307.74 1 20 320.40 1 21 32 82.85 1 22 329.88 1 23 32 89.29 1 24 32 215.10 1 25 321.75 1 26 320.79 1 27 32 15.93 1 28 323.91 1 29 320.27 1 30 320.69 1 31 32 100.58 1 32 32 27.80 1 33 32 13.95 1 34 32 53.24 1 35 340.96 1 36 344.15 1 37 340.19 1 38 340.78 1 39 348.01 1 40 34 31.75 1 41 347.35 1 42 346.50 1 43 348.27 1 44 34 33.91 1 45 34 32.52 1 46 343.16 1 47 344.85 1 48 342.78 1 49 344.67 1 50 341.31 1 51 34 12.06 1 52 34 36.71 1 53 34 72.89 1 54 361.97 1 55 360.59 1 56 362.58 1 57 361.69 1 58 362.71 1 59 36 25.50 1 60 360.35 1 61 360.99 1 62 363.99 1 63 363.67 1 64 362.07 1 65 360.96 1 66 365.35 1 67 362.90 1 68 36 13.77 1 69 380.47 1 70 380.73 1 71 381.40 1 72 380.74 1 73 380.39 1 74 381.13 1 75 380.09 1 76 382.38 1 aftexp-aftreg(Surv(Ti,status) ~ vi, data=data.frame(dataexp), dist=exponentiel) Error in aftreg.fit(X, Y, dist, strats, offset, init, shape, id, control, : exponentiel is not an implemented distribution aftexp-aftreg(Surv(time,status) ~ vi, data=data.frame(dataexp), dist=exponentiel) Error in Surv(time, status) : Time variable is not numeric pleas help me to find a solution to my problem -- View this message in context: http://r.789695.n4.nabble.com/AFT-model-time-dependent-with-weibull-distribution-tp3755079p4639174.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list 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] AFT model time-dependent with weibull distribution
On Aug 4, 2012, at 10:45 AM, hafida wrote: Dear R-community, I have tried to estimate an EXPONENTIEL accelerated failure time(AFT) power rule model with time-independent . For that purpose, I have used the eha package. Please, consider this example: vi Ti 1 265.79 2 26 1579.52 3 26 2323.70 snipped 74 381.13 75 380.09 76 382.38 aftexp-aftreg(Surv(time,status) ~ vi, data=data.frame(dataexp), dist=exponentiel) Error in Surv(time, status) : Time variable is not numeric Your variables being given to the `Surv` function do not match those in you data object. aftexp-aftreg(Surv(Ti,status) ~ vi, data=data.frame(dataexp), dist=exponentiel) Error in Surv(Ti, status) : object 'status' not found status- rep(1, 76) status [1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 [39] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 cbind(dataexp, status) vi Ti status 1 265.79 1 2 26 1579.52 1 snipped 72 380.74 1 73 380.39 1 74 381.13 1 75 380.09 1 76 382.38 1 aftexp-aftreg(Surv(Ti,status) ~ vi, data=data.frame(dataexp), dist=exponentiel) Error in aftreg.fit(X, Y, dist, strats, offset, init, shape, id, control, : exponentiel is not an implemented distribution You are spelling 'exponential' incorrectly (in English anyway.) Bon chance; David. aftexp-aftreg(Surv(time,status) ~ vi, data=data.frame(dataexp), dist=exponentiel) Error in Surv(time, status) : Time variable is not numeric pleas help me to find a solution to my problem -- David Winsemius, MD Alameda, CA, USA __ R-help@r-project.org mailing list 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] AFT model time-dependent with weibull distribution
Hello,
Vous êtes française?
It shows, in english it would be 'exponential', with an 'a'.
Worked with me, after reading the manual.
dataexp - read.table(text=
vi Ti
1 265.79
2 26 1579.52
3 26 2323.70
4 28 68.85
[...]
73 380.39
74 381.13
75 380.09
76 382.38
, header=TRUE)
# Better in a post to R-Help is the output of dput()
# It looks like the following line, without the assignment.
dput(dataexp)
dataexp -
structure(list(vi = c(26L, 26L, 26L, 28L, 28L, 28L, 28L, 28L,
30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L, 32L, 32L,
32L, 32L, 32L, 32L, 32L, 32L, 32L, 32L, 32L, 32L, 32L, 32L, 32L,
34L, 34L, 34L, 34L, 34L, 34L, 34L, 34L, 34L, 34L, 34L, 34L, 34L,
34L, 34L, 34L, 34L, 34L, 34L, 36L, 36L, 36L, 36L, 36L, 36L, 36L,
36L, 36L, 36L, 36L, 36L, 36L, 36L, 36L, 38L, 38L, 38L, 38L, 38L,
38L, 38L, 38L), Ti = c(5.79, 1579.52, 2323.7, 68.85, 426.07,
110.29, 108.29, 1067.6, 17.05, 22.66, 21.02, 175.88, 139.07,
144.12, 20.46, 43.4, 194.9, 47.3, 7.74, 0.4, 82.85, 9.88, 89.29,
215.1, 1.75, 0.79, 15.93, 3.91, 0.27, 0.69, 100.58, 27.8, 13.95,
53.24, 0.96, 4.15, 0.19, 0.78, 8.01, 31.75, 7.35, 6.5, 8.27,
33.91, 32.52, 3.16, 4.85, 2.78, 4.67, 1.31, 12.06, 36.71, 72.89,
1.97, 0.59, 2.58, 1.69, 2.71, 25.5, 0.35, 0.99, 3.99, 3.67, 2.07,
0.96, 5.35, 2.9, 13.77, 0.47, 0.73, 1.4, 0.74, 0.39, 1.13, 0.09,
2.38), status = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1)), .Names = c(vi,
Ti, status), row.names = c(1, 2, 3, 4, 5, 6,
7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17,
18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28,
29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39,
40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50,
51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61,
62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72,
73, 74, 75, 76), class = data.frame)
##Not run
#install.packages('eha')
library(eha)
library(survival)
dataexp$status - rep(1, 76)
aftexp - aftreg(Surv(Ti,status) ~ vi, data=dataexp,
dist=weibull, shape=1)
str(aftexp)
summary(aftexp)
coef(aftexp)
The exponential can be seen as a special case of the weibull
distribution with shape = 1.
And the help page of aftreg() says precisely to use the weibull with
shape = 1 if we want an exponential.
Em 04-08-2012 18:45, hafida escreveu:
Dear R-community,
I have tried to estimate an EXPONENTIEL accelerated failure time(AFT)
power rule model with time-independent . For that purpose, I have used
the eha package.
Please, consider this example:
vi Ti
1 265.79
2 26 1579.52
3 26 2323.70
4 28 68.85
5 28 426.07
6 28 110.29
7 28 108.29
8 28 1067.60
9 30 17.05
10 30 22.66
11 30 21.02
12 30 175.88
13 30 139.07
14 30 144.12
15 30 20.46
16 30 43.40
17 30 194.90
18 30 47.30
19 307.74
20 320.40
21 32 82.85
22 329.88
23 32 89.29
24 32 215.10
25 321.75
26 320.79
27 32 15.93
28 323.91
29 320.27
30 320.69
31 32 100.58
32 32 27.80
33 32 13.95
34 32 53.24
35 340.96
36 344.15
37 340.19
38 340.78
39 348.01
40 34 31.75
41 347.35
42 346.50
43 348.27
44 34 33.91
45 34 32.52
46 343.16
47 344.85
48 342.78
49 344.67
50 341.31
51 34 12.06
52 34 36.71
53 34 72.89
54 361.97
55 360.59
56 362.58
57 361.69
58 362.71
59 36 25.50
60 360.35
61 360.99
62 363.99
63 363.67
64 362.07
65 360.96
66 365.35
67 362.90
68 36 13.77
69 380.47
70 380.73
71 381.40
72 380.74
73 380.39
74 381.13
75 380.09
76 382.38
aftexp-aftreg(Surv(time,status) ~ vi, data=data.frame(dataexp),
dist=exponentiel)
Error in Surv(time, status) : Time variable is not numeric
aftexp-aftreg(Surv(Ti,status) ~ vi, data=data.frame(dataexp),
dist=exponentiel)
Error in Surv(Ti, status) : object 'status' not found
status- rep(1, 76)
status
[1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1
[39] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1
cbind(dataexp, status)
vi Ti status
1 265.79 1
2 26 1579.52 1
3 26 2323.70 1
4 28 68.85 1
5 28 426.07 1
6 28 110.29 1
7 28 108.29 1
8 28 1067.60 1
9 30 17.05 1
10 30 22.66 1
11 30 21.02 1
12 30 175.88 1
13 30 139.07 1
14 30 144.12 1
15 30 20.46 1
16 30 43.40 1
17 30 194.90 1
18 30 47.30 1
19 307.74 1
20 320.40 1
21 32 82.85 1
22 329.88 1
23 32 89.29 1
24 32 215.10 1
25 321.75 1
26 320.79 1
27 32 15.93 1
28 323.91 1
29 320.27 1
30 320.69 1
31 32 100.58 1
32 32 27.80 1
33 32 13.95 1
34 32 53.24 1
35 340.96 1
36 344.15 1
37 340.19 1
38
Re: [R] Partial Likelihood
Joshua Wiley jwiley.psych at gmail.com writes: In addition to Bert's suggestion of r sig mixed models (which I second), I would encourage you to create a more detailed example and explanation of what you hope to accomplish. Sounds a bit like an auto regressive structure, but more details would be good. Cheers, Josh On Aug 4, 2012, at 9:34, Bert Gunter gunter.berton at gene.com wrote: Sounds like generalized linear mixed modeling (glmm) to me. Try posting to the r-sig-mixed-models list rather than here to increase the likelihood of a useful response. -- Bert On Sat, Aug 4, 2012 at 3:55 AM, doctoratza mammas_k at live.com wrote: Hello everyone, i would like to ask if everyone knows how to perfom a glm partial likelihood estimation in a time series whrere dependence exists. lets say that i want to perform a logistic regression for binary data (0, 1) with binary responses which a re the previous days. for example: logistic-glm(dat$Day~dat$Day1+dat$Day2, family=binomial(link=logit)) where dat$Day (0 or 1) is the current day and dat$Day1 is one day before (0 or 1). ... and presumably Day2 is 2 days before? is it possible that R performs partial likelihood estimation automatically? Since it's plausible in this case that the responses are all observed without error, I don't necessarily see why you need GLMMs, or anything beyond a regular GLM fit to do this ... you just need up to set the lagged variables correctly. As I interpret this question, dat - data.frame(Day=c(Day,rep(NA,2)),Day1=c(NA,Day,NA),Day2=c(NA,NA,Day)) glm(Day~Day1+Day2,na.action=na.exclude,data=dat,family=binomial) should work just fine (na.action=na.exclude isn't really necessary -- the default behavior is to omit NAs -- but this way if you do something like predictions it will automatically give you NA values for the beginning and end of the series). Autoregression is only hard when the process is observed with error ... __ R-help@r-project.org mailing list 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] Getting unknown error trying to plot spatial data
Hi there! I'm following an awesome guide to working with spatial data (http://www.frankdavenport.com/blog/2012/6/19/notes-from-a-recent-spatial-r-class-i-gave.html) and am running into an error that I can't figure out how to fix. Disclaimer: I am very much an R n00b Here is the r script I am running: https://dl.dropbox.com/u/28231177/This%20Should%20Work.R data: https://dl.dropbox.com/u/28231177/my_data.csv shapefile: https://dl.dropbox.com/u/28231177/sfzipcodes.zip I am getting two errors: pds - fortify(sf_map) *Using OBJECTID to define regions.* pds$OBJECTID - as.integer(pds$OBJECTID) *Error in `$-.data.frame`(`*tmp*`, OBJECTID, value = integer(0)) : replacement has 0 rows, data has 16249* ## Make the map p1 - ggplot(my_data, aes(map_id = zip)) p1 - p1 + geom_map(aes(fill=vol, map_id = zip), map = pds) p1 - p1 + expand_limits(x = pds$lon, y = pds$lat) + coord_equal() p1 + xlab(Basic Map with Default Elements) *Error in unit(x, default.units) : 'x' and 'units' must have length 0* Anybody have any idea what is happening here or how to resolve this? Thanks -- View this message in context: http://r.789695.n4.nabble.com/Getting-unknown-error-trying-to-plot-spatial-data-tp4639179.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to coerce the result of sweep to be an array if result of FUN is a string?
Hi,
I would like to use sweep to sweep out proportions and confidence intervals
for an array, however when I supply a function which returns a string
(containing something like 9% (3-18%)) I get back a list instead of an array,
here is a simplified example:
# example showing that sweep does not return an array with same dimensions as
STATS as advertised
string.fun - function(a, b) {
paste(a, -, b, sep=)
}
num.fun - function(a, b) {
a+b/100
}
m - array(seq(1:24), dim = c(2,3,4))
stat - array(seq(1:12), dim = c(3,4))
string.ans - sweep(m, c(2,3), stat, string.fun)
dim(string.ans)
length(string.ans)
num.ans - sweep(m, c(2,3), stat, num.fun)
dim(num.ans)
Although the contents are correct, strings.ans is a list while num.ans is an
array. Help(sweep) says that the result of sweep should have the same
dimensions as x, however when the result of the function is a string this is
not the case. How can I coerce the result of the function to cause sweep to
return an array as advertised?
thanks,
Peter
Peter Young, MPH
Surveillance Epidemiologist
CDC Mozambique
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
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] AFT model time-dependent with weibull distribution
thanks a lot sorry for the mistake that i do in exponential, i am francophone and for the programme if we want to apply the power rule condition we use log(vi). it works thank yo -- View this message in context: http://r.789695.n4.nabble.com/AFT-model-time-dependent-with-weibull-distribution-tp3755079p4639184.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list 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] Partial Likelihood
Thank you for your comment. I suspected that a model with well defined predictors should work fine with a glm procedure. Thanks again K -- View this message in context: http://r.789695.n4.nabble.com/Partial-Likelihood-tp4639159p4639185.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list 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] how to coerce the result of sweep to be an array if result of FUN is a string?
Hello,
You can coerce the result to the input shape by assigning the dim
attribute by hand.
string.ans - sweep(m, c(2,3), stat, string.fun)
dim(string.ans) - dim(m)
string.ans
Or put this in a function.
# An abbreviation and the complete function name
sweep.ch - sweep.character -
function(x, MARGIN, STATS, FUN=-, check.margin=TRUE, ...){
result - sweep(x, MARGIN, STATS, FUN, check.margin, ...)
dim(result) - dim(x)
result
}
string.ans2 - sweep.ch(m, c(2,3), stat, string.fun)
string.ans2
Hope this helps,
Rui Barradas
Em 04-08-2012 20:35, Peter Young escreveu:
Hi,
I would like to use sweep to sweep out proportions and confidence intervals for an
array, however when I supply a function which returns a string (containing something like 9%
(3-18%)) I get back a list instead of an array, here is a simplified example:
# example showing that sweep does not return an array with same dimensions as
STATS as advertised
string.fun - function(a, b) {
paste(a, -, b, sep=)
}
num.fun - function(a, b) {
a+b/100
}
m - array(seq(1:24), dim = c(2,3,4))
stat - array(seq(1:12), dim = c(3,4))
string.ans - sweep(m, c(2,3), stat, string.fun)
dim(string.ans)
length(string.ans)
num.ans - sweep(m, c(2,3), stat, num.fun)
dim(num.ans)
Although the contents are correct, strings.ans is a list while num.ans is an
array. Help(sweep) says that the result of sweep should have the same
dimensions as x, however when the result of the function is a string this is
not the case. How can I coerce the result of the function to cause sweep to
return an array as advertised?
thanks,
Peter
Peter Young, MPH
Surveillance Epidemiologist
CDC Mozambique
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
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
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] Questionnaire Analysis virtually without continuous Variables
You may be able to get around zero cells using a an MCMC approach such as with MCMCglmm. On Aug 4, 2012, at 15:30, Sacha Viquerat dawa.ya.m...@googlemail.com wrote: On 08/04/2012 07:57 PM, Joshua Wiley wrote: Hi Sacha, You're right that this is not an R related question really (would be better somewhere like crossvalidated.com). If basically everyone catches 0/1 birds, then I would consider dichotomizing: Y - as.integer(caught = 1) then check cross tabs to make sure there are no zero cells between predictors and outcome: xtabs(~Y + dogs + guns, data=yourdata) then use the glmer() function to model the nested random effects. m - glmer(Y ~ dog + gun + (1 | household) + (1 | village) + (1 | district), data = yourdata, family=binomial) summary(m) Cheers, Josh On Aug 4, 2012, at 7:12, Sacha Viquerat dawa.ya.m...@googlemail.com wrote: Hello! I am doing an analysis on a questionnaire of hunters taken in 4 different districts of some mysterious foreign country. The aim of the study was to gather info on the factors that determine the hunting success of a peculiarly beautiful bird in that area. All variables are factors, i.e. they are variables such as Use of Guns - yes / no, Use of Dogs - yes / no and the likes. The response is upposed to be number of Birds caught, which was designed to be the only continuous variable. However, in reality the number of caught birds is between 0 and 1, sometimes hunters answered with 2. Unfortunately, it is not the questioner who is burdened with the analysis, but me. I am struggling to find an appropriate approach to the analysis. I don't really consider this as count data, since it would be very vulnerable to overinflation (and a steep decline for counts above 0). I can't really suggest binomial models either, since the lack of explanatory, continuous data renders such ! an approach quite vague. I also struggle with the random design of the survey (households nested within villages nested within districts). Adding to that, hunters don't even target the bird as their prime objective. The bird is essentially a by-catch, most often used for instant consumption on the hunting trip. I therefore doubt that any analysis makes more than a little sense, but I will not yet succumb to failure. Any ideas? Thanks in advance! PS: I just realized that this is not a question related to R but to statistics in general. Apologies for that! __ R-help@r-project.org mailing list 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 did exactly what you proposed already (since the binomial model seemed obvious to me), however, of course there are zero cells. I was thinking someone more accustomed to doing questionnaire analysis could unveil some mysterious approach common to sociologists but occluded from the naturalists eyes (hardened after years of dealing with exact science ;) I think I will expand the binomial approach and just try to find fancy graphics that make up for the low value of the actual results (maybe with colours). :D Thank you for the reply (do they really give such tasks for homework these days? These kids must be awesome statisticians!) __ R-help@r-project.org mailing list 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] ggplot2 boxplot help
Hello, I have a data set that looks like this: name G-ID test_id g-id g 1 00077464 C_068131 C_068131 OC_068131- 2 00051728 C_044461 C_044461 OC_044461- 3 00058738 C_050343 C_050343 OC_050343- 4 00059239 C_050649 C_050649 OC_050649- 5 1761 C_000909 C_000909 OC_000909- 6 5119 C_002752 C_002752 OC_002752- locssample_1 sample_2 value_1 value_2 1 37316550-37317847 N C 1.9268400 36.77590 2 27058468-27060176 N C 0.1817890 5.58835 3 4761739-4763268N C0.2309000 7.54035 4 14565311-14567393 N C 0.0294559 1.50886 5 38670994-38675694 N C 0.4678610 14.75560 6 48362804-48380794 N C 10.7258000 92.13150 In this dataset, sample_1 corresponds to value_1 and sample_2 corresponds to value_2. How can I graph this in ggplot2's boxplot function? I am not quite sure how to tell R that sample_1 and sample_2 columns correspond to value_1 and value_2 using ggplot2. Can anyone shed some light on this? Thanks. -- View this message in context: http://r.789695.n4.nabble.com/ggplot2-boxplot-help-tp4639187.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list 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] evaluate sum of sum
Hi
can any body please help me to programme this formula:
a[j]= E[j]-sum from l=i to i-1 (exp{(B0 B1*row matrix*) (z[l]*column
matrix*) } x[l]) / sum from l=i to n
it s complicate for me ; hope you can help me
thank you a lot
--
View this message in context:
http://r.789695.n4.nabble.com/evaluate-sum-of-sum-tp898262p4639191.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
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] sapply and matrix command
Thanks again for the help looks like this will be useful for what I'm doing. Is there any way to use combn to return combinations of values with themselves: e.g. combn(1:3,2) [,1] [,2] [,3] [,4] [,5] [,6] [1,]111 2 2 3 [2,]1 232 3 3 ___ If you reply to this email, your message will be added to the discussion below: http://r.789695.n4.nabble.com/sapply-and-matrix-command-tp4637769p4639190.html To unsubscribe from sapply and matrix command, visit http://r.789695.n4.nabble.com/template/NamlServlet.jtp?macro=unsubscribe_by_codenode=4637769code=ci1oZWxwQHItcHJvamVjdC5vcmd8NDYzNzc2OXwtNzg0MjM1NTA4 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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] sapply and matrix command
Take a look at ?expand.grid Michael On Aug 4, 2012, at 5:03 PM, alijk1989 [via R] ml-node+s789695n463919...@n4.nabble.com wrote: Thanks again for the help looks like this will be useful for what I'm doing. Is there any way to use combn to return combinations of values with themselves: e.g. combn(1:3,2) [,1] [,2] [,3] [,4] [,5] [,6] [1,]111 2 2 3 [2,]1 232 3 3 ___ If you reply to this email, your message will be added to the discussion below: http://r.789695.n4.nabble.com/sapply-and-matrix-command-tp4637769p4639190.html To unsubscribe from sapply and matrix command, visit http://r.789695.n4.nabble.com/template/NamlServlet.jtp?macro=unsubscribe_by_codenode=4637769code=ci1oZWxwQHItcHJvamVjdC5vcmd8NDYzNzc2OXwtNzg0MjM1NTA4 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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 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] lme4 / HLM question
Jeff r at jp.pair.com writes: I'm hoping that this is a relatively easy question for someone familiar with the lme4 package. I'm accustomed to using HLM software and writing a simple 2 level [null] equation like this: L1 - Yij = b0 + e L2 - b0 = B00 + u0 The following command in R provides results that are identical to the HLM program. results - lmer( Y ~ 1 |id , PanelData4) I can't seem to find any examples on-line nor in the help about how to write the lmer4 formula that contains two predictor variables at level 1 with fixed slopes. L1 - Yij = b0 + b1(x) + b2(z) + e L2 - b0 = B00 + u0 b1 = B10 b2 = B20 In general you'd probably be better off asking this question at r-sig-mixed-mod...@r-project.org ... but it's very easy to put fixed effects into a model -- lmer (Y ~ x + z + (1|id), PanelData4) or in lme (nlme package): lme ( Y~ x + z, random = ~ 1|id, PanelData4) Neither lmer nor lme need to be told explicitly which level the fixed effects vary at (the data are always provided in long form). Ben Bolker __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to put barchart and line chart in the same plot in ggplot2
dear userR: I am trying to plot two dependent variables in the same plot in ggplot2. because these two variables have very different magnitude, I have to use a second Y axis. I hope one variable to be line and the other to be barchart. The x axis is continuous. Yet since I have to make barchart, I guess I have to treat it as discrete or categorical. I have been google searching for the whole afternoon but do not have any clue. Can anyone give me a direction (not have to be a complete answer...)? many thanks -- View this message in context: http://r.789695.n4.nabble.com/how-to-put-barchart-and-line-chart-in-the-same-plot-in-ggplot2-tp4639194.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list 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] ptproc package
Dear all I came across ptproc package on following website: http://www.biostat.jhsph.edu/~rpeng/software/index.html Actually I downloaded it on the contributors website and tried to install it manual but R wont unzip it. It is not available on CRAN project. I use R 2.15.1 and windows vista on my computer. Any help would be appreciated. Thanks. Amir Zadeh. -- View this message in context: http://r.789695.n4.nabble.com/ptproc-package-tp4639196.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list 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] Possible bug with MCMCpack metropolis sampler
Hi,
I'm having issues with what I believe is a bug in the MCMCpack's
MCMCmetrop1R function. I have code that basically looks like this:
posterior.sampler - function(data, prior.mu){
log.posterior - function(theta) log.likelihood(data, theta) +
log.prior(prior.mu, theta)
post.samples - MCMCmetrop1R(log.posterior, theta.init=prior.mu,
burnin=100, mcmc=1000, thin=40, tune=1, verbose=0, logfun=T,
optim.method=BFGS)
return(post.samples)
}
x - c(1,1,1)
posterior.sampler(mydata, x)
After calling posterior.sampler, the value for x is different from what I
started with. Perhaps even more interesting, is that if I create a copy of
x, say x2, before running posterior.sampler, x2 is modified as well. And I
have checked that calling log.posterior is not what alters the value of x,
so that leads me to believe it must be MCMCmetrop1R. Also, if I use
theta.init = c(1,1,1) instead of theta.init = prior.mu then x is not
altered. All of the behavior I described persists if I just run the
MCMCmetorp1R call (with the variables renamed of course) on its own and not
as part of my posterior.sampler function.
Any idea what's going on?
Thanks,
Dan Cervone
--
View this message in context:
http://r.789695.n4.nabble.com/Possible-bug-with-MCMCpack-metropolis-sampler-tp4639199.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
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] Package to remove collinear variables
Hi, I need to remove collinear variables to my Near-Infrared table of spectra. What package can I use? Something simple, because I am a novice about statistic. Thank you. Best regards, Roberto -- View this message in context: http://r.789695.n4.nabble.com/Package-to-remove-collinear-variables-tp4639200.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list 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.
