[R] legend not appearing in Word document
I need help with using graphics in Word 2007 that will later be converted into a pdf document. I have tried several formats and found that I get the best quality of graphics using .wmf, .eps format, but when I convert it to .pdf I get a bunch of lines across the figures. I also tried .tiff and .png but they give me much lower quality. The best quality that converts to pdf appears to be .eps. However, I have now come across a problem with my figure legends. For some reason the legend is visible in R but not in Word. Does anyone know why a legend in .eps format won't work in Word, or how I can get it to work? I have made an example of the legend below that you should be able to save as .eps and paste into Word as an example. I would appreciate any help you can offer on getting the legend in .eps format to work, or on other formats that may be better for Word and pdf files. Thanks, Tim library(plotrix) Satelite.Palette - colorRampPalette(c(blue3,cyan,aquamarine,yellow,orange,red)) mycol-Satelite.Palette(ceiling(5000+1))#Max relative angle multiplied by 100 to give larger range. Max is 3.1415, rounded up to 3.15 plus one. col.labels-round((seq(0,5000,length.out=5)/1000),1) plot(0, 0, type=n, axes=F, xlab=, ylab=) #New plot for legend color.legend(0,0,1,1,col.labels, mycol, align=rb, gradient=y) #Adds legend Tim Clark Marine Ecologist National Park of American Samoa Pago Pago, AS 96799 __ 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] legend not appearing in Word document
Thanks, I will take it up with MS. I just downloaded their latest converter and that hasn't fixed the issue. Hopefully they will have additional advice. Aloha, Tim Tim Clark Marine Ecologist National Park of American Samoa Pago Pago, AS 96799 From: Schalk Heunis schalk.heu...@enerweb.co.za Cc: r help r-help r-help@r-project.org; Tim Clark tim_cl...@nps.gov Sent: Sun, December 12, 2010 6:05:23 AM Subject: Re: [R] legend not appearing in Word document Tim Works in Word 2002 on Windows XP with PDF-xchange 3.0 to convert to pdf. Saw reponse from Duncan - agree might be problem with Word 2007 PDF converter. HTH Schalk I need help with using graphics in Word 2007 that will later be converted into a pdf document. I have tried several formats and found that I get the best quality of graphics using .wmf, .eps format, but when I convert it to .pdf I get a bunch of lines across the figures. I also tried .tiff and .png but they give me much lower quality. The best quality that converts to pdf appears to be .eps. However, I have now come across a problem with my figure legends. For some reason the legend is visible in R but not in Word. Does anyone know why a legend in .eps format won't work in Word, or how I can get it to work? I have made an example of the legend below that you should be able to save as .eps and paste into Word as an example. I would appreciate any help you can offer on getting the legend in .eps format to work, or on other formats that may be better for Word and pdf files. Thanks, Tim library(plotrix) Satelite.Palette - colorRampPalette(c(blue3,cyan,aquamarine,yellow,orange,red)) mycol-Satelite.Palette(ceiling(5000+1))#Max relative angle multiplied by 100 to give larger range. Max is 3.1415, rounded up to 3.15 plus one. col.labels-round((seq(0,5000,length.out=5)/1000),1) plot(0, 0, type=n, axes=F, xlab=, ylab=) #New plot for legend color.legend(0,0,1,1,col.labels, mycol, align=rb, gradient=y) #Adds legend Tim Clark Marine Ecologist National Park of American Samoa Pago Pago, AS 96799 __ 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.
[R] Count continuous dates
Dear List, I have a series of dates and I am needing to know the greatest length of continuous dates - i.e. the number of dates before a break in the series. For example, below there are three continuous series of dates with lengths 4, 6, and 8. How can I count the number of continuous dates? rle() will do it for constant integers, but I can't figure a way to modify it for continuous dates. x-c(1/10/10,2/10/10,3/10/10,4/10/10, 6/10/10,7/10/10,8/10/10,9/10/10,10/10/10,11/10/10, 13/10/10,14/10/10,15/10/10,16/10/10,17/10/10,18/10/10,19/10/10,20/10/10) x-strptime(x,%d/%m/%y) Thanks! Tim Tim Clark Marine Ecologist National Park of American Samoa Pago Pago, AS 96799 __ 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] Count continuous dates
Josh, Great use of rle()! I knew it had to fit in there somewhere! Thanks for the help! Tim Tim Clark Marine Ecologist National Park of American Samoa Pago Pago, AS 96799 - Original Message From: Joshua Wiley jwiley.ps...@gmail.com To: Tim Clark mudiver1...@yahoo.com Cc: r help r-help r-help@r-project.org Sent: Wed, November 10, 2010 7:45:15 PM Subject: Re: [R] Count continuous dates Hi Tim, Thanks for providing nice sample data! It made this super easy :) ## use diff() to find the differences ## use rle() to find the run lengths ## use max to find the highest one max(rle(as.vector(diff(x)))$lengths) Cheers, Josh On Wed, Nov 10, 2010 at 10:07 PM, Tim Clark mudiver1...@yahoo.com wrote: Dear List, I have a series of dates and I am needing to know the greatest length of continuous dates - i.e. the number of dates before a break in the series. For example, below there are three continuous series of dates with lengths 4, 6, and 8. How can I count the number of continuous dates? rle() will do it for constant integers, but I can't figure a way to modify it for continuous dates. x-c(1/10/10,2/10/10,3/10/10,4/10/10, 6/10/10,7/10/10,8/10/10,9/10/10,10/10/10,11/10/10, 13/10/10,14/10/10,15/10/10,16/10/10,17/10/10,18/10/10,19/10/10,20/10/10) ) x-strptime(x,%d/%m/%y) Thanks! Tim Tim Clark Marine Ecologist National Park of American Samoa Pago Pago, AS 96799 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joshua Wiley Ph.D. Student, Health Psychology University of California, Los Angeles http://www.joshuawiley.com/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Different results in FFT analysis
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[R] (no subject)
Dear List, I am trying to plot date vs. time, but am having problems getting my y-axis labels how I want them. When left on its own R plots time at 6 hour intervals from 03:00 to 23:00. I am wanting 6 hour intervals from 2:00 to 22:00. I realize yaxp doesn't work in plot(), so I am trying to get it to work in par(). However, now I get the ticks where I want them but the time is output as a very big number (serial time?). I have also tried axis() using at= and also get seriel time numbers. Any suggestions on how to format time on an axis? mydat$Date-as.POSIXct(as.character(mydat$Date), format=c(%m/%d/%Y)) mydat$Time-as.POSIXct(as.character(mydat$Time), format=c(%H:%M:%S)) plot(mydat$DateTime,mydat$Time, xlab=c(Date), ylab=c(Time), xlim=c(min(mydat$DateTime),max(mydat$DateTime)), ylim=c(min(mydat$Time),max(mydat$Time)), yaxt=n, yaxs=i, pch=19, cex=.4, type=n) par(yaxp=c(as.POSIXct(as.character(02:00), format=c(%H:%M)),as.POSIXct(as.character(22:00), format=c(%H:%M)),5)) axis(2) Thanks, Tim Tim Clark Marine Ecologist National Park of American Samoa __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Memory limit problem
Dear List, I am trying to plot bathymetry contours around the Hawaiian Islands using the package rgdal and PBSmapping. I have run into a memory limit when trying to combine two fairly small objects using cbind(). I have increased the memory to 4GB, but am being told I can't allocate a vector of size 240 Kb. I am running R 2.11.1 on a Dell Optiplex 760 with Windows XP. I have pasted the error message and summaries of the objects below. Thanks for your help. Tim xyz-cbind(hi.to.utm,z=b.depth$z) Error: cannot allocate vector of size 240 Kb memory.limit() [1] 4000 memory.size() [1] 1971.68 summary(hi.to.utm) Object of class SpatialPoints Coordinates: min max x 708745.5 923406.7 y 2046153.1 2327910.9 Is projected: TRUE proj4string : [+proj=utm +zone=4 +datum=NAD83 +ellps=GRS80 +towgs84=0,0,0] Number of points: 15328 str(hi.to.utm) Formal class 'SpatialPoints' [package sp] with 3 slots ..@ coords : num [1:15328, 1:2] 708746 710482 712218 713944 715681 ... .. ..- attr(*, dimnames)=List of 2 .. .. ..$ : NULL .. .. ..$ : chr [1:2] x y ..@ bbox : num [1:2, 1:2] 708746 2046153 923407 2327911 .. ..- attr(*, dimnames)=List of 2 .. .. ..$ : chr [1:2] x y .. .. ..$ : chr [1:2] min max ..@ proj4string:Formal class 'CRS' [package sp] with 1 slots .. .. ..@ projargs: chr +proj=utm +zone=4 +datum=NAD83 +ellps=GRS80 +towgs84=0,0,0 summary(b.depth) x y z Min. :-157.0 Min. :18.50 Min. :-5783 1st Qu.:-156.6 1st Qu.:18.98 1st Qu.:-4565 Median :-156.1 Median :19.80 Median :-3358 Mean :-156.1 Mean :19.73 Mean :-3012 3rd Qu.:-155.5 3rd Qu.:20.41 3rd Qu.:-1601 Max. :-155.0 Max. :21.00 Max. : 0 str(b.depth) 'data.frame': 15328 obs. of 3 variables: $ x: num -157 -157 -157 -157 -157 ... $ y: num 21 21 21 21 21 ... $ z: num -110 -114 -110 -88 -76 -122 -196 -224 -240 -238 ... Tim Clark Marine Ecologist National Park of American Samoa __ 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] Changing x-axis on boxplot
Dear List, I am creating a boxplot with two subsets, very similar to the example by Roger Bivand at ?boxplot (reproduced below). I am trying to change the labels on the x-axis to have one number to cover both subsets. I can do this in other plots by using axis=FALSE followed by a separate axis() command. I have also tried variations in the names= argument but can't get it to work. Ideally I would like tickmarks on either side of each factor with the number for the level centered between the two tick marks. Any suggestions? Thanks, Tim Example: boxplot(len ~ dose, data = ToothGrowth, boxwex = 0.25, at = 1:3 - 0.2, subset = supp == VC, col = yellow, main = Guinea Pigs' Tooth Growth, xlab = Vitamin C dose mg, ylab = tooth length, xlim = c(0.5, 3.5), ylim = c(0, 35), yaxs = i) boxplot(len ~ dose, data = ToothGrowth, add = TRUE, boxwex = 0.25, at = 1:3 + 0.2, subset = supp == OJ, col = orange) legend(2, 9, c(Ascorbic acid, Orange juice), fill = c(yellow, orange)) Tim Clark Marine Ecologist National Park of American Samoa __ 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] Adding row name to dataframe
Dear all, I am trying to add a value to a dataframe and name the row with a number. I have tried row.name, rowname, and attr(x,row.names) but none seem to work. It seems like it should be simple, so not sure why I can't get it to work. Any suggestions? Thanks, Tim x-seq(1,20,2) y-seq(20,1,-2) xy-data.frame(x,y) xy-rbind(xy,c(0,0)) #Threeattempts that fail row.names(xy[11,])-c(12) rownames(xy[11,])-c(12) attr(xy[11,], row.names)-c(12) Tim Clark Marine Ecologist National Park of American Samoa __ 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] Yet another memory limit problem
Thanks for everyones help - mydat was not what I thought it was. str(mydat) showed: $ Date : Factor w/ 1504 levels ,2002-11-22,..: 295 295 295 295 295 295 295 295 295 295 ... $ Time : Factor w/ 72447 levels ,00:00:00,..: 15423 15470 15509 17112 17136 17159 17209 17239 17266 21527 ... I didn't properly convert date and time using chron. Not sure how R plots factor levels, but obviously it has problems with it! Aloha, Tim Tim Clark Department of Zoology University of Hawaii - Original Message From: Peter Dalgaard pda...@gmail.com To: Tim Clark mudiver1...@yahoo.com Cc: r-help@r-project.org Sent: Wed, August 18, 2010 8:34:56 AM Subject: Re: [R] Yet another memory limit problem On 08/18/2010 11:19 AM, Tim Clark wrote: Dear List, ... I would appreciate some help. I think I set the target field correctly, both memory.size and memory.limit indicate I have over 2G of memory, yet I can't allocate 831.3Mb? It just doesn't make sense to me. That shouldn't by itself cause concern. It is always the _last_ allocation that triggers the message, so it could happen due to allocation of 3 vectors of size 800M. Straw, camel, back... Another matter is that memory.size is the amount _used_, not amount _available_, so you would seem to have more like 1500M left. However, it does look odd that mydat with dimensions 245x9 should generate an 800M memory request. As others have suggested, perhaps mydat is not what you think it should be... -- Peter Dalgaard Center for Statistics, Copenhagen Business School Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.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] Yet another memory limit problem
Dear List, I have read and read and still don't get why I am getting a memory issue. I am using a Samsung PC running Windows 7. I have set memory in the target field: C:\Program Files (x86)\R\R-2.11.1\bin\Rgui.exe --max-mem-size=3G But when I try a simple plot I get: plot(mydat$Date,mydat$Time) Error: cannot allocate vector of size 831.3 Mb memory.size() [1] 2099.61 memory.limit() [1] 3583 dim(mydat) [1] 245 9 I would appreciate some help. I think I set the target field correctly, both memory.size and memory.limit indicate I have over 2G of memory, yet I can't allocate 831.3Mb? It just doesn't make sense to me. Thanks, Tim __ 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] Reading files with varying number of columns
Dear R list, I am trying to read files with a varying number of columns and can't figure out how to get them in the proper format. Some rows have five value and some have seven. Is there a way to read them in so that two empty columns are added to every row that has only five values? An example of the data follows. Thanks! Tim 46,B,00301,2004-02-03,11:59:16 46,B,00301,2004-02-03,12:03:43 46,B,00301,2004-02-03,12:39:17 46,B,00301,2004-02-03,23:07:28 43,C,00232,2004-02-04,07:39:57,-1.5,meters 44,A,00230,2004-02-04,07:44:59 44,A,00230,2004-02-04,07:46:19 44,A,00230,2004-02-04,07:48:24 44,A,00230,2004-02-04,07:49:12 43,C,00232,2004-02-04,07:53:08,2.5,meters 44,A,00230,2004-02-04,07:54:34 44,A,00230,2004-02-04,07:55:08 43,C,00232,2004-02-04,07:56:18,2.9,meters 43,C,00232,2004-02-04,07:56:39,2.5,meters 46,B,00301,2004-02-04,08:00:49 43,C,00232,2004-02-04,08:02:12,-1.1,meters 43,C,00232,2004-02-04,08:04:01,2.2,meters 43,C,00232,2004-02-04,08:04:26,3.3,meters 43,C,00232,2004-02-04,08:40:26,2.9,meters 43,C,00232,2004-02-04,08:41:04,-2.2,meters 43,C,00232,2004-02-04,08:42:44,2.2,meters 43,C,00232,2004-02-04,08:44:31,2.9,meters 44,A,00231,2004-02-04,09:04:43 44,A,00231,2004-02-04,09:10:30 44,A,00231,2004-02-04,09:12:08 44,A,00231,2004-02-04,09:13:32 44,A,00231,2004-02-04,09:14:06 43,C,00232,2004-02-04,09:18:24,0.4,meters 43,C,00232,2004-02-04,09:20:13,2.5,meters 44,A,00231,2004-02-04,09:21:04 44,A,00231,2004-02-04,09:22:53 44,A,00231,2004-02-04,09:25:32 44,A,00231,2004-02-04,09:29:31 44,A,00231,2004-02-04,09:30:05 43,C,00232,2004-02-04,09:53:25,2.5,meters 43,C,00232,2004-02-04,09:53:53,1.5,meters 46,B,00301,2004-02-04,22:16:24 46,B,00301,2004-02-04,22:19:32 46,B,00258,2004-02-04,23:44:01 46,B,00258,2004-02-04,23:44:28 Tim Clark Marine Ecologist National Park of American Samoa Pago Pago, American Samoa 96799 __ 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] Reading files with varying number of columns
Thanks for the suggestion. I had read fill but didn't pick up on the details. It only uses the first five columns to determine column dimensions, so it ignored my last two columns. Specifying col.names with fill=TRUE did the trick! Aloha, Tim - Original Message From: Uwe Ligges lig...@statistik.tu-dortmund.de To: Tim Clark mudiver1...@yahoo.com Cc: r-help@r-project.org Sent: Sun, July 18, 2010 6:03:47 AM Subject: Re: [R] Reading files with varying number of columns See argument fill in ?read.table Uwe Ligges On 18.07.2010 12:14, Tim Clark wrote: Dear R list, I am trying to read files with a varying number of columns and can't figure out how to get them in the proper format. Some rows have five value and some have seven. Is there a way to read them in so that two empty columns are added to every row that has only five values? An example of the data follows. Thanks! Tim 46,B,00301,2004-02-03,11:59:16 46,B,00301,2004-02-03,12:03:43 46,B,00301,2004-02-03,12:39:17 46,B,00301,2004-02-03,23:07:28 43,C,00232,2004-02-04,07:39:57,-1.5,meters 44,A,00230,2004-02-04,07:44:59 44,A,00230,2004-02-04,07:46:19 44,A,00230,2004-02-04,07:48:24 44,A,00230,2004-02-04,07:49:12 43,C,00232,2004-02-04,07:53:08,2.5,meters 44,A,00230,2004-02-04,07:54:34 44,A,00230,2004-02-04,07:55:08 43,C,00232,2004-02-04,07:56:18,2.9,meters 43,C,00232,2004-02-04,07:56:39,2.5,meters 46,B,00301,2004-02-04,08:00:49 43,C,00232,2004-02-04,08:02:12,-1.1,meters 43,C,00232,2004-02-04,08:04:01,2.2,meters 43,C,00232,2004-02-04,08:04:26,3.3,meters 43,C,00232,2004-02-04,08:40:26,2.9,meters 43,C,00232,2004-02-04,08:41:04,-2.2,meters 43,C,00232,2004-02-04,08:42:44,2.2,meters 43,C,00232,2004-02-04,08:44:31,2.9,meters 44,A,00231,2004-02-04,09:04:43 44,A,00231,2004-02-04,09:10:30 44,A,00231,2004-02-04,09:12:08 44,A,00231,2004-02-04,09:13:32 44,A,00231,2004-02-04,09:14:06 43,C,00232,2004-02-04,09:18:24,0.4,meters 43,C,00232,2004-02-04,09:20:13,2.5,meters 44,A,00231,2004-02-04,09:21:04 44,A,00231,2004-02-04,09:22:53 44,A,00231,2004-02-04,09:25:32 44,A,00231,2004-02-04,09:29:31 44,A,00231,2004-02-04,09:30:05 43,C,00232,2004-02-04,09:53:25,2.5,meters 43,C,00232,2004-02-04,09:53:53,1.5,meters 46,B,00301,2004-02-04,22:16:24 46,B,00301,2004-02-04,22:19:32 46,B,00258,2004-02-04,23:44:01 46,B,00258,2004-02-04,23:44:28 Tim Clark Marine Ecologist National Park of American Samoa Pago Pago, American Samoa 96799 __ 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.
[R] Generating all possible models from full model
(mantas~year+cosmonth+sinmonth+coslunar+sinlunar, data=mydata)) m23457-glm.convert(glm.nb(mantas~year+cosmonth+sinmonth+coslunar+plankton, data=mydata)) m23467-glm.convert(glm.nb(mantas~year+cosmonth+sinmonth+sinlunar+plankton, data=mydata)) m23567-glm.convert(glm.nb(mantas~year+cosmonth+coslunar+sinlunar+plankton, data=mydata)) m24567-glm.convert(glm.nb(mantas~year+sinmonth+coslunar+sinlunar+plankton, data=mydata)) m34567-glm.convert(glm.nb(mantas~cosmonth+sinmonth+coslunar+sinlunar+plankton, data=mydata)) #Six terms - 7 models m123456-glm.convert(glm.nb(mantas~site*year+cosmonth+sinmonth+coslunar+sinlunar, data=mydata)) m123457-glm.convert(glm.nb(mantas~site*year+cosmonth+sinmonth+coslunar+plankton, data=mydata)) m123467-glm.convert(glm.nb(mantas~site*year+cosmonth+sinmonth+sinlunar+plankton, data=mydata)) m123567-glm.convert(glm.nb(mantas~site*year+cosmonth+coslunar+sinlunar+plankton, data=mydata)) m124567-glm.convert(glm.nb(mantas~site*year+sinmonth+coslunar+sinlunar+plankton, data=mydata)) m134567-glm.convert(glm.nb(mantas~site+cosmonth+sinmonth+coslunar+sinlunar+plankton, data=mydata)) m234567-glm.convert(glm.nb(mantas~year+cosmonth+sinmonth+coslunar+sinlunar+plankton, data=mydata)) #Seven terms - 1 model m1234567-glm.convert(glm.nb(mantas~site*year+cosmonth+sinmonth+coslunar+sinlunar+plankton, data=mydata)) Tim Clark Department of Zoology 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] Generating all possible models from full model
Thanks to everyone for the replies and functions. For some reason my emails
are not going to the thread. I just switched to the newer mail version for my
yahoo account, so hopefully this one will get posted.
Aloha,
Tim
Tim Clark
Department of Zoology
University of Hawaii
- Original Message
From: Xiaogang Su xiaogan...@gmail.com
To: Tim Clark mudiver1...@yahoo.com
Cc: r-help@r-project.org
Sent: Wed, May 19, 2010 9:42:40 AM
Subject: Re: [R] Generating all possible models from full model
A couple of years ago, I wrote a function for doing this. It can be
found in the following file:
http://pegasus.cc.ucf.edu/~xsu/CLASS/STA4164/mt4-2009.doc
I also pasted a copy below. Hope you find it useful. -XG
#
# ALL POSSIBLE REGRESSIONS
#
# The parameter k is the total number of predictors. To apply the function,
# you need to prepare the data so that all predictors for selection are named
# as x1, x2, ..., and the response is called y.
all.possible.regressions - function(dat, k){
n - nrow(dat)
regressors - paste(x, 1:k, sep=)
lst - rep(list(c(T, F)), k)
regMat - expand.grid(lst);
names(regMat) - regressors
formular - apply(regMat, 1, function(x)
as.character(paste(c(y ~ 1, regressors[x]), collapse=+)))
allModelsList - apply(regMat, 1, function(x)
as.formula(paste(c(y ~ 1, regressors[x]),collapse= + )) )
allModelsResults - lapply(allModelsList,
function(x, data) lm(x, data=data), data=dat)
n.models - length(allModelsResults)
extract - function(fit) {
df.sse - fit$df.residual
p - n - df.sse -1
sigma - summary(fit)$sigma
MSE - sigma^2
R2 - summary(fit)$r.squared
R2.adj - summary(fit)$adj.r.squared
sse - MSE*df.sse
aic - n*log(sse) + 2*(p+2)
bic - n*log(sse) + log(n)*(p+2)
out - data.frame(df.sse=df.sse, p=p, SSE=sse, MSE=MSE,
R2=R2, R2.adj=R2.adj, AIC=aic, BIC=bic)
return(out)
}
result - lapply(allModelsResults, extract)
result - as.data.frame(matrix(unlist(result), nrow=n.models, byrow=T))
result - cbind(formular, result)
rownames(result) - NULL
colnames(result) - c(model, df.sse, p, SSE, MSE, R2,
R2.adj, AIC, BIC)
return(result)
}
all.possible.regressions(dat=quasar, k=5)
On Tue, May 18, 2010 at 11:38 PM, Tim Clark mudiver1...@yahoo.com wrote:
Is there a function that will allow me to run all model iterations if I
specify a full model? I am using information criteria to choose between
possible candidate models. I have been writing out all possible model
combinations by hand, and I am always worried that I am missing models or
have made a mistake somewhere. It is also difficult to alter models if I
want to change a term. For example, below are the set of models I would like
to run. Is there a way to specify the full model and have R generate the
rest? I.e. specify
m1234567-glm.convert(glm.nb(mantas~site*year+cosmonth+sinmonth+coslunar+sinlunar+plankton,
data=mydata))
and have R run all the other models.
library(MASS)
#Intercept only
m0-glm.convert(glm.nb(mantas~1,data=mydata))
#One term - 7 models
#Manta abundance is greater at one of the two sites
m1-glm.convert(glm.nb(mantas~site,data=mydata))
#Manta abundance increases each year as the population increases in size due
to births or immigration being greater than deaths and emmigration
m2-glm.convert(glm.nb(mantas~year,data=mydata))
#Manta abundances increases during part of the year due to seasonal cycles
in resources (mates, food)
m3-glm.convert(glm.nb(mantas~cosmonth,data=mydata))
m4-glm.convert(glm.nb(mantas~sinmonth,data=mydata))
#Manta abundance decreases with increased lunar phase
m5-glm.convert(glm.nb(mantas~coslunar, data=mydata))
m6-glm.convert(glm.nb(mantas~sinlunar, data=mydata))
#Manta abundance increases with increased levels of plankton
m7-glm.convert(glm.nb(mantas~plankton,data=mydata))
#Two terms - 21 models
m12-glm.convert(glm.nb(mantas~site*year, data=mydata)) #Interaction term
to account for hotel being closed at Keauhou for some years
m13-glm.convert(glm.nb(mantas~site+cosmonth,data=mydata))
m14-glm.convert(glm.nb(mantas~site+sinmonth,data=mydata))
m15-glm.convert(glm.nb(mantas~site+coslunar,data=mydata))
m16-glm.convert(glm.nb(mantas~site+sinlunar,data=mydata))
m17-glm.convert(glm.nb(mantas~site+plankton,data=mydata)) #Should this have
an interaction term? Plankton may varry by site
m23-glm.convert(glm.nb(mantas~year+cosmonth,data=mydata))
m24-glm.convert(glm.nb(mantas~year+sinmonth,data=mydata))
m25-glm.convert(glm.nb(mantas~year+coslunar,data=mydata))
m26-glm.convert(glm.nb(mantas~year+sinlunar,data=mydata))
m27-glm.convert(glm.nb(mantas~year+plankton,data=mydata))
m34-glm.convert(glm.nb(mantas~cosmonth+sinmonth,data=mydata))
m35-glm.convert(glm.nb(mantas~cosmonth+coslunar,data
Re: [R] bar order using lattice barchart()
Greg, Thanks for the great explanation. Knowing the philosophy behind these kind of things really helps avoid problems in the future. Aloha, Tim Tim Clark Department of Zoology University of Hawaii --- On Thu, 5/6/10, Greg Snow greg.s...@imail.org wrote: From: Greg Snow greg.s...@imail.org Subject: RE: [R] bar order using lattice barchart() To: Tim Clark mudiver1...@yahoo.com, r-help@r-project.org r-help@r-project.org Date: Thursday, May 6, 2010, 7:51 AM The short answer to your query is ?reorder The longer answer (or a longer answer) gets into a bit of philosophy (so feel free to go back to the short answer and skip this if you don't want to get into the philosophy, you have been warned). Let's start with the question: is the order of the bars a property of the graph/analysis? The dataset as a whole? Or the individual variable? Some programs take the first approach, the order is a property of the graph or analysis, these programs have you specify things like the order at the time of creating the graph or doing the analysis. I don't like this approach because it seems to me that this should be more inherent to the data than the output, also this means that you have to keep specifying the order for every graph/analysis and I am too lazy to like that. Your attempt was number 2, the order should be a property of the dataset. For your example this makes a lot of sense, you want x ordered by y. But this does not generalize to some other situations, so is not the standard for R. My favorite (and what is used in R, so apparently I'm not the only one) is to have things like the order be a property of the individual variable (x in this case). Your data frame will have x as a factor by default (and even if you specifically told R to not convert it to a factor, then many plotting/analysis functions will still convert it to a factor). If you don't specify the order of the factor levels then the default is to do it alphabetically and this will be used by plotting/analysis functions regardless of the order in the data set. One of the easier ways to change the order of the factor levels, especially for a case like yours is using the reoder function, try something like: xy$x - reorder( xy$x, xy$y ) You won't see any obvious differences when you print the data frame, but if you print just x then you should and your plots should come out the way that you want now. You did ask why? and were warned. Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of Tim Clark Sent: Wednesday, May 05, 2010 8:57 PM To: r-help@r-project.org Subject: [R] bar order using lattice barchart() Dear List, I am want to plot my data in increasing order using the lattice barchart() function. I used order() to put my data in the order I want, but when I plot it I get the original order of the data. I think this has to do with the row index number since order() does not re- number the rows in the new order but instead keeps the original row numbers and puts them in a different order. For example: xy-data.frame(x=letters[1:5],y=c(3,9,2,1,10)) #This produces a dataframe in alphebitical order with row numbers in #numerical order. xy x y 1 a 8.921657 2 b 10.314625 3 c 9.531537 4 d 10.818563 5 e 9.084872 #If I re-order the data based on the value of y xy-xy[order(y),] #I get a dataframe ordered by y but the row numbers are still in #alphebetical order based on x xy x y 1 a 8.921657 5 e 9.084872 3 c 9.531537 2 b 10.314625 4 d 10.818563 #I then try to plot the data and it plots it in alphabetical instead of #numeric order library(lattice) barchart(y~x, data=xy) Why are the rows not re-indexed, and how do I get barchart() to plot my data in increasing numeric order? Thanks, Tim Tim Clark Department of Zoology 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. __ 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] Estimating theta for negative binomial model
Brian, glm.convert() does exactly what I am wanting. Thanks for putting it in MASS. Sorry about not giving credit. It is a great package and I keep finding functions that make statistical programming easy! Aloha, Tim Tim Clark Department of Zoology University of Hawaii --- On Mon, 5/3/10, Prof Brian Ripley rip...@stats.ox.ac.uk wrote: From: Prof Brian Ripley rip...@stats.ox.ac.uk Subject: Re: [R] Estimating theta for negative binomial model To: Tim Clark mudiver1...@yahoo.com Cc: r-help@r-project.org Date: Monday, May 3, 2010, 7:50 PM On Mon, 3 May 2010, Tim Clark wrote: Dear List, I am trying to do model averaging for a negative binomial model using the package AICcmodavg. I need to use glm() since the package does not accept glm.nb() models. I can get glm() to work if I first run glm.nb and take theta from that model, but is there a simpler way to estimate theta for the glm model? The two models are: mod.nb-glm.nb(mantas~site,data=mydata) mod.glm-glm(mantas~site,data=mydata, family=negative.binomial(mod.nb$theta)) How else can I get theta for the family=negative.binomial(theta=???) library(MASS) is missing here -- do give credit where credit is due. It contains functions glm.convert theta.ml and the first does what you seem to want and the second answers the actual question you asked. Thanks! Tim Tim Clark Department of Zoology University of Hawaii -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UK Fax: +44 1865 272595 __ 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] Estimating theta for negative binomial model
Dear List, I am trying to do model averaging for a negative binomial model using the package AICcmodavg. I need to use glm() since the package does not accept glm.nb() models. I can get glm() to work if I first run glm.nb and take theta from that model, but is there a simpler way to estimate theta for the glm model? The two models are: mod.nb-glm.nb(mantas~site,data=mydata) mod.glm-glm(mantas~site,data=mydata, family=negative.binomial(mod.nb$theta)) How else can I get theta for the family=negative.binomial(theta=???) Thanks! Tim Tim Clark Department of Zoology 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.
[R] Periodic regression - lunar percent cover
Dear List, I am trying include a lunar variable in a model and am having problems figuring out the correct way to include it. I want to convert the percent lunar illumination (fraction of moon showing) to a combination of sin and cos variables to account for the periodic nature of the lunar cycle. Would someone let me know if I am doing this correctly? I have included the first 20 variables from my dataset as an example. Y is count data and lp is the lunar percent cover. The lunar period is 29.53. y-c(1, 3, 0, 0, 0, 0, 2, 4, 0, 1, 0, 5, 3, 2, 4, 2, 0, 1, 3, 5) lp-c(0.80, 0.88, 0.62, 0.19, 0.21, 0.01, 0.70, 1.00, 0.88, 0.04, 0.70, 0.93, 0.23, 0.99, 0.19, 0.79, 1.00, 0.03, 0.01, 0.00) g1-glm(y~cos((2*pi*lp)/29.530589)+sin((2*pi*lp)/29.530589)) Thanks, Tim Tim Clark Department of Zoology 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.
[R] Best subset of models for glm.nb()
Dear List, I am looking for a function that will find the best subset of negative binomial models. I have a large data set with 15 variables that I am interested in. I want an easy way to run all possible models and find a subset of the best models that I can then look at in more detail. I have found two functions that seem to provide what I am looking for, but am not sure which one (if either) are appropriate. glmulti() in package glmulti does an exhaustive search of all models and gives a number of candidate models to choose from based on your choice of Information Criterion. This seems to be exactly what I am after, but I found nothing about it on this list which makes me think there is some reason no one is using it. gl1ce() in package lasso2 uses the least absolute shrinkage and selection operator (lasso) to do something. I found it at another thread: http://tolstoy.newcastle.edu.au/R/help/05/03/0121.html I did not understand the paper it was based on, and want to know if it even does what I am interested in before investing a lot of time in trying to understand it. Yes, I have read about the problems with stepwise algorithms and am looking for a valid alternative to narrowing down models when you have a lot of data and a large number of variables your interested in. Any thoughts on either of these methods? Or should I be doing something else? Thanks for your help, Tim Tim Clark Department of Zoology 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.
[R] r help date format changes with c() vs. rbind()
Dear List, I am having a problem with dates and I would like to understand what is going on. Below is an example. I can produce a date/time using as.POSIXct, but I am trying to combine two as.POSIXct objects and keep getting strange results. I thought I was using the wrong origin, but according to structure(0,class=Date) I am not (see below). In my example a is a simple date/time object, b combines it using rbind(), c converts b to a date/time object again using as.POSIXct and gives the incorrect time, and d combines a using c() and gives the correct time. Why doesn't c give me the correct answer? Thanks, Tim a-as.POSIXct(2000-01-01 12:00:00) a [1] 2000-01-01 12:00:00 HST b-rbind(a,a) b [,1] a 946764000 a 946764000 c-as.POSIXct(b,origin=1970-01-01) c [1] 2000-01-01 22:00:00 HST [2] 2000-01-01 22:00:00 HST d-c(a,a) d [1] 2000-01-01 12:00:00 HST [2] 2000-01-01 12:00:00 HST structure(0,class=Date) [1] 1970-01-01 Tim Clark Department of Zoology 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] r help date format changes with c() vs. rbind()
Glad to know it isn't just me! I couldn't use Phil's data.frame method since my real problem went from a POSIXct object to a large matrix where I used rbind and then back to POSIXct. Jim's function worked great on converting the final product back to the proper date. Thanks! Tim Tim Clark Department of Zoology University of Hawaii --- On Fri, 2/19/10, jim holtman jholt...@gmail.com wrote: From: jim holtman jholt...@gmail.com Subject: Re: [R] r help date format changes with c() vs. rbind() To: Tim Clark mudiver1...@yahoo.com Cc: r-help@r-project.org Date: Friday, February 19, 2010, 12:19 PM I have used the following function to convert to POSIXct from a numeric without any problems: unix2POSIXct - function (time) structure(time, class = c(POSIXt, POSIXct)) unix2POSIXct(946764000) [1] 2000-01-01 17:00:00 EST On Fri, Feb 19, 2010 at 4:07 PM, Tim Clark mudiver1...@yahoo.com wrote: Dear List, I am having a problem with dates and I would like to understand what is going on. Below is an example. I can produce a date/time using as.POSIXct, but I am trying to combine two as.POSIXct objects and keep getting strange results. I thought I was using the wrong origin, but according to structure(0,class=Date) I am not (see below). In my example a is a simple date/time object, b combines it using rbind(), c converts b to a date/time object again using as.POSIXct and gives the incorrect time, and d combines a using c() and gives the correct time. Why doesn't c give me the correct answer? Thanks, Tim a-as.POSIXct(2000-01-01 12:00:00) a [1] 2000-01-01 12:00:00 HST b-rbind(a,a) b [,1] a 946764000 a 946764000 c-as.POSIXct(b,origin=1970-01-01) c [1] 2000-01-01 22:00:00 HST [2] 2000-01-01 22:00:00 HST d-c(a,a) d [1] 2000-01-01 12:00:00 HST [2] 2000-01-01 12:00:00 HST structure(0,class=Date) [1] 1970-01-01 Tim Clark Department of Zoology 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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ 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] Formatting question for separate polygons
Thanks Uwe! And Peter for the correction. I would never have come up with that! Tim Tim Clark Department of Zoology University of Hawaii --- On Fri, 2/12/10, Uwe Ligges lig...@statistik.tu-dortmund.de wrote: From: Uwe Ligges lig...@statistik.tu-dortmund.de Subject: Re: [R] Formatting question for separate polygons To: Peter Ehlers ehl...@ucalgary.ca Cc: Tim Clark mudiver1...@yahoo.com, r-help@r-project.org Date: Friday, February 12, 2010, 1:57 AM On 12.02.2010 12:54, Peter Ehlers wrote: Nice, Uwe. Small correction: make that nrow=4: x1 - as.numeric(rbind(matrix(rep(x, each=2), nrow=4), NA)) Whoops, thanks! Uwe -Peter Ehlers Uwe Ligges wrote: On 11.02.2010 22:38, Tim Clark wrote: Dear List, I am trying to plot several separate polygons on a graph. I have figured out how to do it by manually, but have too much data to use such a tedious method. I would appreciate your help. I have made a simple example to illustrate the problem. How can I get x into the proper format (x1)? #Sample data x-c(1,2,3,4,5,6) y-c(1,2,2,1) #I need to format the data like this x1-c(1,1,2,2,NA,3,3,4,4,NA,5,5,6,6,NA) y1-rep(c(1,2,2,1,NA),length(x)/2) x1 - as.numeric(rbind(matrix(rep(x, each=2), nrow=2), NA)) Uwe Ligges #Final plot plot(c(1,6), 1:2, type=n) polygon(x1,y1,density=c(40)) Thanks, Tim Tim Clark Department of Zoology 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. __ 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] Find each time a value changes
It was brought to my attention that the rle() answer to this question was not posted. The following gives the correct answer once the last value is deleted. x-seq(1:100) y1-rep(1,10) y2-rep(2,10) y-c(y1,y2,y1,y1,y1,y2,y1,y2,y1,y2) xy-cbind(x,y) print(xy) print(str(xy)) # SEE WHAT RLE GIVES test - rle(xy[,2]) print(str(test) # USE JIMS TRICK OF CUMULATIVE SUMMING # TO GET THE LOCATIONS result - cumsum(c(1,rle(xy[,2])$lengths)) Tim Clark Department of Zoology University of Hawaii --- On Wed, 2/10/10, Ben Tupper ben.bigh...@gmail.com wrote: From: Ben Tupper ben.bigh...@gmail.com Subject: Re: [R] Find each time a value changes To: r-help@r-project.org Cc: Tim Clark mudiver1...@yahoo.com Date: Wednesday, February 10, 2010, 4:16 PM Hi, On Feb 10, 2010, at 8:58 PM, Tim Clark wrote: Dear List, I am trying to find each time a value changes in a dataset. The numbers are variables for day vs. night values, so what I am really getting is the daily sunrise and sunset. A simplified example is the following: x-seq(1:100) y1-rep(1,10) y2-rep(2,10) y-c(y1,y2,y1,y1,y1,y2,y1,y2,y1,y2) xy-cbind(x,y) I would like to know each time the numbers change. Correct answer should be: x=1,11,21,51,61,71,81,91 I think this gets close... which(diff(y) != 0) [1] 10 20 50 60 70 80 90 You'll need to fiddle to get exactly what you want. Cheers, Ben I would appreciate any help or suggestions. It seems like it should be simple but I’m stuck! Thanks, Tim Tim Clark Department of Zoology 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. __ 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] Formatting question for separate polygons
Dear List, I am trying to plot several separate polygons on a graph. I have figured out how to do it by manually, but have too much data to use such a tedious method. I would appreciate your help. I have made a simple example to illustrate the problem. How can I get x into the proper format (x1)? #Sample data x-c(1,2,3,4,5,6) y-c(1,2,2,1) #I need to format the data like this x1-c(1,1,2,2,NA,3,3,4,4,NA,5,5,6,6,NA) y1-rep(c(1,2,2,1,NA),length(x)/2) #Final plot plot(c(1,6), 1:2, type=n) polygon(x1,y1,density=c(40)) Thanks, Tim Tim Clark Department of Zoology 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.
[R] Find each time a value changes
Dear List, I am trying to find each time a value changes in a dataset. The numbers are variables for day vs. night values, so what I am really getting is the daily sunrise and sunset. A simplified example is the following: x-seq(1:100) y1-rep(1,10) y2-rep(2,10) y-c(y1,y2,y1,y1,y1,y2,y1,y2,y1,y2) xy-cbind(x,y) I would like to know each time the numbers change. Correct answer should be: x=1,11,21,51,61,71,81,91 I would appreciate any help or suggestions. It seems like it should be simple but I’m stuck! Thanks, Tim Tim Clark Department of Zoology 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] Find each time a value changes
Thanks everyone! I would have been banging my head around for quite a while and still wouldn't have come up with either solution. The function rle() is a good one to know! Aloha, Tim Tim Clark Department of Zoology University of Hawaii --- On Wed, 2/10/10, Ben Tupper ben.bigh...@gmail.com wrote: From: Ben Tupper ben.bigh...@gmail.com Subject: Re: [R] Find each time a value changes To: r-help@r-project.org Cc: Tim Clark mudiver1...@yahoo.com Date: Wednesday, February 10, 2010, 4:16 PM Hi, On Feb 10, 2010, at 8:58 PM, Tim Clark wrote: Dear List, I am trying to find each time a value changes in a dataset. The numbers are variables for day vs. night values, so what I am really getting is the daily sunrise and sunset. A simplified example is the following: x-seq(1:100) y1-rep(1,10) y2-rep(2,10) y-c(y1,y2,y1,y1,y1,y2,y1,y2,y1,y2) xy-cbind(x,y) I would like to know each time the numbers change. Correct answer should be: x=1,11,21,51,61,71,81,91 I think this gets close... which(diff(y) != 0) [1] 10 20 50 60 70 80 90 You'll need to fiddle to get exactly what you want. Cheers, Ben I would appreciate any help or suggestions. It seems like it should be simple but I’m stuck! Thanks, Tim Tim Clark Department of Zoology 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. __ 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] Plotting color.legend() outside of plot region
Dear List, I am trying to plot a color.legend() in the right outer margin of my device region. I have read multiple threads on the subject and still can't get it right. I have stolen an example from one of the threads to demonstrate my problem. I have extended the outer margin using par(oma()), and have used par(xpd=NA) to tell it to plot in the device region. I can get the legend to plot in the figure region but it will not plot in the outer margin. What am I doing wrong? Example follows with the legend in the figure region. I would like it more to the right in the device region. Thank, Tim library(TeachingDemos) op - par(mfrow = c(3,3), ## split region oma = c(0,0,4,12) + 0.1, ## create outer margin mar = c(5,4,2,2) + 0.1) ## shrink some margins plot(1:10, main = a, pch = 1:2, col= 1:2) plot(1:10, main = b, pch = 1:2, col= 1:2) tmp1 - cnvrt.coords( 0.5, 0, input='plt' )$tdev # save location for mtext plot(1:10, main = c, pch = 1:2, col= 1:2) plot(1:10, main = d, pch = 1:2, col= 1:2) plot(1:10, main = e, pch = 1:2, col= 1:2) plot(1:10, main = f, pch = 1:2, col= 1:2) plot(1:10, main = g, pch = 1:2, col= 1:2) plot(1:10, main = h, pch = 1:2, col= 1:2) plot(1:10, main = i, pch = 1:2, col= 1:2) ## title mtext(My Plots, side = 3, outer = TRUE, font = 2, line = 1, cex = 1.2,at=tmp1$x) ## draw legend #Set colors Satelite.Palette - colorRampPalette(c(blue3,cyan,aquamarine,yellow,orange,red)) mycol-Satelite.Palette(101) #Add legend BL-unlist(corner.label(x=-1,y=-1,figcorner=FALSE))#Coordinates for bottom left corner of plot TR-unlist(corner.label(x=1,y=1,figcorner=FALSE)) #Coordinates for top right corner of plot par(xpd=NA) #Plots to the device region (outer margin) library(plotrix) col.labels-seq(0,100,20) color.legend(TR[1],BL[2],TR[1]+1,TR[2],col.labels,mycol,align=rb,gradient=y) #Reset par par(op) Tim Clark Department of Zoology 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] Plotting color.legend() outside of plot region
Peter, Thanks, layout() does exactly what I want. Now I also realize why I wasn't able to get the legend where I wanted it. I didn't realize par(xpd=...) was only for the last plot. That explains why I was getting weird results when trying to move my legend around! Aloha, Tim Tim Clark Department of Zoology University of Hawaii --- On Mon, 11/30/09, Peter Ehlers ehl...@ucalgary.ca wrote: From: Peter Ehlers ehl...@ucalgary.ca Subject: Re: [R] Plotting color.legend() outside of plot region To: Tim Clark mudiver1...@yahoo.com Cc: r-help@r-project.org Date: Monday, November 30, 2009, 12:28 PM Tim Clark wrote: Dear List, I am trying to plot a color.legend() in the right outer margin of my device region. I have read multiple threads on the subject and still can't get it right. I have stolen an example from one of the threads to demonstrate my problem. I have extended the outer margin using par(oma()), and have used par(xpd=NA) to tell it to plot in the device region. I can get the legend to plot in the figure region but it will not plot in the outer margin. What am I doing wrong? Example follows with the legend in the figure region. I would like it more to the right in the device region. Thank, Tim library(TeachingDemos) op - par(mfrow = c(3,3), ## split region oma = c(0,0,4,12) + 0.1, ## create outer margin mar = c(5,4,2,2) + 0.1) ## shrink some margins plot(1:10, main = a, pch = 1:2, col= 1:2) plot(1:10, main = b, pch = 1:2, col= 1:2) tmp1 - cnvrt.coords( 0.5, 0, input='plt' )$tdev # save location for mtext plot(1:10, main = c, pch = 1:2, col= 1:2) plot(1:10, main = d, pch = 1:2, col= 1:2) plot(1:10, main = e, pch = 1:2, col= 1:2) plot(1:10, main = f, pch = 1:2, col= 1:2) plot(1:10, main = g, pch = 1:2, col= 1:2) plot(1:10, main = h, pch = 1:2, col= 1:2) plot(1:10, main = i, pch = 1:2, col= 1:2) ## title mtext(My Plots, side = 3, outer = TRUE, font = 2, line = 1, cex = 1.2,at=tmp1$x) ## draw legend #Set colors Satelite.Palette - colorRampPalette(c(blue3,cyan,aquamarine,yellow,orange,red)) mycol-Satelite.Palette(101) #Add legend BL-unlist(corner.label(x=-1,y=-1,figcorner=FALSE)) #Coordinates for bottom left corner of plot TR-unlist(corner.label(x=1,y=1,figcorner=FALSE)) #Coordinates for top right corner of plot par(xpd=NA) #Plots to the device region (outer margin) library(plotrix) col.labels-seq(0,100,20) color.legend(TR[1],BL[2],TR[1]+1,TR[2],col.labels,mycol,align=rb,gradient=y) #Reset par par(op) The par(xpd=...) only applies to the last plot. I can't think offhand how to avoid that. I would use Paul Murrell's handy layout() function. Something like this (you'll have to adjust margins, etc) m - matrix(c(1:9,10,10,10), 3, 4) nf - layout(m) layout.show(nf) plot(1:10, main = a, pch = 1:2, col= 1:2) plot(1:10, main = b, pch = 1:2, col= 1:2) plot(1:10, main = c, pch = 1:2, col= 1:2) plot(1:10, main = d, pch = 1:2, col= 1:2) plot(1:10, main = e, pch = 1:2, col= 1:2) plot(1:10, main = f, pch = 1:2, col= 1:2) plot(1:10, main = g, pch = 1:2, col= 1:2) plot(1:10, main = h, pch = 1:2, col= 1:2) plot(1:10, main = i, pch = 1:2, col= 1:2) plot(0:1, 0:1, type=n, axes=F, xlab=, ylab=) color.legend(.2,.2,.8,.8, col.labels, mycol, align=rb, gradient=y) See ?layout -Peter Ehlers Tim Clark Department of Zoology 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. __ 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] Removing objects from a list based on nrow
Dear List, I have a list containing data frames of various numbers of rows. I need to remove any data frame that has less than 3 rows. For example: df1-data.frame(letter=c(A,B,C,D,E),number=c(1,2,3,4,5)) df2-data.frame(letter=c(A,B),number=c(1,2)) df3-data.frame(letter=c(A,B,C,D,E),number=c(1,2,3,4,5)) df4-data.frame(letter=c(A,B,C,D,E),number=c(1,2,3,4,5)) lst-list(df1,df2,df3,df4) How can I determine that the second object (df2) has less than 3 rows and remove it from the list? Thanks! Tim Tim Clark Department of Zoology 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] Removing objects from a list based on nrow
Linlin, Thanks! That works great! Tim Tim Clark Department of Zoology University of Hawaii --- On Sat, 11/28/09, Linlin Yan yanlinli...@gmail.com wrote: From: Linlin Yan yanlinli...@gmail.com Subject: Re: [R] Removing objects from a list based on nrow To: Tim Clark mudiver1...@yahoo.com Cc: r-help@r-project.org Date: Saturday, November 28, 2009, 10:43 PM Try these: sapply(lst, nrow) # get row numbers which(sapply(lst, nrow) 3) # get the index of rows which has less than 3 rows lst - lst[-which(sapply(lst, nrow) 3)] # remove the rows from the list On Sun, Nov 29, 2009 at 4:36 PM, Tim Clark mudiver1...@yahoo.com wrote: Dear List, I have a list containing data frames of various numbers of rows. I need to remove any data frame that has less than 3 rows. For example: df1-data.frame(letter=c(A,B,C,D,E),number=c(1,2,3,4,5)) df2-data.frame(letter=c(A,B),number=c(1,2)) df3-data.frame(letter=c(A,B,C,D,E),number=c(1,2,3,4,5)) df4-data.frame(letter=c(A,B,C,D,E),number=c(1,2,3,4,5)) lst-list(df1,df2,df3,df4) How can I determine that the second object (df2) has less than 3 rows and remove it from the list? Thanks! Tim Tim Clark Department of Zoology 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. __ 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] Removing objects from a list based on nrow
Jim, Good catch! I know in my current problem there are objects with less than 3 rows, but will make sure to modify the function for future use. Thanks, Tim Tim Clark Department of Zoology University of Hawaii --- On Sun, 11/29/09, Linlin Yan yanlinli...@gmail.com wrote: From: Linlin Yan yanlinli...@gmail.com Subject: Re: [R] Removing objects from a list based on nrow To: jim holtman jholt...@gmail.com Cc: Tim Clark mudiver1...@yahoo.com, r-help@r-project.org Date: Sunday, November 29, 2009, 4:35 AM Thank Jim! You are right. I didn't notice the case of none of rows match the condition. On Sun, Nov 29, 2009 at 10:10 PM, jim holtman jholt...@gmail.com wrote: One thing to be careful of is if no dataframe have less than 3 rows: df1-data.frame(letter=c(A,B,C,D,E),number=c(1,2,3,4,5)) df2-data.frame(letter=c(A,B),number=c(1,2)) df3-data.frame(letter=c(A,B,C,D,E),number=c(1,2,3,4,5)) df4-data.frame(letter=c(A,B,C,D,E),number=c(1,2,3,4,5)) lst-list(df1,df3,df4) lst [[1]] letter number 1 A 1 2 B 2 3 C 3 4 D 4 5 E 5 [[2]] letter number 1 A 1 2 B 2 3 C 3 4 D 4 5 E 5 [[3]] letter number 1 A 1 2 B 2 3 C 3 4 D 4 5 E 5 lst[-which(sapply(lst, nrow) 3)] list() Notice the list is now empty. Instead use: lst[sapply(lst, nrow) =3] [[1]] letter number 1 A 1 2 B 2 3 C 3 4 D 4 5 E 5 [[2]] letter number 1 A 1 2 B 2 3 C 3 4 D 4 5 E 5 [[3]] letter number 1 A 1 2 B 2 3 C 3 4 D 4 5 E 5 On Sun, Nov 29, 2009 at 3:43 AM, Linlin Yan yanlinli...@gmail.com wrote: Try these: sapply(lst, nrow) # get row numbers which(sapply(lst, nrow) 3) # get the index of rows which has less than 3 rows lst - lst[-which(sapply(lst, nrow) 3)] # remove the rows from the list On Sun, Nov 29, 2009 at 4:36 PM, Tim Clark mudiver1...@yahoo.com wrote: Dear List, I have a list containing data frames of various numbers of rows. I need to remove any data frame that has less than 3 rows. For example: df1-data.frame(letter=c(A,B,C,D,E),number=c(1,2,3,4,5)) df2-data.frame(letter=c(A,B),number=c(1,2)) df3-data.frame(letter=c(A,B,C,D,E),number=c(1,2,3,4,5)) df4-data.frame(letter=c(A,B,C,D,E),number=c(1,2,3,4,5)) lst-list(df1,df2,df3,df4) How can I determine that the second object (df2) has less than 3 rows and remove it from the list? Thanks! Tim Tim Clark Department of Zoology 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. __ 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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ 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] Continuous legend colors
Dear List, I am trying to get a basic plot to show a continuous range of fill colors. It is probably easiest to demonstrate. I would like a legend like in the following example: Satellite.Palette -colorRampPalette(c(blue3,cyan,aquamarine,yellow,orange,red)) require(fields) image.plot(volcano, col = Satellite.Palette (500), legend.lab=Scale) contour(volcano, levels = seq(90, 200, by = 5), add = TRUE) However, I am using the basic plot function. So far I have figured out how to remove any space between the colors using the y.intersp call in legend(). Now I need to somehow plot the legend so that it 1) fits on the plotting region, 2) has fewer labels, and 3) doesn't have black lines between each color. The example I am trying to get to work is: Sat.Pal-Satellite.Palette (101) x-rnorm(100, mean = 50, sd = 50) y-rnorm(100, mean = 50, sd = 50) z-seq(1,100, by=1) plot(x,y,pch=16,col=Sat.Pal[z]) legend(topleft, legend=seq(0,100, by=1), fill=Sat.Pal[seq(1,101, by=1)], bty=n, y.intersp=.5) I would appreciate any help or suggestions on how to get this to produce a legend with continuous colors. Thanks, Tim Tim Clark Department of Zoology 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] Continuous legend colors
Thanks Jim, that is exactly what I was looking for! Tim Tim Clark Department of Zoology University of Hawaii --- On Sun, 11/29/09, Jim Lemon j...@bitwrit.com.au wrote: From: Jim Lemon j...@bitwrit.com.au Subject: Re: [R] Continuous legend colors To: Tim Clark mudiver1...@yahoo.com Cc: r-help@r-project.org Date: Sunday, November 29, 2009, 8:13 PM On 11/30/2009 03:22 PM, Tim Clark wrote: Dear List, I am trying to get a basic plot to show a continuous range of fill colors. It is probably easiest to demonstrate. I would like a legend like in the following example: Satellite.Palette-colorRampPalette(c(blue3,cyan,aquamarine,yellow,orange,red)) require(fields) image.plot(volcano, col = Satellite.Palette (500), legend.lab=Scale) contour(volcano, levels = seq(90, 200, by = 5), add = TRUE) However, I am using the basic plot function. So far I have figured out how to remove any space between the colors using the y.intersp call in legend(). Now I need to somehow plot the legend so that it 1) fits on the plotting region, 2) has fewer labels, and 3) doesn't have black lines between each color. The example I am trying to get to work is: Sat.Pal-Satellite.Palette (101) x-rnorm(100, mean = 50, sd = 50) y-rnorm(100, mean = 50, sd = 50) z-seq(1,100, by=1) plot(x,y,pch=16,col=Sat.Pal[z]) legend(topleft, legend=seq(0,100, by=1), fill=Sat.Pal[seq(1,101, by=1)], bty=n, y.intersp=.5) I would appreciate any help or suggestions on how to get this to produce a legend with continuous colors. Hi Tim, Have a look at the color.legend function in the plotrix package. Jim __ 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] Adding columns to lower level of list
Dear List,
I have very little experience with lists and am having some very basic
problems. I don't know how to add columns to the lower levels of a list, or
how to take something from the upper level and add it as a column to the lower
level. I am analyzing animal movement data in the package Adehabitat. I have
a list of animal movements called cut.ltr (class ltraj) that have been
divided into a series of burst - i.e. movements with no gaps in time over a
given threashold. I would like to
1. Add the speed to each item in the list, and also the burst. I can
calculate speed as:
sp-lapply(cut.ltr,function(l){l$dist/l$dt})
This creates a list of the correct size. But I don't know how to add this to
my original list. I.e. add a column to the lower levels of the list called
speed.
2. Add the burst to each lower level of the list. It is in the upper level,
but I don't know how to access it.
I have tried attribute(), attr(), cut.ltr$burst, and several other creative
guesses.
The first five items in the upper level are below - cut.ltr[1:5], along with
head(cut.ltr[[1]]). I would like my final result to have two more columns in
cut.ltr[[1]]. One with speed, and the second with burst.
Thanks in advance for your help.
Tim
cut.ltr[1:5]
*** List of class ltraj ***
Type of the traject: Type II (time recorded)
Irregular traject. Variable time lag between two locs
Characteristics of the bursts:
id burst nb.reloc NAs date.begindate.end
1 Abigail Abigail.1 47 0 2003-05-31 13:29:59 2003-06-01 00:59:56
2 Abigail Abigail.2 288 0 2003-06-18 17:28:11 2003-06-21 17:14:59
3 Abigail Abigail.3 10 0 2003-08-03 23:33:00 2003-08-04 01:43:58
4 Abigail Abigail.4 43 0 2003-08-04 08:15:25 2003-08-04 18:59:58
5 Abigail Abigail.5 78 0 2003-08-05 00:44:19 2003-08-05 20:15:00
head(cut.ltr[[1]])
x ydate dx dy dist dt
R2n abs.angle rel.angle
1 809189.8 2189722 2003-05-31 13:29:59 81.87136 315.3389 325.7937 901
0.0 1.316775 NA
2 809271.6 2190037 2003-05-31 13:45:00 13.00097 258.7351 259.0616 901
106141.5 1.520590 0.20381526
3 809284.6 2190296 2003-05-31 14:00:01 250.52656 669.2065 714.5634 898
338561.8 1.212584 -0.30800666
4 809535.2 2190965 2003-05-31 14:14:59 -171.14372 791.1522 809.4516 902
1665046.9 1.783836 0.57125215
5 809364.0 2191756 2003-05-31 14:30:01 302.26979 707.0157 768.9202 900
4169281.4 1.166785 -0.61705039
6 809666.3 2192463 2003-05-31 14:45:01 284.40962 725.2169 778.9919 900
7742615.6 1.197057 0.03027109
Tim Clark
Department of Zoology
University of Hawaii
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Bezier interpolation
Dear List, I am trying to interpolate animal tracking data using Bezier curves. I need a function similar to spline() or approx() but that has a method Bezier. I have tried xspline() but it does not allow you to set the number of points to interpolate between a given interval (n points between min(x) and max(x)). Mark Hindell asked the same question in 2006 (http://tolstoy.newcastle.edu.au/R/e2/help/06/12/7034.html). I contacted him and he never found a workable function. Has one been developed since then? Thanks, Tim Tim Clark Department of Zoology 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.
[R] Satellite ocean color palette?
Dear List, Is there a color palette avaliable similar to what is used in satellite ocean color imagery? I.e. a gradient with blue on one end and red on the other, with yellow in the middle? I have tried topo.colors(n) but that comes out more yellow on the end. I am looking for something similar to what is found on the CoastWatch web page: http://oceanwatch.pifsc.noaa.gov/imagery/GA2009281_2009282_sst_2D_eddy.jpg Thanks! Tim Tim Clark Department of Zoology 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] Satellite ocean color palette?
Thanks! The colorRampPalette() did just what I need. Tim Tim Clark Department of Zoology University of Hawaii --- On Fri, 10/9/09, Barry Rowlingson b.rowling...@lancaster.ac.uk wrote: From: Barry Rowlingson b.rowling...@lancaster.ac.uk Subject: Re: [R] Satellite ocean color palette? To: Tim Clark mudiver1...@yahoo.com Cc: r-help@r-project.org Date: Friday, October 9, 2009, 9:06 AM On Fri, Oct 9, 2009 at 7:51 PM, Tim Clark mudiver1...@yahoo.com wrote: Dear List, Is there a color palette avaliable similar to what is used in satellite ocean color imagery? I.e. a gradient with blue on one end and red on the other, with yellow in the middle? I have tried topo.colors(n) but that comes out more yellow on the end. I am looking for something similar to what is found on the CoastWatch web page: http://oceanwatch.pifsc.noaa.gov/imagery/GA2009281_2009282_sst_2D_eddy.jpg Thanks! You could build one yourself with the colorRamp function: satRampP = colorRampPalette(c(black,blue,cyan,yellow,orange,red,black)) that looks roughly like the one in the jpg, but I'm not sure about the black at the far end...anyway, let's see: image(matrix(seq(0,1,len=100),100,1),col=satRampP(100)) Or you could try my colour schemes package: https://r-forge.r-project.org/projects/colourscheme/ Barry __ 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] Paste a character to an object
David,
Thanks! You just gave me the answer. All I had to do was:
xx-c()
for (i in c('100', '75', '50') )
{
x-homerange[[1]]$polygons[[i]] ; xx-rbind(x,xx)
}
xx
I didn't know you could use characters as index values in a for loop, or that
you could use characters in double brackets instead of using the $ symbol.
homerange[[1]]$polygons[['100']]
is the same as
homerange[[1]]$polygons$'100
The list is actually the output of the NNCH function in Adehabitat. I thought
about changing the function first, but looked at the code and couldn't figure
it out. I knew there had to be an easier way.
I greatly appreciate all your help,
Tim
Tim Clark
Department of Zoology
University of Hawaii
--- On Sat, 10/3/09, David Winsemius dwinsem...@comcast.net wrote:
From: David Winsemius dwinsem...@comcast.net
Subject: Re: [R] Paste a character to an object
To: Tim Clark mudiver1...@yahoo.com
Cc: r-help@r-project.org
Date: Saturday, October 3, 2009, 5:43 PM
On Oct 3, 2009, at 11:14 PM, Tim Clark wrote:
David,
Thanks, that helps me in making an example of what I
am trying to do. Given the following example, I would
like to run through a for loop and obtain a vector of the
data only for the 100, 75, and 50 percent values. Is
there a way to get this to work, either using paste as in
the example below or some other method?
homerange - list()
homerange[[1]] - test
homerange[[1]]$polygons - test2
homerange[[1]]$polygons$`100` - rnorm(20,10,1)
homerange[[1]]$polygons$`90` - rnorm(20,10,1)
homerange[[1]]$polygons$`75` - rnorm(20,10,1)
homerange[[1]]$polygons$`50` - rnorm(20,10,1)
xx-c()
percent-c(100,75,50)
for (i in 1:length(percent))
{
x-paste(homerange[[1]]$polygons$
, percent[i]) #This does not work!!!
^?^
And why _would_ you expect an expression ending in a $ to
be acceptable to the parser? You did not put quotes around
it so the interpreter tried to evaluate it.
You are probably looking for the capabilities of the
functions get and assign which take string variable and
either get the object named by a sstring or assign a vlaue
to an object so named.
But why are you intent in causing yourself all this
pain? (Not to mention asking questions I cannot
answer.) Working with expressions involving backquotes
is a recipe for hair-pulling and frustration for us normal
mortals. Why not call your lists p100, p90, p75,
p50? Then everything is simple:
xx-c()
percent-c(100, 75, 50)
for (i in c(p100, p75, p50) )
+ {
+ x-homerange[[1]]$polygons[[i]] ;
xx-rbind(x,xx) # could have simplified this
+ }
xx
[,1]
[,2] [,3]
[,4] [,5]
[,6] [,7]
[,8] [,9]
x 9.660935 10.46526 10.75813 8.866064
9.967950 9.987941 10.757160 10.180826 9.992162
x 11.674645 10.51753 10.88061 10.515120 9.440838 11.460845
12.033612 9.318392 9.592026
x 10.057021 10.14339 10.29757 9.164233 8.977280
9.733971 9.965002 9.693649 9.430043
[,10]
[,11] [,12]
[,13] [,14]
[,15] [,16]
[,17] [,18]
x 11.78904 9.437353 11.910747 10.996167
11.631264 9.386944 9.602160 10.498921
9.09349
x 9.11036 9.546378 11.030323
9.715164 9.500268 11.762440 9.101104
9.610251 10.56210
x 9.62574 12.738020 9.146863 10.497626
10.485520 11.644503 10.303581 11.340263 11.34873
[,19] [,20]
x 10.146955 9.640136
x 9.334912 10.101603
x 8.710609 11.265633
The x-paste(...) in this function does not work,
and that is what I am stuck on. The result should be a
vector the values for the 100,75,and 50 levels, but
not the 90 level.
Aloha,
Tim Clark
Department of Zoology
University of Hawaii
--- On Sat, 10/3/09, David Winsemius dwinsem...@comcast.net
wrote:
From: David Winsemius dwinsem...@comcast.net
Subject: Re: [R] Paste a character to an object
To: Tim Clark mudiver1...@yahoo.com
Cc: r-help@r-project.org
Date: Saturday, October 3, 2009, 4:45 PM
On Oct 3, 2009, at 10:26 PM, Tim Clark wrote:
Dear List,
I can't seem to get a simple paste function to
work
like I need. I have an object I need to call
but it
ends in a character string. The object is a
list of
home range values for a range of percent
isopleths. I
need to loop through a vector of percent values,
so I need
to paste the percent as a character on the end of
the object
variable. I have no idea why the percent is
in
character form, and I can't use a simple index
value
(homerange[[1]]$polygons[100]) because there are a
variable
number of isopleths that are calculated and [100]
will not
always correspond to 100. So I am stuck.
What I want is:
homerange[[1]]$polygons$100
What I need is something like the following,
but that
works:
percent-c(100,75,50)
p=1
paste(homerange[[1]]$polygons$,percent[p],sep=)
Not a reproducible example, but here is some code
[R] Paste a character to an object
Dear List, I can't seem to get a simple paste function to work like I need. I have an object I need to call but it ends in a character string. The object is a list of home range values for a range of percent isopleths. I need to loop through a vector of percent values, so I need to paste the percent as a character on the end of the object variable. I have no idea why the percent is in character form, and I can't use a simple index value (homerange[[1]]$polygons[100]) because there are a variable number of isopleths that are calculated and [100] will not always correspond to 100. So I am stuck. What I want is: homerange[[1]]$polygons$100 What I need is something like the following, but that works: percent-c(100,75,50) p=1 paste(homerange[[1]]$polygons$,percent[p],sep=) Thanks for the help, Tim Tim Clark Department of Zoology University of Hawaii __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Paste a character to an object
David,
Thanks, that helps me in making an example of what I am trying to do. Given
the following example, I would like to run through a for loop and obtain a
vector of the data only for the 100, 75, and 50 percent values. Is there a way
to get this to work, either using paste as in the example below or some other
method?
homerange - list()
homerange[[1]] - test
homerange[[1]]$polygons - test2
homerange[[1]]$polygons$`100` - rnorm(20,10,1)
homerange[[1]]$polygons$`90` - rnorm(20,10,1)
homerange[[1]]$polygons$`75` - rnorm(20,10,1)
homerange[[1]]$polygons$`50` - rnorm(20,10,1)
xx-c()
percent-c(100,75,50)
for (i in 1:length(percent))
{
x-paste(homerange[[1]]$polygons$,percent[i]) #This does not work!!!
xx-rbind(x,xx)
}
The x-paste(...) in this function does not work, and that is what I am stuck
on. The result should be a vector the values for the 100,75,and 50
levels, but not the 90 level.
Aloha,
Tim
Tim Clark
Department of Zoology
University of Hawaii
--- On Sat, 10/3/09, David Winsemius dwinsem...@comcast.net wrote:
From: David Winsemius dwinsem...@comcast.net
Subject: Re: [R] Paste a character to an object
To: Tim Clark mudiver1...@yahoo.com
Cc: r-help@r-project.org
Date: Saturday, October 3, 2009, 4:45 PM
On Oct 3, 2009, at 10:26 PM, Tim Clark wrote:
Dear List,
I can't seem to get a simple paste function to work
like I need. I have an object I need to call but it
ends in a character string. The object is a list of
home range values for a range of percent isopleths. I
need to loop through a vector of percent values, so I need
to paste the percent as a character on the end of the object
variable. I have no idea why the percent is in
character form, and I can't use a simple index value
(homerange[[1]]$polygons[100]) because there are a variable
number of isopleths that are calculated and [100] will not
always correspond to 100. So I am stuck.
What I want is:
homerange[[1]]$polygons$100
What I need is something like the following, but that
works:
percent-c(100,75,50)
p=1
paste(homerange[[1]]$polygons$,percent[p],sep=)
Not a reproducible example, but here is some code that
shows that it is possible to construct names that would
otherwise be invalid due to having numerals as a first
character by using back-quotes:
percent-c(100,75,50)
p=1
paste(homerange[[1]]$polygons$,percent[p],sep=)
Error: syntax error
homerange - list()
homerange[[1]] - test
homerange[[1]]$polygons - test2
Warning message:
In homerange[[1]]$polygons - test2 : Coercing LHS to
a list
homerange
[[1]]
[[1]][[1]]
[1] test
[[1]]$polygons
[1] test2
homerange[[1]]$polygons$`100` - percent[1]
Warning message:
In homerange[[1]]$polygons$`100` - percent[1] :
Coercing LHS to a list
homerange[[1]]$polygons$`100`
[1] 100
--David Winsemius
Thanks for the help,
Tim
Tim Clark
Department of Zoology
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.
__
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] bwplot scales in alphabetical order
Dear List, I know this has been covered before, but I don't seem to be able to get it right. I am constructing a boxplot in lattice and can't get the scales in the correct alphebetical order. I have already read that this is due to the way factors are treated, and I have to redefine the levels of the factors. However, I have failed. As a simple example: library(lattice) id-rep(letters[1:9], each=20) x-rep(seq(1:10),each=18) y-rnorm(180,50,20) #Reverse alphebetical order bwplot(y~x|id, horizontal=FALSE) #alphebetical order reading right to left id-factor(id,levels = sort(id,decreasing = TRUE)) bwplot(y~x|id, horizontal=FALSE) It appears that bwplot plots scales from the bottom left to the top right. If so my factor levels would need to be levels=c(7,8,9,4,5,6,1,2,3). I tried that but can't seem to get the factor function to work. #Did not work! id-factor(id,levels=c(7,8,9,4,5,6,1,2,3),lables=letters[1:9]) Your help would be greatly appreciated. Tim Tim Clark Department of Zoology 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] bwplot scales in alphabetical order
Peter, Thanks, that did it! Tim Tim Clark Department of Zoology University of Hawaii --- On Wed, 9/30/09, Peter Ehlers ehl...@ucalgary.ca wrote: From: Peter Ehlers ehl...@ucalgary.ca Subject: Re: [R] bwplot scales in alphabetical order To: Tim Clark mudiver1...@yahoo.com Cc: r-help@r-project.org Date: Wednesday, September 30, 2009, 2:43 AM Tim, Add the argument as.table=TRUE to your call: bwplot(y~x|id, horizontal=FALSE, as.table=TRUE) Peter Ehlers Tim Clark wrote: Dear List, I know this has been covered before, but I don't seem to be able to get it right. I am constructing a boxplot in lattice and can't get the scales in the correct alphebetical order. I have already read that this is due to the way factors are treated, and I have to redefine the levels of the factors. However, I have failed. As a simple example: library(lattice) id-rep(letters[1:9], each=20) x-rep(seq(1:10),each=18) y-rnorm(180,50,20) #Reverse alphebetical order bwplot(y~x|id, horizontal=FALSE) #alphebetical order reading right to left id-factor(id,levels = sort(id,decreasing = TRUE)) bwplot(y~x|id, horizontal=FALSE) It appears that bwplot plots scales from the bottom left to the top right. If so my factor levels would need to be levels=c(7,8,9,4,5,6,1,2,3). I tried that but can't seem to get the factor function to work. #Did not work! id-factor(id,levels=c(7,8,9,4,5,6,1,2,3),lables=letters[1:9]) Your help would be greatly appreciated. Tim Tim Clark Department of Zoology 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. __ 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] xyplot help - colors and break in plot
Felix, Thanks, that did the trick! Lattice is a lot less intuitive than basic plotting! Also, another person suggested using gap.plot from the plotrix package to put a break in the graph. I am surprised Lattice doesn't have something similar since it seems like a common problem when you have data that groups in clusters separated by a large range. Aloha, Tim Tim Clark Department of Zoology University of Hawaii --- On Mon, 9/28/09, Felix Andrews fe...@nfrac.org wrote: From: Felix Andrews fe...@nfrac.org Subject: Re: [R] xyplot help - colors and break in plot To: Tim Clark mudiver1...@yahoo.com Cc: r-help@r-project.org Date: Monday, September 28, 2009, 1:50 PM 2009/9/29 Tim Clark mudiver1...@yahoo.com: Dear List, I am new to lattice plots, and am having problems with getting my plot to do what I want. Specifically: 1. I would like the legend to have the same symbols as the plot. I tried simpleKey but can't seem to get it to work with autoKey. Right now my plot has dots (pch=19) and my legend shows circles. Rather than the pch = 19 argument, use par.settings = simpleTheme(pch = 19, cex = .4) 2. I have nine groups but xyplot seems to only be using seven colors, so two groups have the same color. How do I get a range of nine colors? Yes, in the default theme, there are seven colours: see trellis.par.get(superpose.symbol) You can change the set of colours yourself by modifying that list (via trellis.par.set). An easier option is to use one of the predefined ColorBrewer palettes, with custom.theme() from the latticeExtra package, or just simpleTheme(). See ?brewer.pal (RColorBrewer package) You will see there are a few qualitative color palettes with 9 or more colours: e.g. brewer.pal(9, Set1) brewer.pal(12, Set3) 3. I have one group who's y range is much greater than all the others. I would like to split the plot somehow so that the bottom part shows ylim=c(0,200) and the top shows ylim=c(450,550). Is this possible? Yes... in the absence of a reproducible example, maybe something like xyplot(Area.km2 ~ DataPoint | (Area.km2 200), m.dp.area, groups = Manta, scales = list(y = free)) or AreaRange - shingle(Area.km2, rbind(c(0,200),c(450,550))) xyplot(Area.km2 ~ DataPoint | AreaRange, m.dp.area, groups = Manta, scales = list(y = free)) What I have so far is: library(lattice) xyplot(m.dp.area$Area.km2 ~ m.dp.area$DataPoint, m.dp.area, groups = m.dp.area$Manta, main = Cummulative area of 100% MCP, xlab = Data Point, ylab = MCP Area, ylim = c(0,150), scales = list(tck = c(1, 0)), #Removes tics on top and r-axis pch=19,cex=.4, auto.key = list(title = Mantas, x = .05, y=.95, corner = c(0,1),border = TRUE)) #Legend Thanks, Tim Tim Clark Department of Zoology University of Hawaii __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Felix Andrews / 安福立 Postdoctoral Fellow Integrated Catchment Assessment and Management (iCAM) Centre Fenner School of Environment and Society [Bldg 48a] The Australian National University Canberra ACT 0200 Australia M: +61 410 400 963 T: + 61 2 6125 1670 E: felix.andr...@anu.edu.au CRICOS Provider No. 00120C -- http://www.neurofractal.org/felix/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] xyplot help - colors and break in plot
Dear List, I am new to lattice plots, and am having problems with getting my plot to do what I want. Specifically: 1. I would like the legend to have the same symbols as the plot. I tried simpleKey but can't seem to get it to work with autoKey. Right now my plot has dots (pch=19) and my legend shows circles. 2. I have nine groups but xyplot seems to only be using seven colors, so two groups have the same color. How do I get a range of nine colors? 3. I have one group who's y range is much greater than all the others. I would like to split the plot somehow so that the bottom part shows ylim=c(0,200) and the top shows ylim=c(450,550). Is this possible? What I have so far is: library(lattice) xyplot(m.dp.area$Area.km2 ~ m.dp.area$DataPoint, m.dp.area, groups = m.dp.area$Manta, main = Cummulative area of 100% MCP, xlab = Data Point, ylab = MCP Area, ylim = c(0,150), scales = list(tck = c(1, 0)), #Removes tics on top and r-axis pch=19,cex=.4, auto.key = list(title = Mantas, x = .05, y=.95, corner = c(0,1),border = TRUE)) #Legend Thanks, Tim Tim Clark Department of Zoology 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] Data formatting for matplot
Henrique, Thanks for the suggestion. I think I may not understand matplot() because the graph did not come out like it should have. Gabor suggested: library(lattice) xyplot(y ~ x, mydat, groups = id) Which gave what I was looking for. Is there a way to get matplot() to give the same graph? I don't have to use matplot(), but would like to understand its use. Thanks, Tim Tim Clark Department of Zoology University of Hawaii --- On Sun, 9/27/09, Henrique Dallazuanna www...@gmail.com wrote: From: Henrique Dallazuanna www...@gmail.com Subject: Re: [R] Data formatting for matplot To: Tim Clark mudiver1...@yahoo.com Cc: r-help@r-project.org Date: Sunday, September 27, 2009, 4:47 PM You can try this: matplot(do.call(cbind, split.dat)) On Sun, Sep 27, 2009 at 11:42 PM, Tim Clark mudiver1...@yahoo.com wrote: Dear List, I am wanting to produce a multiple line plot, and know I can do it with matplot but can't get my data in the format I need. I have a dataframe with three columns; individuals ID, x, and y. I have tried split() but it gives me a list of matrices, which is closer but not quite what I need. For example: id-rep(seq(1,5,1),length.out=100) x-rnorm(100,5,1) y-rnorm(100,20,5) mydat-data.frame(id,x,y) split.dat-split(mydat[,2:3],mydat[,1]) I would appreciate your help in either how to get this into a format acceptable to matplot or other options for creating a multiple line plot. Thanks, Tim Tim Clark Department of Zoology 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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ 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] Data formatting for matplot
Thanks for everyones help. It is great to have a number of options that result in the same graph. Aloha, Tim Tim Clark Department of Zoology University of Hawaii --- On Mon, 9/28/09, Henrique Dallazuanna www...@gmail.com wrote: From: Henrique Dallazuanna www...@gmail.com Subject: Re: [R] Data formatting for matplot To: Tim Clark mudiver1...@yahoo.com Cc: r-help@r-project.org Date: Monday, September 28, 2009, 1:43 AM Tim, With Gabor examples, I understand this, You can get a similar graph with plot: with(mydat, plot(x, y, col = id)) On Mon, Sep 28, 2009 at 3:01 AM, Tim Clark mudiver1...@yahoo.com wrote: Henrique, Thanks for the suggestion. I think I may not understand matplot() because the graph did not come out like it should have. Gabor suggested: library(lattice) xyplot(y ~ x, mydat, groups = id) Which gave what I was looking for. Is there a way to get matplot() to give the same graph? I don't have to use matplot(), but would like to understand its use. Thanks, Tim Tim Clark Department of Zoology University of Hawaii --- On Sun, 9/27/09, Henrique Dallazuanna www...@gmail.com wrote: From: Henrique Dallazuanna www...@gmail.com Subject: Re: [R] Data formatting for matplot To: Tim Clark mudiver1...@yahoo.com Cc: r-help@r-project.org Date: Sunday, September 27, 2009, 4:47 PM You can try this: matplot(do.call(cbind, split.dat)) On Sun, Sep 27, 2009 at 11:42 PM, Tim Clark mudiver1...@yahoo.com wrote: Dear List, I am wanting to produce a multiple line plot, and know I can do it with matplot but can't get my data in the format I need. I have a dataframe with three columns; individuals ID, x, and y. I have tried split() but it gives me a list of matrices, which is closer but not quite what I need. For example: id-rep(seq(1,5,1),length.out=100) x-rnorm(100,5,1) y-rnorm(100,20,5) mydat-data.frame(id,x,y) split.dat-split(mydat[,2:3],mydat[,1]) I would appreciate your help in either how to get this into a format acceptable to matplot or other options for creating a multiple line plot. Thanks, Tim Tim Clark Department of Zoology 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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ 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] Data formatting for matplot
Dear List, I am wanting to produce a multiple line plot, and know I can do it with matplot but can't get my data in the format I need. I have a dataframe with three columns; individuals ID, x, and y. I have tried split() but it gives me a list of matrices, which is closer but not quite what I need. For example: id-rep(seq(1,5,1),length.out=100) x-rnorm(100,5,1) y-rnorm(100,20,5) mydat-data.frame(id,x,y) split.dat-split(mydat[,2:3],mydat[,1]) I would appreciate your help in either how to get this into a format acceptable to matplot or other options for creating a multiple line plot. Thanks, Tim Tim Clark Department of Zoology 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.
[R] Trying to rename spatial pts data frame slot that isn't a slot()
Dear List, I am analyzing the home range area of fish and seem to have lost the individuals ID names during my manipulations, and can't find out how to rename them. I calculated the MCP of the fish using mcp() in Adehabitat. MCP's were converted to spatial points data frame and exported to qGIS for manipulations. At this point the ID names were lost. I brought the manipulated shapefiles back into qGIS, but can't figure out how to rename the individuals. #Calculate MCP and save as a shapefile my.mcp-mcp(xy, id=id, percent=100) spol-area2spol(my.mcp) spdf - SpatialPolygonsDataFrame(spol, data=data.frame +(getSpPPolygonsLabptSlots(spol), +row.names=getSpPPolygonsIDSlots(spol)), match.ID = TRUE) writeOGR(spdf,dsn=mcp.dir,layer=All Mantas MCP, driver=ESRI +Shapefile) #Read shapefile manipulated in qGIS mymcp-readOGR(dsn=mcp.dir,layer=All mantas MCP land differenc) My spatial points data frame has a number of Slots, including one that contained the original names called Slot ID. However, I can not access this slot using slot() or slotNames(). slotNames(mymcp) [1] datapolygonsplotOrder bbox proj4string What am I missing here? Is Slot ID not a slot? Can I export the ID's with the shapefiles to qGIS? Can I rename the ID's when I bring them back into R? When is a slot not a slot()? Thanks, TIm Tim Clark Department of Zoology 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] Finding minimum of time subset
Jim, That works great! However, would you please explain what the '[' and the 1 do in the sapply function? I understand that you are cutting x by quarter, then creating a list of x that is split based on those cuts. I just don't understand what [ means in this contex, or what the number one at the end does. Thanks for you help, Tim Tim Clark Department of Zoology University of Hawaii --- On Fri, 8/14/09, jim holtman jholt...@gmail.com wrote: From: jim holtman jholt...@gmail.com Subject: Re: [R] Finding minimum of time subset To: Tim Clark mudiver1...@yahoo.com Cc: r-help@r-project.org Date: Friday, August 14, 2009, 6:18 AM Here is one way to do it: mytime-c(12:00:00,12:00:05,12:15:05,12:15:06,12:20:00,12:30:01,12:45:01,13:00:00,13:15:02) # you might want a date on your data x - as.POSIXct(mytime, format=%H:%M:%S) # create quarter hour intervals for the data range quarter - seq(trunc(min(x), 'days'), trunc(max(x) + 86400, 'days'), by='15 min') # add 86400 to add a day for truncation # cut the data by quarter hours and then take the first value in each group x.s - sapply(split(x, cut(x, breaks=quarter), drop=TRUE), '[', 1) # lost the 'class' for some reason; put it back class(x.s) - c(POSIXt, POSIXct) # the answer x.s 2009-08-14 12:00:00 2009-08-14 12:15:00 2009-08-14 12:30:00 2009-08-14 12:45:00 2009-08-14 13:00:00 2009-08-14 12:00:00 EDT 2009-08-14 12:15:05 EDT 2009-08-14 12:30:01 EDT 2009-08-14 12:45:01 EDT 2009-08-14 13:00:00 EDT 2009-08-14 13:15:00 2009-08-14 13:15:02 EDT On Thu, Aug 13, 2009 at 4:10 PM, Tim Clarkmudiver1...@yahoo.com wrote: Dear List, I have a data frame of data taken every few seconds. I would like to subset the data to retain only the data taken on the quarter hour, and as close to the quarter hour as possible. So far I have figured out how to subset the data to the quarter hour, but not how to keep only the minimum time for each quarter hour. For example: mytime-c(12:00:00,12:00:05,12:15:05,12:15:06,12:20:00,12:30:01,12:45:01,13:00:00,13:15:02) subtime-grep(pattern=[[:digit:]]+[[:punct:]]00[[:punct:]][[:digit:]]+|[[:digit:]]+[[:punct:]]15[[:punct:]][[:digit:]]+|[[:digit:]]+[[:punct:]]30[[:punct:]][[:digit:]]+|[[:digit:]]+[[:punct:]]45[[:punct:]][[:digit:]]+,mytime) mytime[subtime] [1] 12:00:00 12:00:05 12:15:05 12:15:06 12:30:01 12:45:01 13:00:00 13:15:02 This gives me the data taken at quarter hour intervals (removes 12:20:00) but I am still left with multiple values at the quarter hours. I would like to obtain: 12:00:00 12:15:05 12:30:01 12:45:01 13:00:00 13:15:02 Thanks! Tim Tim Clark Department of Zoology 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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ 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] Finding minimum of time subset
Jim, Got it! Thanks for the explanation and the example. Always nice to learn new tricks on R. Aloha, Tim Tim Clark Department of Zoology University of Hawaii --- On Fri, 8/14/09, jim holtman jholt...@gmail.com wrote: From: jim holtman jholt...@gmail.com Subject: Re: [R] Finding minimum of time subset To: Tim Clark mudiver1...@yahoo.com Cc: r-help@r-project.org Date: Friday, August 14, 2009, 7:51 AM sapply(mylist, '[', 1) is equivalent to sapply(mylist, function(x) x[1]) # select just the first element [ is an function that is called with a object and an index. Using it the way I did in the email was a shorthand way of doing it. Here is an example: x - list(1,2,3) x[1] [[1]] [1] 1 `[`(x, 1) [[1]] [1] 1 Notice the function call `[`(x,1). This is what is being done in the sapply and passing the 1 as the second parameter. On Fri, Aug 14, 2009 at 1:30 PM, Tim Clarkmudiver1...@yahoo.com wrote: Jim, That works great! However, would you please explain what the '[' and the 1 do in the sapply function? I understand that you are cutting x by quarter, then creating a list of x that is split based on those cuts. I just don't understand what [ means in this contex, or what the number one at the end does. Thanks for you help, Tim Tim Clark Department of Zoology University of Hawaii --- On Fri, 8/14/09, jim holtman jholt...@gmail.com wrote: From: jim holtman jholt...@gmail.com Subject: Re: [R] Finding minimum of time subset To: Tim Clark mudiver1...@yahoo.com Cc: r-help@r-project.org Date: Friday, August 14, 2009, 6:18 AM Here is one way to do it: mytime-c(12:00:00,12:00:05,12:15:05,12:15:06,12:20:00,12:30:01,12:45:01,13:00:00,13:15:02) # you might want a date on your data x - as.POSIXct(mytime, format=%H:%M:%S) # create quarter hour intervals for the data range quarter - seq(trunc(min(x), 'days'), trunc(max(x) + 86400, 'days'), by='15 min') # add 86400 to add a day for truncation # cut the data by quarter hours and then take the first value in each group x.s - sapply(split(x, cut(x, breaks=quarter), drop=TRUE), '[', 1) # lost the 'class' for some reason; put it back class(x.s) - c(POSIXt, POSIXct) # the answer x.s 2009-08-14 12:00:00 2009-08-14 12:15:00 2009-08-14 12:30:00 2009-08-14 12:45:00 2009-08-14 13:00:00 2009-08-14 12:00:00 EDT 2009-08-14 12:15:05 EDT 2009-08-14 12:30:01 EDT 2009-08-14 12:45:01 EDT 2009-08-14 13:00:00 EDT 2009-08-14 13:15:00 2009-08-14 13:15:02 EDT On Thu, Aug 13, 2009 at 4:10 PM, Tim Clarkmudiver1...@yahoo.com wrote: Dear List, I have a data frame of data taken every few seconds. I would like to subset the data to retain only the data taken on the quarter hour, and as close to the quarter hour as possible. So far I have figured out how to subset the data to the quarter hour, but not how to keep only the minimum time for each quarter hour. For example: mytime-c(12:00:00,12:00:05,12:15:05,12:15:06,12:20:00,12:30:01,12:45:01,13:00:00,13:15:02) subtime-grep(pattern=[[:digit:]]+[[:punct:]]00[[:punct:]][[:digit:]]+|[[:digit:]]+[[:punct:]]15[[:punct:]][[:digit:]]+|[[:digit:]]+[[:punct:]]30[[:punct:]][[:digit:]]+|[[:digit:]]+[[:punct:]]45[[:punct:]][[:digit:]]+,mytime) mytime[subtime] [1] 12:00:00 12:00:05 12:15:05 12:15:06 12:30:01 12:45:01 13:00:00 13:15:02 This gives me the data taken at quarter hour intervals (removes 12:20:00) but I am still left with multiple values at the quarter hours. I would like to obtain: 12:00:00 12:15:05 12:30:01 12:45:01 13:00:00 13:15:02 Thanks! Tim Tim Clark Department of Zoology 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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ 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] Finding minimum of time subset
Thanks for everyones help and for the alternate ways of doing this. I am always amazed at how many solutions this list comes up with for things I get stuck on! It really helps us non-programmers learn R! Aloha, Tim Tim Clark Department of Zoology University of Hawaii --- On Fri, 8/14/09, Henrique Dallazuanna www...@gmail.com wrote: From: Henrique Dallazuanna www...@gmail.com Subject: Re: [R] Finding minimum of time subset To: Tim Clark mudiver1...@yahoo.com Cc: r-help@r-project.org Date: Friday, August 14, 2009, 7:19 AM Try this also: times - as.POSIXlt(mytime, format = %H:%M:%S) subTimes - times[times[['min']] %in% c(0,15,30,45)] format(subTimes[!duplicated(format(subTimes, %H:%M))], %H:%M:%S) On Thu, Aug 13, 2009 at 5:10 PM, Tim Clark mudiver1...@yahoo.com wrote: Dear List, I have a data frame of data taken every few seconds. I would like to subset the data to retain only the data taken on the quarter hour, and as close to the quarter hour as possible. So far I have figured out how to subset the data to the quarter hour, but not how to keep only the minimum time for each quarter hour. For example: mytime-c(12:00:00,12:00:05,12:15:05,12:15:06,12:20:00,12:30:01,12:45:01,13:00:00,13:15:02) subtime-grep(pattern=[[:digit:]]+[[:punct:]]00[[:punct:]][[:digit:]]+|[[:digit:]]+[[:punct:]]15[[:punct:]][[:digit:]]+|[[:digit:]]+[[:punct:]]30[[:punct:]][[:digit:]]+|[[:digit:]]+[[:punct:]]45[[:punct:]][[:digit:]]+,mytime) mytime[subtime] [1] 12:00:00 12:00:05 12:15:05 12:15:06 12:30:01 12:45:01 13:00:00 13:15:02 This gives me the data taken at quarter hour intervals (removes 12:20:00) but I am still left with multiple values at the quarter hours. I would like to obtain: 12:00:00 12:15:05 12:30:01 12:45:01 13:00:00 13:15:02 Thanks! Tim Tim Clark Department of Zoology 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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ 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] Finding minimum of time subset
Dear List, I have a data frame of data taken every few seconds. I would like to subset the data to retain only the data taken on the quarter hour, and as close to the quarter hour as possible. So far I have figured out how to subset the data to the quarter hour, but not how to keep only the minimum time for each quarter hour. For example: mytime-c(12:00:00,12:00:05,12:15:05,12:15:06,12:20:00,12:30:01,12:45:01,13:00:00,13:15:02) subtime-grep(pattern=[[:digit:]]+[[:punct:]]00[[:punct:]][[:digit:]]+|[[:digit:]]+[[:punct:]]15[[:punct:]][[:digit:]]+|[[:digit:]]+[[:punct:]]30[[:punct:]][[:digit:]]+|[[:digit:]]+[[:punct:]]45[[:punct:]][[:digit:]]+,mytime) mytime[subtime] [1] 12:00:00 12:00:05 12:15:05 12:15:06 12:30:01 12:45:01 13:00:00 13:15:02 This gives me the data taken at quarter hour intervals (removes 12:20:00) but I am still left with multiple values at the quarter hours. I would like to obtain: 12:00:00 12:15:05 12:30:01 12:45:01 13:00:00 13:15:02 Thanks! Tim Tim Clark Department of Zoology 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.
[R] Maximizing values in subsetted dataframe
Dear List, I am trying to sub-sample some data by taking a data point every x minutes. The data contains missing values, and I would like to take the sub-sample that maximizes the number of valid points in the sample. I.e. minimizes the number of NA's in the data set. For example, given the following: da-seq(Sys.time(),by=1,length.out=10) x-c(1,2,NA,4,NA,6,NA,8,9,10) mydata-data.frame(da,x) If I wanted to take a subsample every 2 seconds, I would have the following two possible answers: answer1: 2,4,NA,8 answer2: 1,NA,NA,7 I would like a function that would choose between these and obtain the one with the fewest missing values. In my real dataset I have multiple variables collected every second and I would like to subsample it every 5, 10, and 15 minutes. I appreciate your help. Tim Tim Clark Department of Zoology 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] Duplicated date values aren't duplicates
Don and Jim, Thanks! I got it! Duplicated is only returning one of the two duplicated dates (the second date). It all makes sense now! Tim Tim Clark Department of Zoology University of Hawaii --- On Fri, 7/24/09, Don MacQueen m...@llnl.gov wrote: From: Don MacQueen m...@llnl.gov Subject: Re: [R] Duplicated date values aren't duplicates To: Tim Clark mudiver1...@yahoo.com, r-help@r-project.org Date: Friday, July 24, 2009, 4:00 AM Look at results of table( mydata$DateTime ) and I think you will see that some are duplicated. Specifically, the two in your dupes object. -Don At 5:50 PM -0700 7/23/09, Tim Clark wrote: Dear list, I just had a function (as.ltraj in Adehabitat) give me the following error: Error in as.ltraj(xy, id, date = da) : non unique dates for a given burst I checked my dates and got the following: dupes-mydata$DateTime[duplicated(mydata$DateTime)] dupes [1] (07/30/02 00:00:00) (08/06/03 17:45:00) Is there a reason different dates would come up as duplicate values? I would prefer not to have to delete them if I don't have to. Any suggestions on how to get R to realize they are different? Thanks, Tim Tim Clark Department of Zoology University of Hawaii __ R-help@r-project.org mailing list https://*stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://*www.*R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- -- Don MacQueen Environmental Protection Department Lawrence Livermore National Laboratory Livermore, CA, USA 925-423-1062 -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Duplicated date values aren't duplicates
Dear list, I just had a function (as.ltraj in Adehabitat) give me the following error: Error in as.ltraj(xy, id, date = da) : non unique dates for a given burst I checked my dates and got the following: dupes-mydata$DateTime[duplicated(mydata$DateTime)] dupes [1] (07/30/02 00:00:00) (08/06/03 17:45:00) Is there a reason different dates would come up as duplicate values? I would prefer not to have to delete them if I don't have to. Any suggestions on how to get R to realize they are different? Thanks, Tim Tim Clark Department of Zoology University of Hawaii __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Calculating distance between spatial points
Dear List, I am trying to determine the speed an animal is traveling on each leg of a track. My data is in longitude and latitude, so I am using the package rgdal to convert it into a spatial points data frame and transform it to UTM. I would then like to find the difference between successive longitudes and latitudes, find the euclidean distance between points, and compute the speed of the animal on each leg. My problem is that once I convert the lat and long into a spatial points data frame I can not access the lat and long individually. As far as I know I need to convert them in order to transform the lat and long to UTM. Is there a way I can call each variable separately in the sp dataframe? My code with example data is below. Any suggestions would be appreciated. library(rgdal) date.diff-(20,30,10,30) Long-c(-156.0540 ,-156.0541 ,-156.0550 ,-156.0640) Lat-c(19.73733,19.73734,19.73743,19.73833) SP-data.frame(Long,Lat) SP-SpatialPoints(SP,proj4string=CRS(+proj=longlat +ellps=WGS84)) SP.utm-spTransform(SP, CRS(+proj=utm +zone=4 +ellps=WGS84)) long.diff-diff(SP.utm$Long) lat.diff-diff(SP.utm$Lat) d=(long.diff^2+lat.diff^2)^.5 speed=d/date.diff Aloha, Tim Tim Clark Department of Zoology 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.
[R] PBSmapping problems with importGSHHS
Dear List, I am trying to use the PBSmapping package to import and plot the shoreline of Hawaii. I am having problems importing and plotting the specific regions that I would like to plot. Specifically, I can't get it to import the range of x variables that I would like. I think my problem is with a variable called xoff, but I have no idea what it is, what it does, or how I am supposed to use it. It's default value is -360, but I have found examples in the PBSmapping web publication of it being set to zero as well. The help files provides the following definitions: xlim range of X-coordinates to clip. Range should match the transform xoff. xoff transform the X-coordinates by specified amount. I can import certain regions of GSHH, but I don't seem to be able to limit them like I want using xlim. If I change xoff to something besides -360 I get different sizes of plots that are all the color I am using for land. I have the following: Hawaii.GSHH- importGSHHS(GSHHfile, xlim=c(-156.3,-155.92), ylim=c(19.45,19.5), xoff=-360) The GSHHS shoreline data is found at: http://www.soest.hawaii.edu/wessel/gshhs/gshhs.html Any suggestions on what is going on? Thanks, Tim Tim Clark Department of Zoology 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] Need a faster function to replace missing data
Many thanks to Jim, Bill, and Carl. Using indexes instead of the for loop gave me my answer in minutes instead of hours! Thanks for all of your great suggestions! Aloha, Tim Tim Clark Department of Zoology University of Hawaii --- On Fri, 5/22/09, jim holtman jholt...@gmail.com wrote: From: jim holtman jholt...@gmail.com Subject: Re: [R] Need a faster function to replace missing data To: Tim Clark mudiver1...@yahoo.com Cc: r-help@r-project.org Date: Friday, May 22, 2009, 4:59 PM Here is a modification that should now find the closest: myvscan-data.frame(c(1,NA,1.5),as.POSIXct(c(12:00:00,12:14:00,12:20:00), + format=%H:%M:%S)) # convert to numeric names(myvscan)-c(Latitude,DateTime) myvscan$tn - as.numeric(myvscan$DateTime) # numeric for findInterval mygarmin-data.frame(c(20,30,40),as.POSIXct(c(12:00:00,12:10:00,12:15:00), + format=%H:%M:%S)) names(mygarmin)-c(Latitude,DateTime) mygarmin$tn - as.numeric(mygarmin$DateTime) # use 'findInterval' na.indx - which(is.na(myvscan$Latitude)) # find NAs # create matrix of values to test the range indices - findInterval(myvscan$tn[na.indx],mygarmin$tn) x - cbind(indices, + abs(myvscan$tn[na.indx] - mygarmin$tn[indices]), # lower + abs(myvscan$tn[na.indx] - mygarmin$tn[indices + 1])) #higher # now determine which index is closer closest - x[,1] + (x[,2] x[,3]) # determine the proper index # replace with garmin latitude myvscan$Latitude[na.indx] - mygarmin$Latitude[closest] myvscan Latitude DateTime tn 1 1.0 2009-05-23 12:00:00 124308 2 40.0 2009-05-23 12:14:00 1243080840 3 1.5 2009-05-23 12:20:00 1243081200 On Fri, May 22, 2009 at 7:39 PM, Tim Clark mudiver1...@yahoo.com wrote: Jim, Thanks! I like the way you use indexing instead of the loops. However, the find.Interval function does not give the right result. I have been playing with it and it seems to give the closest number that is less than the one of interest. In this case, the correct replacement should have been 40, not 30, since 12:15 from mygarmin is closer to 12:14 in myvscan than 12:10. Is there a way to get the function to find the closest in value instead of the next smaller value? I was trying to use which.min to get the closet date but can't seem to get it to work right either. Aloha, Tim Tim Clark Department of Zoology University of Hawaii --- On Fri, 5/22/09, jim holtman jholt...@gmail.com wrote: From: jim holtman jholt...@gmail.com Subject: Re: [R] Need a faster function to replace missing data To: Tim Clark mudiver1...@yahoo.com Cc: r-help@r-project.org Date: Friday, May 22, 2009, 7:24 AM I think this does what you want. It uses 'findInterval' to determine where a possible match is: myvscan-data.frame(c(1,NA,1.5),as.POSIXct(c(12:00:00,12:14:00,12:20:00), format=%H:%M:%S)) # convert to numeric names(myvscan)-c(Latitude,DateTime) myvscan$tn - as.numeric(myvscan$DateTime) # numeric for findInterval mygarmin-data.frame(c(20,30,40),as.POSIXct(c(12:00:00,12:10:00,12:15:00), format=%H:%M:%S)) names(mygarmin)-c(Latitude,DateTime) mygarmin$tn - as.numeric(mygarmin$DateTime) # use 'findInterval' na.indx - which(is.na(myvscan$Latitude)) # find NAs # replace with garmin latitude myvscan$Latitude[na.indx] - mygarmin$Latitude[findInterval(myvscan$tn[na.indx], mygarmin$tn)] myvscan Latitude DateTime tn 1 1.0 2009-05-22 12:00:00 1243008000 2 30.0 2009-05-22 12:14:00 1243008840 3 1.5 2009-05-22 12:20:00 1243009200 On Fri, May 22, 2009 at 12:45 AM, Tim Clark mudiver1...@yahoo.com wrote: Dear List, I need some help in coming up with a function that will take two data sets, determine if a value is missing in one, find a value in the second that was taken at about the same time, and substitute the second value in for where the first should have been. My problem is from a fish tracking study. We put acoustic tags in fish and track them for several days. Location data is supposed to be automatically recorded every time we detect a ping from the fish. Unfortunately the GPS had some problems and sometimes the fishes depth was recorded but not its location. I fortunately had a back-up GPS that was taking location data every five minutes. I would like to merge the two files, replacing the missing value in the vscan (automatic) file with the location from the garmin file. Since we were getting vscan records every 1-2 seconds and garmin records every 5 minutes, I need to find the right place in the vscan file to place the garmin record - i.e. the closest in time, but not greater than 5 minutes. I have written a function
[R] Creating multiple graphs based on one variable
Dear List, I would like to create several graphs of similar data. I have x and y values for several different individuals (in this case fish). I would like to plot the x and y values for each fish separately. I can do it using a for loop, but I think I should be using apply. Please let me know what I am doing wrong, or if there is a better way to do this. What I have is: #Test data dat-data.frame(c(rep(1:10,4)),c(rep(1:10,4)),c(rep(c(Tony,Mike,Vicky,Fred),each=10))) names(dat)-c(x,y,Name) #Create function to plot x and y myplot-function() plot(dat$x,dat$y) #Apply the function to each of the names par(mfcol=c(2,2)) apply(dat,2,myplot,by=dat$Name) #Does not work - tried various versions I would like separate plots for Tony, Mike, and Vicky. What is the best way to do this? Thank! Tim Tim Clark Department of Zoology 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.
[R] Problem with fractional seconds
Dear List, I am having problems converting a file with fractional seconds to class POSIXct. I have set my options to include digits.secs and my format to just time, but my output is the current date with my time lacking the fractions of a second. For example: options(digits.secs=3) t-c(06:00:00.100,06:00:01.231) myt-as.POSIXct(t,format=%H:%M:%S) myt [1] 2009-05-26 06:00:00 HST 2009-05-26 06:00:01 HST I would like the output to be just time with fractional seconds. I.e. 06:00:00.100,06:00:01.231 I have also tried Chron times() which did not work either. Interestingly, Sys.time() does produce fractional seconds, so I know the options are working. I would appreciate your help and suggestions. Aloha, Tim Tim Clark Department of Zoology 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] Creating multiple graphs based on one variable
Luc, Thanks! I was not aware of that package. It looks a lot easier than what I have been trying to do! Aloha, Tim Tim Clark Department of Zoology University of Hawaii --- On Tue, 5/26/09, Luc Villandre villa...@dms.umontreal.ca wrote: From: Luc Villandre villa...@dms.umontreal.ca Subject: Re: [R] Creating multiple graphs based on one variable To: Tim Clark mudiver1...@yahoo.com Cc: r-help@r-project.org Date: Tuesday, May 26, 2009, 4:01 AM Tim Clark wrote: Dear List, I would like to create several graphs of similar data. I have x and y values for several different individuals (in this case fish). I would like to plot the x and y values for each fish separately. I can do it using a for loop, but I think I should be using apply. Please let me know what I am doing wrong, or if there is a better way to do this. What I have is: #Test data dat-data.frame(c(rep(1:10,4)),c(rep(1:10,4)),c(rep(c(Tony,Mike,Vicky,Fred),each=10))) names(dat)-c(x,y,Name) #Create function to plot x and y myplot-function() plot(dat$x,dat$y) #Apply the function to each of the names par(mfcol=c(2,2)) apply(dat,2,myplot,by=dat$Name) #Does not work - tried various versions I would like separate plots for Tony, Mike, and Vicky. What is the best way to do this? Thank! Tim Tim Clark Department of Zoology 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. Hi Tim, I'm now rather fond of Hadley Wickham's ggplot2 package. Its structure is most of the times intuitive and it does yield nice-looking output. In order to solve your problem, taking advantage of the ggplot2 framework, you can simply use the following: library(ggplot2) ; ## If you want all the curves to be on the same plotting grid ; p - ggplot(dat, aes(x=x,y=y, group=Name)) ; p + geom_line(aes(colour=Name)) ; ## Only one curve will be visible since they are all superposed. ## If you want the curves to be on separate plotting grids ; p - ggplot(dat, aes(x=x,y=y, group=Name)) ; p - p + geom_line(aes(colour=Name)) ; p+facet_grid(. ~ Name) ; Hope this helps, -- *Luc Villandré* /Biostatistician McGill University Health Center - Montreal Children's Hospital Research Institute/ __ 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] Creating multiple graphs based on one variable
Stefan, Thanks for the suggestion. Lattice works great. You might also want to check out the ggplot2 package that Luc suggested. They both seem to provide quite a few more options than the basic graphics package in R. Aloha, Tim Tim Clark Department of Zoology University of Hawaii --- On Mon, 5/25/09, Stefan Grosse singularit...@gmx.net wrote: From: Stefan Grosse singularit...@gmx.net Subject: Re: [R] Creating multiple graphs based on one variable To: Tim Clark mudiver1...@yahoo.com Cc: r-help@r-project.org Date: Monday, May 25, 2009, 11:55 PM On Tue, 26 May 2009 02:34:55 -0700 (PDT) Tim Clark mudiver1...@yahoo.com wrote: TC I would like separate plots for Tony, Mike, and Vicky. What is the TC best way to do this? use the lattice package: library(lattice) xyplot(y~x|Name,data=dat) Mr. Sarkar (the author of the package) has written an excellent book on his package I recommend it. hth Stefan __ 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] Problem with fractional seconds
Gabor, Thanks, that worked. However, is there is way to just get the time and not have the date added? I assume the date is added since POSIX is based on seconds since 1970. I can't seem to convert the POSIX value to Chron times(), and Chron won't take fractional seconds. Are there other ways to deal with just time and not date? Thanks, Tim Tim Clark Department of Zoology University of Hawaii --- On Tue, 5/26/09, Gabor Grothendieck ggrothendi...@gmail.com wrote: From: Gabor Grothendieck ggrothendi...@gmail.com Subject: Re: [R] Problem with fractional seconds To: Tim Clark mudiver1...@yahoo.com Cc: r-help@r-project.org Date: Tuesday, May 26, 2009, 11:59 AM Try this: as.POSIXct(c(06:00:00.100,06:00:01.231), format = %H:%M:%S%OS) [1] 2009-05-26 06:00:00.100 EDT 2009-05-26 06:00:00.231 EDT On Tue, May 26, 2009 at 5:52 PM, Tim Clark mudiver1...@yahoo.com wrote: Dear List, I am having problems converting a file with fractional seconds to class POSIXct. I have set my options to include digits.secs and my format to just time, but my output is the current date with my time lacking the fractions of a second. For example: options(digits.secs=3) t-c(06:00:00.100,06:00:01.231) myt-as.POSIXct(t,format=%H:%M:%S) myt [1] 2009-05-26 06:00:00 HST 2009-05-26 06:00:01 HST I would like the output to be just time with fractional seconds. I.e. 06:00:00.100,06:00:01.231 I have also tried Chron times() which did not work either. Interestingly, Sys.time() does produce fractional seconds, so I know the options are working. I would appreciate your help and suggestions. Aloha, Tim Tim Clark Department of Zoology 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. __ 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] Need a faster function to replace missing data
Jim,
Thanks! I like the way you use indexing instead of the loops. However, the
find.Interval function does not give the right result. I have been playing
with it and it seems to give the closest number that is less than the one of
interest. In this case, the correct replacement should have been 40, not 30,
since 12:15 from mygarmin is closer to 12:14 in myvscan than 12:10. Is there a
way to get the function to find the closest in value instead of the next
smaller value? I was trying to use which.min to get the closet date but can't
seem to get it to work right either.
Aloha,
Tim
Tim Clark
Department of Zoology
University of Hawaii
--- On Fri, 5/22/09, jim holtman jholt...@gmail.com wrote:
From: jim holtman jholt...@gmail.com
Subject: Re: [R] Need a faster function to replace missing data
To: Tim Clark mudiver1...@yahoo.com
Cc: r-help@r-project.org
Date: Friday, May 22, 2009, 7:24 AM
I think this does what you
want. It uses 'findInterval' to determine where a
possible match is:
myvscan-data.frame(c(1,NA,1.5),as.POSIXct(c(12:00:00,12:14:00,12:20:00),
format=%H:%M:%S))
# convert to numeric
names(myvscan)-c(Latitude,DateTime)
myvscan$tn - as.numeric(myvscan$DateTime) #
numeric for findInterval
mygarmin-data.frame(c(20,30,40),as.POSIXct(c(12:00:00,12:10:00,12:15:00),
format=%H:%M:%S))
names(mygarmin)-c(Latitude,DateTime)
mygarmin$tn - as.numeric(mygarmin$DateTime)
# use 'findInterval'
na.indx - which(is.na(myvscan$Latitude)) # find
NAs
# replace with garmin latitude
myvscan$Latitude[na.indx] -
mygarmin$Latitude[findInterval(myvscan$tn[na.indx],
mygarmin$tn)]
myvscan
Latitude DateTime
tn
1 1.0 2009-05-22 12:00:00 1243008000
2 30.0 2009-05-22 12:14:00 1243008840
3 1.5 2009-05-22 12:20:00 1243009200
On Fri, May 22, 2009 at 12:45 AM,
Tim Clark mudiver1...@yahoo.com
wrote:
Dear List,
I need some help in coming up with a function that will
take two data sets, determine if a value is missing in one,
find a value in the second that was taken at about the same
time, and substitute the second value in for where the first
should have been. My problem is from a fish tracking
study. We put acoustic tags in fish and track them for
several days. Location data is supposed to be
automatically recorded every time we detect a
ping from the fish. Unfortunately the GPS had
some problems and sometimes the fishes depth was recorded
but not its location. I fortunately had a back-up GPS that
was taking location data every five minutes. I would like
to merge the two files, replacing the missing value in the
vscan (automatic) file with the location from the garmin
file. Since we were getting vscan records every 1-2
seconds and garmin records every 5 minutes, I need to find
the right place in the vscan file to place the garmin record
- i.e. the
closest in time, but not greater than 5 minutes. I have
written a function that does this. However, it works with my
test data but locks up my computer with my real data. I
have several million vscan records and several thousand
garmin records. Is there a better way to do this?
My function and test data:
myvscan-data.frame(c(1,NA,1.5),times(c(12:00:00,12:14:00,12:20:00)))
names(myvscan)-c(Latitude,DateTime)
mygarmin-data.frame(c(20,30,40),times((12:00:00,12:10:00,12:15:00)))
names(mygarmin)-c(Latitude,DateTime)
minute.diff-1/24/12 #Time diff is in days, so this
is 5 minutes
for (k in 1:nrow(myvscan))
{
if (is.na(myvscan$Latitude[k]))
{
if ((min(abs(mygarmin$DateTime-myvscan$DateTime[k])))
minute.diff )
{
index.min.date-which.min(abs(mygarmin$DateTime-myvscan$DateTime[k]))
myvscan$Latitude[k]-mygarmin$Latitude[index.min.date]
}}}
I appreciate your help and advice.
Aloha,
Tim
Tim Clark
Department of Zoology
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.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] help with gsub and date pattern
Dear List, I am having a problem using gsub to remove dates from a date/time string. For example: x-c(5/31/2009 12:34:00,6/1/2009 1:14:00) I would like to remove the date and have just the time. I have tried: gsub([0-9+]/[0-9+]/[0-9+],,x) and various versions. I think my problem is that the / is a special character and is telling it something that I don't mean. I would appreciate any suggestions on how to proceed. Thanks, Tim Tim Clark Department of Zoology 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] help with gsub and date pattern
Thanks for the help and the various options! Putting the + outside the brackets worked, but I like the strsplit option. Always nice to learn new functions! Aloha, Tim --- On Thu, 5/21/09, Marc Schwartz marc_schwa...@me.com wrote: From: Marc Schwartz marc_schwa...@me.com Subject: Re: [R] help with gsub and date pattern To: Tim Clark mudiver1...@yahoo.com Cc: r-help@r-project.org Date: Thursday, May 21, 2009, 11:34 AM On May 21, 2009, at 4:13 PM, Tim Clark wrote: Dear List, I am having a problem using gsub to remove dates from a date/time string. For example: x-c(5/31/2009 12:34:00,6/1/2009 1:14:00) I would like to remove the date and have just the time. I have tried: gsub([0-9+]/[0-9+]/[0-9+],,x) and various versions. I think my problem is that the / is a special character and is telling it something that I don't mean. I would appreciate any suggestions on how to proceed. Thanks, Tim Switch the '+' to outside the brackets: gsub([0-9]+/[0-9]+/[0-9]+ ,,x) [1] 12:34:00 1:14:00 A few other options: # Use strsplit sapply(strsplit(x, split = ), [, 2) [1] 12:34:00 1:14:00 # Return the pattern contained within the parens # See ?regex gsub(^.* (.*)$, \\1, x) [1] 12:34:00 1:14:00 # Replace the characters up to the space with an empty vector gsub(^.* , , x) [1] 12:34:00 1:14:00 HTH, Marc Schwartz __ 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] Need a faster function to replace missing data
Dear List,
I need some help in coming up with a function that will take two data sets,
determine if a value is missing in one, find a value in the second that was
taken at about the same time, and substitute the second value in for where the
first should have been. My problem is from a fish tracking study. We put
acoustic tags in fish and track them for several days. Location data is
supposed to be automatically recorded every time we detect a ping from the
fish. Unfortunately the GPS had some problems and sometimes the fishes depth
was recorded but not its location. I fortunately had a back-up GPS that was
taking location data every five minutes. I would like to merge the two files,
replacing the missing value in the vscan (automatic) file with the location
from the garmin file. Since we were getting vscan records every 1-2 seconds
and garmin records every 5 minutes, I need to find the right place in the vscan
file to place the garmin record - i.e. the
closest in time, but not greater than 5 minutes. I have written a function
that does this. However, it works with my test data but locks up my computer
with my real data. I have several million vscan records and several thousand
garmin records. Is there a better way to do this?
My function and test data:
myvscan-data.frame(c(1,NA,1.5),times(c(12:00:00,12:14:00,12:20:00)))
names(myvscan)-c(Latitude,DateTime)
mygarmin-data.frame(c(20,30,40),times((12:00:00,12:10:00,12:15:00)))
names(mygarmin)-c(Latitude,DateTime)
minute.diff-1/24/12 #Time diff is in days, so this is 5 minutes
for (k in 1:nrow(myvscan))
{
if (is.na(myvscan$Latitude[k]))
{
if ((min(abs(mygarmin$DateTime-myvscan$DateTime[k]))) minute.diff )
{
index.min.date-which.min(abs(mygarmin$DateTime-myvscan$DateTime[k]))
myvscan$Latitude[k]-mygarmin$Latitude[index.min.date]
}}}
I appreciate your help and advice.
Aloha,
Tim
Tim Clark
Department of Zoology
University of Hawaii
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