[R] legend with mixed boxes and lines (not both)
Hi, I seem to be unable to get a mixed legend that has lines *or* polygons (not both). For example: ppi - seq(0,2*pi,length.out=21)[-21] frame() plot.window(ylim=c(-5,5),xlim=c(-5,5),asp=1) polygon(cos(ppi)*4+rnorm(20,sd=.2),sin(ppi)*4+rnorm(20,sd=.2), col=green,border=FALSE) polygon(cos(ppi)*2+rnorm(20,sd=.1),sin(ppi)*2+rnorm(20,sd=.1), col=blue,border=FALSE) abline(0,2,col=red) legend(topleft,legend=c(out,in,line),bty=n, fill=c(green,blue,NA),col=c(NA,NA,red), lwd=c(NA,NA,1)) I'm really guessing the behaviour in the legend() call, by setting fill to NA for the item, etc. I also tried fill=c(green,blue,FALSE), but that didn't go over too well either. I also tried adding merge=TRUE, but that just puts the line into the box. I also tried using box.lwd=c(1,1,0), but that also did not work Is there either a way to do this or a clean workaround? Thanks in advance. +mt __ R-help@stat.math.ethz.ch 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] sequential for loop
Hi all,
I'm usually comfortable using the *apply functions for vectorizing loops
in R. However, my particular problem now is using it in a sequential
operation, which uses values evaluated in an offset of the loop vector.
Here is my example using a for loop approach:
dat - data.frame(year=rep(1970:1980,each=365),yday=1:365)
dat$value - sin(dat$yday*2*pi/365)+rnorm(nrow(dat),sd=0.5)
dat$ca - dat$cb - 0 # consecutive above and below 0
for(n in 2:nrow(dat)){
if(dat$value[n] 0)
dat$ca[n] - dat$ca[n-1] + 1
else
dat$cb[n] - dat$cb[n-1] + 1
}
I'm inquiring if there is a straightforward way to vectorize this (or a
similar example) in R, since it gets rather slow with larger data
frames. If there is no straightforward method, no worries.
Thanks in advance.
+mt
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[R] timeDate business day
Those numbers look like ... well, numbers. You want characters! Try converting the integer to a character before trying to do a string parse, e.g.: ymd.int - c(20050104, 20050105, 20050106, 20050107, 20050110, 20050111, 20050113, 20050114) ymd - as.Date(as.character(ymd.int),%Y%m%d) As far as the other functions you are looking at (timeDate, timeRelative) -- I've never seen these, so I'm guessing they are S-PLUS. In R, you can use diff or difftime (which works with Date and POSIXlt-or Date-Time classes) , e.g.: diff(ymd) diff(ymd,2) diff(ymd,3) or do some arithmetic: difftime(ymd[1],ymd[4]) difftime(ymd[1],ymd[4],unit=weeks) Hopefully this is helpful to you! +mt __ R-help@stat.math.ethz.ch 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] timeDate business day
[1] 20050104 20050105 20050106 20050107 20050110 20050111 20050113 20050114 ymd - as.Date(as.character(ymd.int),%Y%m%d) ymd [1] 2005-01-04 2005-01-05 2005-01-06 2005-01-07 2005-01-10 [6] 2005-01-11 2005-01-13 2005-01-14 class(ymd) [1] Date While the variable ymd is actually of class Date, the format is not mmdd but -mm-dd as one can see in the previous example. As Young, I do not see what I am missing here. Any hint would be appreciated. AA. What happened in the beginning is that I had to parse the character into a Date-Time class (Date, in this case as you correctly pointed out). POSIX is a kind of standard that (mainly Unix) computers use date formatters, such as %Y for a 4-digit year, and others. They are all listed in great detail in ?strptime (which means string parse time). In this case the input parsing format pattern was %Y%m%d. There were no spaces in-between each number. When that class prints out, the default format is ISO 8601 ( see http://en.wikipedia.org/wiki/ISO_8601 ). When R prints the class Date to your screen, it decides to format it ISO 8601-style for you. If you want to see if differently, you can try: format(ymd,%Y/%d/%m) The date is actually stored internally as an ordinal, somewhat like how MS Excel dates work. You can see how it works internally: str(ymd) Hopefully I've demystified some of this .. any other questions? +mt __ R-help@stat.math.ethz.ch 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] timeDate business day
Sadly, I don't know of any tutorials or much help on the web for R ... that doesn't mean it doesn't exist ... you might just have to look around for it (www.rseek.org is a good place to start) I've learned almost everything I know through: ?strptime Also check out the methods for the classes, for example: methods(class=Date) methods(class=POSIXct) And certainly check their help pages ... there is loads of stuff here that I haven't discovered myself. (Note, if you are new to S3 classes .. if it begins with the method, then . class, you only need to type the beginning. For example summary(ymd) ... not summary.Date(ymd) if ymd has `class(ymd) == Date `. I think the fundamental things to know are there are three main DateTimeClasses: 1. POSIXct - has date, time and optionally time-zone info -- very handy for using in data.frame objects (and frankly I think it should be renamed to DateTime since the class POSIXct has nothing really to do directly with date/times) 2. POSIXlt - as far as I'm concerned, this is has the same functionality as POSIXct, but it cannot be used in data.frame objects (and frankly, I think it should be deprecated in favour of #1 to reduce future confusion) 3. Date - use this if you don't care about times or time-zones But it would be nice to track down a good tutorial somewhere. +mt Young Cho wrote: Thanks so Michael! If you know of a tutorial or introductory document about timeDate manipulation or time series manipulation in R, can you share it? It is hard to find by googling... I'd very appreciate any advice. __ R-help@stat.math.ethz.ch 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] pdf device bounding box?
I apologize if I don't fully understand your question, but the pdf device has a MediaBox, which is equivalent to the BoundingBox in EPS file. The PDFs from R are defined nicely using height/width dimensions, and work well with embedding in pdflatex, etc. For example: pdf(test.pdf,height=3,width=3) plot(1:10) and view the (partially binary) output in your shell: less -N test.pdf on line 117 of this file, I see /MediaBox [0 0 216 216] which is a 3in by 3in box measured in PostScript points. I don't understand how you are mixing this in with the epstopdf command. If you want to make both a PDF and EPS, my best advice is to do both directly from R (see ?postscript for EPS file generation .. the same example as above will have %%BoundingBox: 0 0 216 216 on line 10), and your output for both formats should be clean, simple, and good enough for publishers and everyone else to use. Just one caution, if you have a Windows computer and R 2.5.1 (which is most of us), make sure you write EPS files before loading up a PDF device (PR#9517). __ R-help@stat.math.ethz.ch 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] curve of density on histogram
Your are so close ... you just need to specify that you want your histograms to show density, not percent. I've only edited one line of your script, the rest is good and plots nicely: histogram(~ resp | group, col=steelblue,type=density, __ R-help@stat.math.ethz.ch 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] Using logarithmic y-axis (density) in a histogram
You might also want to try density, since it can theoretically have non-zero bins, since it doesn't use bins. For example, take a Weibull distribution, which could look better with a log y-axis: x - rweibull(1000,1,5) par(mfrow=c(2,1)) plot(density(x,from=0)) rug(x) plot(density(x,from=0),log=y) rug(x) you may need to fiddle with the bw (bandwidth) parameter of density, since this controls the smoothness of the kernel (see ?density). +mt __ R-help@stat.math.ethz.ch 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] plot(): I want to display dates on X-axis.
e.g. dat [,1]dat[,2] [1,]300 20060101 [2,]257 20060102 [3,]320 20060103 [4,]311 20060104 [5,]297 20060105 [6,]454 20060106 [7,]360 20060107 [8,]307 20060108 This looks like a matrix ... not a data frame. You defiantly want a data frame. So lets say you have: dat - matrix(c(round(rnorm(8)*100+400),20060101:20060108),ncol=2) #convert it: dat - as.data.frame(dat) #determine the dates: dat$date - as.Date(as.character(dat$V2),%Y%m%d) #plot it: plot(V1 ~ date, dat) __ R-help@stat.math.ethz.ch 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] Named backreferences in replacement patterns
How about turning them into a native date-time class, then re-formatting it. For example, say you have some American dates in a character vector: American.datechar - c(5/15/1976,2/15/1970,1/9/2006) # parse this: American.date - strptime(American.datechar,%m/%d/%Y) # reformat: format(American.date,%d/%m/%Y) __ R-help@stat.math.ethz.ch 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] alpha parameter in function rgb to specify color
Part of the problem is that alpha is a new and undocumented feature (note to developers: add this info into the pdf and rgb documentation). It only works in PDFs (maybe on quartz on Macs?), so you need to write to a PDF file: pdf(out.pdf,version=1.4) plot(1:10,col=rgb(1,0,0,alpha=seq(0,1,length.out=10))) dev.off() Now open up the PDF file... __ R-help@stat.math.ethz.ch 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] exact matching of names in attr
In R 2.5.0 (r40806), one of the change is to allow partial matching of name in the attr function. However, how can I tell if I have an exact match or not? For example, checking to see if an object has a name attribute, then giving it one if it doesn't: dat - data.frame(x=1:10,y=rnorm(10)) if(is.null(attr(dat,name))) attr(dat,name) - Site 1 str(dat) (This example works in R 2.5) Although there is no name attribute to the data.frame, it partially matches to names, resulting in not setting the attribute. (Personally, I think this change in the attr function is not desirable, and much prefer exact matches to avoid unintentional errors). How can I tell if this is an exact match? Is there a way to force an exact match? Thanks. +mt __ R-help@stat.math.ethz.ch 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] looping
Another way is to use an indexed list, which is far more tidier than your method. If you mean about 100 as in an irregular number, then a list is your friend (i.e., a ragged array, that can have sometimes 97 samples, sometime 105 samples, etc.). Similar to your example: dat - runif(10,0,100) # fake dataset smp - list() # need an empty list first for(i in 1:1000) smp[[i]] - sample(dat,100) However, if you are new to R/S, the best advice is to learn to _not_ use the for loop (because it is slow, and there are vectorized ways). For example, if we want to find the mean of each sample, then return a tidy result: sapply(samp,mean) or a crazy new analysis you might be working on: crazy - function(x,y) (sum(xy)^2)/sum(x) sapply(smp,crazy,10) etc. +mt __ R-help@stat.math.ethz.ch 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] setting options in a list
Hi, I have a question regarding setting options that use a list. There are none that I'm aware of in R-base, however, it can make sense for some custom situations. For example: options(myoptions=list(lwd=1,density=NULL,angle=45,col=green)) Getting values is simple: getOption(myoptions)$col However, setting a value from the list is less intuitive; here are a few failed attempts that could make sense from an intuitive level: setOption(myoptions)$col - red # non-existing function options(myoptions)[[1]]$col - red Here is an attempt that almost works, but as documented, it only sets a local .Options object for S-compatibility: .Options[myoptions][[1]]$col - red .Options[myoptions][[1]]$col # works, but ... getOption(myoptions)$col The only method I know of is to modify a copy of the list from the options, then re-set the option: cp - getOption(myoptions) cp$col - red options(myoptions=cp) getOption(myoptions)$col So, my question is if there is a more elegant method of setting an option in a list, that doesn't need copying and multiple commands? Having a 'setOption' function sure could be helpful for this instance, however I'm unsure how to implement this method (or if it is possible in the current version of R). Thanks. +mt __ R-help@stat.math.ethz.ch 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] Log-scale in histogramm
Log y-axis on histograms are conceptually wrong, but aren't a bad idea either. It is conceptually safer to show this using density. Consider an exponential distribution, which could look better with a log y-axis: x - rexp(1,.1) xd - density(x,from=0) par(mfrow=c(2,1)) plot(xd) plot(xd,log=y,ylab=Log Density) Just use caution in the interpretation. The large dropping/osculating features on the right-hand side of the log plot are from sparse data-points, and this can be smoothed out by adjusting 'bw' to get a theoretical interpretation of the true distribution. +mt __ R-help@stat.math.ethz.ch 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] Object attributes in R
Hi, I have questions about object attributes, and how they are handled when subsetted. My examples will use: tm - (1:10)/10 ds - (1:10)^2 attr(tm,units) - sec attr(ds,units) - cm dat - data.frame(tm=tm,ds=ds) attr(dat,id) - test1 When a primitive class object (numeric, character, etc.) is subsetted, the attributes are often stripped away, but the rules change for more complex classes, such as a data.frame, where they 'stick' for the data.frame, but attributes from the members are lost: tm[3:5]# lost ds[-3] # lost str(dat[1:3,]) # only kept for data.frame Is there any way of keeping the attributes when subsetted from primitive classes, like a fictional attr.drop option within the [ braces? The best alternative I have found is to make a new object, and copy the attributes: tm2 - tm[3:5] attributes(tm2) - attributes(tm) However, for the data.frame, how can I copy the attributes over (without using a for loop -- I've tried a few things using sapply but no success)? Also I don't see how this is consistent with an empty index, [], where attributes are always retained (as documented): tm[] I have other concerns about the evaluation of objects with attributes (e.g. ds/tm), where the attributes from the first object are retained for the output, but this evaluation of logic is a whole other can of worms I'd rather keep closed for now. +mt __ R-help@stat.math.ethz.ch 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] Documentation patch for 'match' and 'palette'
Here is a patch to improve documentation for finding useful, yet newish,
functions: 'findInterval' and 'colorRamp'. I think that it is worthwhile
to mention these in the 'seealso' section of the similar 'match' and
'palette' documents. I had difficulty finding these functions at first,
as they have compound names. Modify the patch as needed.
+mt
Index: src/library/base/man/match.Rd
===
--- src/library/base/man/match.Rd (revision 39542)
+++ src/library/base/man/match.Rd (working copy)
@@ -65,6 +65,8 @@
\code{\link{pmatch}} and \code{\link{charmatch}} for (\emph{partial})
string matching, \code{\link{match.arg}}, etc for function argument
matching.
+ \code{\link{findInterval}} similarly returns a vector of positions, but
+ finds numbers within intervals, rather than exact matches.
\code{\link{is.element}} for an S-compatible equivalent of \code{\%in\%}.
}
Index: src/library/grDevices/man/palette.Rd
===
--- src/library/grDevices/man/palette.Rd(revision 39542)
+++ src/library/grDevices/man/palette.Rd(working copy)
@@ -31,7 +31,7 @@
\code{\link{colors}} for the vector of built-in \dQuote{named} colors;
\code{\link{hsv}}, \code{\link{gray}}, \code{\link{rainbow}},
\code{\link{terrain.colors}},\dots to construct colors;
-
+ \code{\link{colorRamp}} to interpolate colors, making custom palettes;
\code{\link{col2rgb}} for translating colors to RGB 3-vectors.
}
\examples{
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