similar to: plot only x- and y-axis with origin, no box()

Dear R-users, Does anyone knows how I can order my serie of boxplots from lowest to highest median (which is much better for visualization purposes). thanks in advance, willem [[alternative HTML version deleted]]

Dear Rs I have a 3x3 multiple plot. I would like to have a overall legend in the outer right margin. From the help archive, I found that it can be done by setting par(xpd=NA). However, I couldn't find the correct values for x and y co-ordinates for the legend. Please find the code snippet below: par(mfrow=c(3,3), mar=c(4,4,0.9,0.5), oma=c(1,2,2,4),cex.main=1.1)

Dear R Users, I would like to plot three time series on the same graph, two axis on the left and one axis on the right. The time series that I am graphing on the left do not share a similar scale: one has a range of 1:100 and the other a range of 25000:70000. How can I display the tick marks for both on the left hand side without superimposing them, i.e. one set of tick marks in black right

Hi, Does anyone know if there is a way to 'reset the plot number' on a traditional graphics device? For instance, I want to have two plots on stacked top of each other (mfrow=c(2,1)) but with underlying grid lines spanning both figures vertically. I can put the grid lines on top if I add them last: par(mfrow=c(2,1)) plot.new() plot.window(c(0,1),c(0,1)) for( i in 1:2) axis(i)

Hello all, I just wanted to share a small perl script that generates a dot file from the result of the R profiler. The dot file can than be used to create a graphical display of the profiling. You can save this file in the bin directory of your R installation and then create a graph, for example an SVG by piping the output of the script to dot: $ R CMD Rprof2dot Rprof.out | dot -Tsvg >

There are still places available on the following courses. For more detailed information and a registration form please contact training@mango-solutions.com, or visit our website at www.mango-solutions.com <http://www.mango-solutions.com/> . R for Finance **************************************************** Dates: 21st to 23rd October 2008. Duration: 3 days. Location: London.

Hi, I'm looking to use the grep function (or something else) to return the matched pattern as opposed to the whole element. For example: x <- c("pjhj24jhjhd") grep("[[:digit:]]{2}", x, value=T) returns "people", whereas I simply want "24". Any help would be appreciated, Thanks, Aidan [[alternative HTML version deleted]]

Dear all, is there a good way to create SVG plots with R whose elements have titles (tooltips) or act as hyperlinks? I am using the RSvgDevice package, which works great - but it doesn't seem to support the notion that plot objects have titles or are act as hyperlinks, so I am helping myself by giving the objects funny unique colors and then postprocessing the .svg file. I wonder

BaselR - The new R meeting We are pleased to announce the new R meeting to be held in Basel, Switzerland. BaselR will be held from 6:30-9:30pm on Tues, Apr 27 at TransBARent: http://transbarent.business.sv-group.ch Doors open at 6:30,pm with the presentations starting at 7:00pm Introduction: What is Basel R? Andreas Krause:... Graphing Pharma Data Yann Abraham: Graphics Charles Roosen: Web

Dear R users, just another little question... are there other ways, I mean more easy to write, to obtain the same result I got with: data[95:length(dati[,1]), ] where data is a data frame to extract the last elements starting from a fixed position? thank you very much best regards Manuele PEsenti -- Manuele Pesenti manuele a inventati.org amicogodzilla a jabber.linux.it

I am relatively new to R, so this may be a much simpler question than it seems to me. I am trying to create a function that includes two summations, and can't figure out how. I am attaching the equation as a pdf file. This function will then be optimized over a chosen range of values, but right now I just need help with the function. Thank you. Zack Darnell Duke University -- Zack

Dear UseRs, I would like to know the way to find classes of each column of data.frame(). Thank you in advance. ========================================================= Dong-hyun Oh Center of Excellence for Science and Innovation Studies Royal Institute or Technology, Sweden e-mail: oh.dongh at gmail.com cel: +46 73 563 45 22

Hi there! I have i.e. v = c(0.43554, -0.22343), and so on. Now I'd like to make a format like format(v, digits=2) but without the leading 0, so 0.43554 becomes .44, -0.22343 becomes -.22 How can I do that? runnable example for copy and paste: v = c(0.43554, -0.22343) format(v, digits=2) Thanks, Martin -- Ihr Partner f?r Webdesign, Webapplikationen und Webspace.

Hello, Appending information to the profiler's output seems to generate problems. Here is a small example of code : <code r> require(boot) Rprof( memory.profiling = TRUE) Rprof(NULL) for(i in 1:2){ Rprof( memory.profiling = TRUE, append = TRUE) example(boot) Rprof(NULL) } </code> The problem is that the file Rprof.out contains more than once the header information: $ grep

Dear all, I would like to take out the values from one vector that are equal to the values in another vector. Example: a <- c(1,2,3,4,5,6,7,8,9) b <- c(3,10,20,5,6) b_noRepeats = c(10,20) So I would like to have the vector b without the same values as vector a. Kind regards, João Fadista [[alternative HTML version deleted]]

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Hi, I am looking for a way to compare if every element of a vector is > 0. i.e. while(abs(U) > 0) { .. } is there a function for this or do I have to write one? I'd appreciate your help! Benjamin

Greetings UseRs, Pls advise if there is a way to write a func that can be supplied to aggregate to compute weighted MeanAbsolute Dev (MAD). I am having trouble passing the correct weights from each group level and cannot see the code behind aggregate. But maybe 'aggregate' is not the best way to do that. > m1 <-

Hi, I have a vector of strings (class character) with 6 elements (length 6). I call it 'names'. "Graham Chapman" "John Cleese" "Terry Gilliam" "Eric Idle" "Terry Jones" "Michael Palin" And I want to turn it into another vector of strings called 'shortnames' with the same length. The new vector should look like:

How does one convert objects c("a","b","c") and "d" into "abcd"? > paste(c("a","b","c"), "d") of course yields [1] "a d" "b d" "c d" -- Ajay Shah http://www.mayin.org/ajayshah ajayshah at mayin.org