similar to: source function help

The following code produces an eps file with the tops of each of the ylabs clipped off. par(mfrow=c(2,2)) plot(runif(10), ylab="Function(Lengthy Expression)",xlab="Prediction") plot(runif(10), ylab=expression(Delta * Beta^2),xlab="Prediction") plot(runif(10), ylab="Function(Lengthy Expression)",xlab="Prediction")

Is anyone using R with postgres? I'd like to do so, but cannot seem to find any reasonable explanation of how to do so. Joshua Gramlich Chicago, IL

Hello, I'd like to do something like: n<-100 zz<-cbind(rnorm(n),rnorm(n),floor(runif(n)*3+1)) colors<-c("red","green","blue") plot(zz,col=colors(zz[3])) and have a matrix of scatterplots colored by class. The above does not work, of course, but I'm not sure exactly what function I'm looking for. Thank you for your time, Dave, -- Dave

Is there a tool for reading a graphics file into an object? I might not be looking with the correct vocabulary, but I'm finding lots of references to producing graphics from R, but not any for inputting graphics into R. I'd like to use a jpeg image and have the data available in R for analysis. I could convert the image to another format, PPM perhaps, and parse that in, GRASS seems like

Hello, Here's a repost on my message of a few days ago, for reasons that I'm not sure if it went through. If it did, my apologies, all suggestions welcome. I'm having an identical problem on slackware linux 7.0 and freebsd 3.4-release and windows 98-2, with sp1, and samba 2.05 and 2.06, depending on which system I've got up. Whenever I restart samba for whatever reason, making a

Hi, I'd like to partition a 2d probability density function into regions of equal probability. It is straightforward in the 1d case, like qnorm(seq(0,1,length=5)) but for 2d I'd need more constraints. Any suggestions for how to approach this? Is seems like a spatial sampling problem but I'm not sure where to look. Thanks for your time, Dave -- Dr. David Forrest drf at

Hello all, It seems that the next improvement to the R Graph Gallery is categorization of the graphics, that way each graph will be easier to find. That step should be done *carefully* if we want to avoid the opposite side-effect : graph not reachable through the categories. That's why the wisdom of the R community is required. Graphics will be classified in : - categories -

Hi, I have an array of 2d node locations and an array triangles, and would like to plot something like a image or persp. An example of doing it with rgl is: library(ncdf) library(rgl) # wget http://www.maplepark.com/~drf5n/extras/teapot.nc teapot<-open.ncdf("teapot.nc") z<-get.var.ncdf(teapot,"tris") l<-get.var.ncdf(teapot,"locations")

Is there a package for stacks with pushing and popping? I'd really like to do something like push(par(no.readonly=TRUE)) some stuff.... par(pop()) It seems like it wouldn't be difficult, and that someone may have already implemented a set of stack functions, and I wouldn't like to duplicate it. Thanks for your time, Dave -- Dave Forrest (434)924-3954w(111B) (804)642-0662h

My friend Veslot proposed me the Raven ! http://www.teteamodeler.com/allopass/images/corbeau.jpg As a matter of fact the intelligence of this bird is comparable to one of a monkey, although its brain is close to a reptile brain. R is quite the same : small, compact, but so clever. This is a canadian website explaining (in french) the R-aven intelligence

Um texto embutido e sem conjunto de caracteres especificado associado... Nome: n?o dispon?vel Url: https://stat.ethz.ch/pipermail/r-help/attachments/20070312/112cd7f9/attachment.pl

I have a problem when using nnet to predict the negative values. For example : X = matrix(c(1,1,0,0,1,0,1,0),4,2) X Y = matrix(c(0,1,1,0)) - 0.5 # XOR - 0.5 Y nn = nnet(X,Y,size=3) val = predict(nn,X) val # this is expected to be close to Y, but it's not ! The 'val' is always positive. I tried to change the options, but the result isn't much better. Could someone give me an

I wish to perform brain surgery this afternoon at 4pm and don't know where to start. My background is the history of great statistician sports legends but I am willing to learn. I know there are courses and numerous books on brain surgery but I don't have the time for those. Please direct me to the appropriate HowTos, and be on standby for solving any problem I may encounter while

Hi, I failed to search for R package providing random number generator of "Park and Miller". Anyone know any R package supporting this kind of function? Thanks, Grace -- View this message in context: http://www.nabble.com/Random-Number-Generator-of-Park-and-Miller-tf3636501.html#a10154554 Sent from the R help mailing list archive at Nabble.com.

even for people who didn't like the outcome of the US elections this year, it must have been a joy to see all the nice (and not so nice) graphs that were shown. As an exercise, I recreated the map shown on the NY-Times website [ http://www.nytimes.com/packages/khtml/2004/11/03/politics/ 20041103_px_ELECT_GRAPHIC.html ] If you're interested,

One quirk to be watched is the use of connections to the DBMS via the Unix sockets vs ports. The PostgreSQL driver bundled with unixODBC will use Unix sockets to `localhost', but this driver seems unreliable (see the ChangeLog). The current driver will only use a TCP/IP port, and to use that needs postmaster started with the -i flag (which is not the default) and with tcp/ip

Is there an element-by-element multiplication in R, like the .* operator in Matlab? eg: A (2x3) B (2x3) C=A.*B C (2x3) C = [[a11*b11 a12*b12 a13*b13]; [a21*b21 a22*b22 a23*b23]] I can't find one... Thanks -Mike Gates

Dear Sir/Madam, I am doing a project related to R. However, it is always difficult find some R functions. The R user guide seems not complete. Is there any free document about all R functions ? Who knows ? please help me. My email is myzhaogong@yahoo.com Thank you very much --------------------------------- [[alternative HTML version deleted]]

Hello, I don't know if the result of acos(0.5) == pi/3 is a bug or not. It looks strange to me. Inaki Murillo

My apologies for the statistical naivete of my question but... Is there an established method or calulating the statistical power of a correlation test? And if so is there a method in R for it? Thank you, Collin Lynch.