similar to: indexing without looping

Hello, In a nutshell, I've got a data.frame like this: > assignation <- data.frame(value=c(6.5,7.5,8.5,12.0),class=c(1,3,5,2)) > assignation value class 1 6.5 1 2 7.5 3 3 8.5 5 4 12.0 2 > and a long vector of classes like this: > x <- c(1,1,2,7,6,5,4,3,2,2,2...) And would like to obtain a vector of length = length(x), with the corresponding

Hi, Is there a way to know which library is giving a warning? Specifically, I'm getting a set of warnings: "Too many open raster files" Thanks and best wishes, -- Javier Garc?a-Pintado Institute of Earth Sciences Jaume Almera (CSIC) Lluis Sole Sabaris s/n, 08028 Barcelona Phone: +34 934095410 Fax: +34 934110012 e-mail:jgarcia at ija.csic.es

Hi all, Let's say I have a long data frame and a short one, both with three colums: $east, $north, $value And I need to fill in the short$value, extracting the corresponding value from long$value, for coinciding $east and $north in both tables. I know the possibility: for (i in 1:length(short$value)){ short$value[i] <- long$value[long$east==short$east & long$north==short$north] } How

Hi, I'm trying to use expression() to write a text to a graphic in the margin. Using: mtext(expression(beta),side=1,line=2) writes a perfect beta greek character, but I need to add a subindex "max", and I'm trying: mtext(paste(expression(beta),"max"),side=1,line=2) simply writes "beta max" in the plot. Please, Could you tell me what I'm doing wrong?

Hello, A brief question that I am not able to resolve by my own, but I presume it's easy for you: I've used apply to find the values closer to 0 in each row of a matrix: closer.to.0 <- apply(abs(myarray),MARGIN=1,"min") My question is how could I use which (or another way) to obtain a vector that indicates which columns contains these "closer to 0" values? (I

Hi! Is there any way of having a greyed ("ghosted") text (i.e, 2006) in the background of a plot? I'm making a dynamic plot and would like to show the year of each time step as a big greyed text in the background. (the idea comes from Hans Rosling video: http://video.google.com/videoplay?docid=4237353244338529080&sourceid=searchfeed ) Thanks Agus -- Dr. Agustin Lobo Institut

Hello, I have a program that used to run well in October, it uses library snow. Since then, one change has ocurred (snow library has been updated) and another could have ocurred (I've unadvertently modified something). Anyway, now when I make the call: parallel.model.results <- clusterApply(cl,processors.struct,MCexe) exactly as I used to do, where MCexe is my function and

Hi! Given a vector (or a factor within a df),i.e. v1 <- c(1,1,1,2,3,4,1,10,3) and a dictionary cbind(c(1,2,3),c(1001,1002,1003)) is there a function (on the same line than recode() in car) to get v2 as c(1001,1001,1001,1002,1003,4,1001,10,1003) ? I'm using myself a function based on match() since long ago (I think that thanks to advice by Prof. B. Ripley), but would like to know if there

Hi! Is it possible to set a file to which both commands and output would get automatically saved? I've tried with sink(), but only get the output. I mean something like a combined history and sink, as you get with File/Save to File.. in the windows GUI. Tis is done with diary filename in Matlab, and you can state diary on and diary off to control what is being saved to the file. Thanks Agus

Which is the proper way to rename a factor? If I do: test$Parc[test$Parc=="Ol?rdola"]<-"Ol?rdola" R complains that Warning message: invalid factor level, NAs generated in: `[<-.factor`(`*tmp*`, test$Parc == "Ol?rdola", value = "Ol?rdola") Thanks Agus -- Dr. Agustin Lobo Institut de Ciencies de la Terra "Jaume Almera" (CSIC) LLuis Sole

I've set layout(matrix(c(1,3,2,4),ncol=2)) and par(mar=c(0, 4, 4, 0) + 0.0) and par(mar=c(0, 0, 4, 3) + 0.0) for the upper-left and upper-right plots, but still get an space between both plots, is there any way to completely eliminate that space? Thanks Agus -- Dr. Agustin Lobo Institut de Ciencies de la Terra "Jaume Almera" (CSIC) LLuis Sole Sabaris s/n 08028 Barcelona Spain

Hi, The argument xaxt="n" for removing the x axis from a image plot does not work for me. I'm really using a function called plot.grassmeta() in library GRASS that is a wrapper for image(), but it seems to me that the problem is not in plot.grassmeta() but in image(). If I'm right could you tell me if there is a way to remove one axis from a call to image()? Thanks and wishes,

Hello; Well, I'm afraid this is the second related problem I report in two days (I'm sorry for this) I've programmed a tcltk interface for a model and it includes a comboBox. The comboBox widget comes with the "BWidgets" library of Tcl/Tk. It works all perfectly with the windows R GUI (with the help of Prof. B. Ripley), following the advices in ?Startup, but I would like

Is there an specific function to eliminate blank space within strings? Thanks -- Dr. Agustin Lobo Institut de Ciencies de la Terra "Jaume Almera" (CSIC) LLuis Sole Sabaris s/n 08028 Barcelona Spain Tel. 34 934095410 Fax. 34 934110012 email: Agustin.Lobo at ija.csic.es http://www.ija.csic.es/gt/obster

Given: > mitest <- matrix(1:16,ncol=4) > dimnames(mitest)[[1]] <- c("a","b","c","d") > dimnames(mitest)[[2]] <- c("a","b","c","d") > mitest a b c d a 1 5 9 13 b 2 6 10 14 c 3 7 11 15 d 4 8 12 16 I can do: > mitest[cbind(c(1,2,3),c(2,3,2))] [1] 5 10 7 but using the names does not

Given char vector delme2: > str(delme2) chr [1:1065] "30-1-08 8:48:21" "30-1-08 8:55:17" "30-1-08 9:00:22" ... I do: > delme3 <- strptime(delme2,format="%d-%m-%y %H:%M:%S") But then: > str(delme3) POSIXlt[1:9], format: "2008-01-30 08:48:21" "2008-01-30 08:55:17" ... > length(delme3) [1] 9 whie >

Hi! If I'm plotting 2 variables with the same units (i.e. wing length of 2 species), how can I force that X and Y axes have the same size for length unit? For example, if X ranges from 0 to 10 and Y ranges from 10 to 30, I want Y axis to be twice as long as X axis. I know I can just put xlim and ylim from 0 to 30, i.e., plot(runif(10,0,10),runif(10,10,30),xlim=c(0,30),ylim=c(0,30)) but this

Hi! I'm confused by this: > as.numeric(as.POSIXct(518400,origin="2001-01-01")) [1] 978822000 I guess the problem is that as.numeric() assumes a different origin, but cannot find any default origin. How can I get back the seconds from the POSIXct format? In other words, which the inverse function of as.POSIXct()? I've tried as.numeric and unclass() using a origin=

This should be theoretically very simple, but I dont get the elegant answer (without looping). I've got a long (thousands of rows) data frame: > E.coor[1:10,] east north dat 1 582650 4248850 0.8316848 2 582750 4248850 0.7230272 3 582850 4248850 0.3250818 4 582950 4248850 0.6144006 5 583050 4248850 0.8706312 6 583150 4248850 0.2149651 7 583250 4248850 0.1659519 8

Hello, I'm using the gstat package within R for an automated procedure that uses ordinary kriging. I can see that there is a logical ("singular") atrtibute of some adjusted model semivariograms: .- attr(*, "singular")= logi TRUE I cannot find documentation about the exact meaning and the implications of this attribute, and I dont know anything about the inner calculations