similar to: sourcecode for the balloonplot function from the gplots package

Hi, I am interested in implementing a special variant of balloonplot.  Let me explain with an example dataset from the reference manual : library(gplots) data(Titanic) dframe<-as.data.frame(Titanic) survived<-dframe[dframe$Survived=="Yes",] attach(survived) balloonplot(x=Class,y=list(Age,Sex),z=Freq,sort=TRUE,show.zeros=TRUE,cum.margins=FALSE,             main="BalloonPlot :

I'm not understanding something. I'm trying to add xlab & ylab to a balloon plot of a table object. From docs I thought following should work: require(gplots) # From balloonplot example: # Create an example using table xnames <- sample( letters[1:3], 50, replace=2) ynames <- sample( 1:5, 50, replace=2) tab <- table(xnames, ynames) balloonplot(tab)

Announcing gplots 2.5.0 --------------------------------- gplots provides additional plotting functions, including several enhanced versions of base R functions. Provided functions include: balloonplot, bandplot, barplot2, boxplot.n, colorpanel, heatmap.2, hist2d, lowess, ooplot, overplot, plot.lm2, plotCI, plotmeans, qqnorm.aov, residplot, rich.color, sinkplot, smartlegend, space,

Announcing gplots 2.5.0 --------------------------------- gplots provides additional plotting functions, including several enhanced versions of base R functions. Provided functions include: balloonplot, bandplot, barplot2, boxplot.n, colorpanel, heatmap.2, hist2d, lowess, ooplot, overplot, plot.lm2, plotCI, plotmeans, qqnorm.aov, residplot, rich.color, sinkplot, smartlegend, space,

Dear R community, I encounter a problem that is counterintuitive to my understanding of the documentation of source and the "local" argument of that function. With the following code, I would expect the content of "test.R" to be evaluated inside the environment of the function "test". This, however, does not seem to be the case as the object "a" can

Hi all, I need help with axis in plot. I want to edit y axis label of my plot. My data is like: x <- c(100,50,10,1,0.1,0.05,0.001) plot(log(x)) axTicks(2) # Label of y axis [1] -6 -4 -2 0 2 4 I'd like that y axis label was like: e^-6, e^-4, etc. (with text "e" superscript -6, -4, etc.) I try to use expression(), but don't work. plot(log(x), yaxt="n")

Hi, I'm currently reworking a report, originating from a MS Access database, but should be implemented in R. Now I'm facing the task to convert a lot of queries to postgreSQL. What I want to do is make a function which takes the MS Access query as an argument and returns the pgSQL version. So: SELECT [public_tblFiche].[Fichenr], [public_tblArtnr].[Artnr] FROM [public_tblFiche],

Hi All, has anyone run into maximum depth of nested JSON arrays in either rjson or RJSONIO ? I seem to be able to get up to 10 depth levels without problem, but crossing over to 11 either causes an error or fails to load the nodes properly. with RJSONIO I tried: a = fromJSON('data/myJSON.json', depth=1000) but I still get this error: Error in fromJSON(content, handler, default.size,

I'm trying to enter a frequency table manually so that I can run a goodness of fit test (I only have the frequencies, I don't have the raw data). So for example, let's say I want to re-create the HorseKicks table: library(vcd) data(HorseKicks) str(HorseKicks) 'table' int [1:5(1d)] 109 65 22 3 1 - attr(*, "dimnames")=List of 1 ..$ nDeaths: chr [1:5]

Dear r-help members, I have a number in the form of a string, say: a<-"-01020.909200" I'd like to extract "1020." as well as ".9092" Front<-grep(pattern="[1-9]+[0-9]*\\.", value=TRUE, x=a, fixed=FALSE) End<-grep(pattern="\\.[0-9]*[1-9]+", value=TRUE, x=a, fixed=FALSE) However, both strings give "-01020.909200", exactly

I would like to efficiently find the first index of each unique value in a very large vector. For example, if I have a vector A<-c(9,2,9,5) I would like to return not only the unique values (2,5,9) but also their first indices (2,4,1). I tried using a for loop with which(A==unique(A)[i])[1] to find the first index of each unique value but it is very slow. What I am trying to do is easily

Hello, We have submitted the updated version of gdata, gmodels, gplots and gtools to CRAN. Summary of the changes is attached at the end. Best, Nitin ______________________ Nitin Jain, PhD <nitin.jain at pfizer.com> Non Clinical Statistics Pfizer, Inc. (Groton, CT) Bldg: 260, # 1451 Ph: (860) 686-2526 (Office) Fax: (860) 686-6170 Brief description of changes: CHANGES IN GDATA 2.1.2

Hello, We have submitted the updated version of gdata, gmodels, gplots and gtools to CRAN. Summary of the changes is attached at the end. Best, Nitin ______________________ Nitin Jain, PhD <nitin.jain at pfizer.com> Non Clinical Statistics Pfizer, Inc. (Groton, CT) Bldg: 260, # 1451 Ph: (860) 686-2526 (Office) Fax: (860) 686-6170 Brief description of changes: CHANGES IN GDATA 2.1.2

Why the error is coming? even though the length of outcome.new$compkey and outcome.new$armkey were exactly same. Can anyone help? setwd("D:/AZ") library("RODBC") cdb_cnct <- odbcConnectExcel("AZIF_DC_GVK_NSCLC_MSALL_287papers_02072012_141450_v1_4.xls") outcomes <- sqlFetch(cdb_cnct, "Outcomes_info") odbcClose(cdb_cnct) rm(cdb_cnct)

Hi, My data.frame "A" has FID like this FID a a b b b c c d d d d Now my second data.frame "B" has age value for a, b, c, d like FID Age a      5 b      7 c      9 d      3 How can search for the Age column in "B" and replace the values in "A" so that my new "A" looks like this FID Age a      5 a      5 b      7 b      7 b      7

Full_Name: Cesar Henrique Torres Version: 2.2.0 OS: winXP-pro Submission from: (NULL) (201.1.196.50) Hi developer's! I'd like to know were gtools package is, because in the link below it isn't: http://cran.r-project.org/bin/windows/contrib/r-release/ This package is used in gplots and I'm trying to make a balloonplot. Thanks a lot, Cesar Torres.

I am having difficulty installing packages. For example, here I have tried to install "gplots" but to no avail. I have tried multiple mirrors and different packages (errbar) but after installation I seem to be unable to access any of the commands or help pages. I am a little concerned about the 'lib' missing message but this has been present for previous successful package

I would like to know whether there is a faster way to do the below operation (updating vec1). My objective is to update the elements of a vector (vec1), where a particular element i is dependent on the previous one. I need to do this on vectors that are 1 million or longer and need to repeat that process several hundred times. The for loop works but is slow. If there is a faster way, please let

Dear Group, I have the following data matrix which is a timeseries. > dput(tData) structure(list(A = c(0.2, 0.13, 0.05, 0.1, 0.02, 0.18, 0.09, 0.06, 0.13), B = c(0.15, 0.06, 0.09, 0.02, 0.03, 0.12, 0.01, 0.15, 0.06), C = c(-0.1, 0, -0.07, -0.06, -0.05, -0.05, -0.06, -0.08, -0.07), D = c(-0.15, -0.05, -0.1, -0.03, -0.13, -0.04, -0.1, -0.04, -0.15), E = c(-0.17, -0.16, -0.08, -0.07, -0.09,

Is there a way to create a 'bubble plot' in R? For example, if we define the following data frame containing the level of y observed for 5 patients at three time points: time<-c(rep('time 1',5),rep('time 2',5),rep('time 3',5))