similar to: Changing the x-axis labels in plot()

Quite a simple one really! When I run boxplot(), the labels on the X axis are horizontal, and I want them vertical. So I did: par(las=3) boxplot(...) And my labels just aren't there anymore.... Any help???

Hi When I use plot(hclust(dist..)...)...) etc to create a dendrogram of a hierarchial cluster analysis, I end up with a vertical tree. What do I need to do to get a horizontal tree? Also, my users are used to seeing trees who's leaves all "end" at the same place (eg. Like in minitab). Is this possible in R? Thanks Mick Michael Watson Head of Informatics Institute for Animal

Hi First a simple question to do with plot(). How do I change the x-axis labels on a plot? For example, I am plotting each row of a matrix, and I want each row to be a line on my graph. Simple really. Eg: plot(sg[1,], type="l") When I do this, the x-axis is labelled 1:38, as I have 38 columns in my matrix. When I do: plot(sg[1,order(sg[1,])], type="l") Ideally I would

Hi I have a function which draws a plot, using the layout() function to divide the screen into two parts. The function works fine, but then my next call to plot() draws the plot in the first section of the plot I've just drawn using layout() - whereas what I want it to do is create a new plot. I tried using dev.off() but that just closes the layout plot window, which is not what I want. So

Hi I want to draw a rectangle behind a barplot such that it highlights those particular bars from the rest of the plot. I have figured out how to draw a barplot(), and then how to draw a rectangle of the correct shape and size (using rect()), but when I use rect() it draws over the top of the bars, and then when I re-draw the bars, it draws with a white background, thus eliminating my rectangle!

Hi Sorry to ask such a basic question. I have a list, each element of which is a vector of two values. What I actually want is a matrix with two columns, and one row per element of the list. Obviously I have tried as.matrix(), and as.vector() but I didn't expect the latter to work. I feel so lame asking this. Any suggestions? Mick

Hi Thanks once again for your help, I do appreciate it..... however.... Here is what I get with your test.... (under tcsh - i normally use bash, but I will keep everything the same) users/mwatson> env http_proxy=http://wwwcache.bbsrc.ac.uk:8080/ R >options(internet.info=0) >update.packages() trying URL `http://cran.r-project.org/src/contrib/PACKAGES' unable to connect to

Hi When I give the command: > heatmap.2(as.matrix(d),Rowv=as.dendrogram(hc.gene),Colv=FALSE,scale="row ",trace="none",col=greenred.colors(79)) The resulting heatmap has re-ordered my columns! This is time-course data, and I don't want my columns re-ordered! Note from the help: Rowv: determines if and how the _row_ dendrogram should be reordered.

Hi Forgive my ignorance. I am selecting a column of a data.frame using the column name, and I want to know what the resulting column "is". My data frame is called "submin" and the column name is held in a variable called "display.gname" Eg: > is.data.frame(submin) [1] TRUE > is.data.frame(submin[,display.gname]) [1] FALSE >

Hi I did a google search on this and came up with nothing. OK, in the latest release of R for windows, I want to change the gui to SDI mode. Edit -> GUI preferences Choose SDI Apply I get the message about properties not being able to be changed on a running console, and that I need to save. So I click Save, and am met with a "Save As" dialogue box with the default file called

Hi I have a complicated plot which has a potentially large legend. What I want to do is actually create the legend in a new window. Has anyone done this before? I'd like to be able to create a window with just the legend in it, and have it so the window is just the right size etc. I'm sure someone must have done this already? If not, any tips would be welcome. Thanks Mick

Hi We have touched on this before, but I don't think I quite got it right. So I have a list, each element of which is a a vector of 2 numbers: > l2 $cat000_a01 [1] 0.3429944 4.5138244 $cat000_a02 [1] 0.1929336 4.3064944 $cat000_a03 [1] -0.2607796 4.1551591 What I actually want to convert this into is a matrix with the names (cat000_a01 etc) as row names, the first element of each of

Hi I have data from 12 subjects. The measurement is log(expression) of a particular gene and can be assumed to be normally distributed. The 12 subjects are divided into the following groups: Infected, Vaccinated, Lesions - 3 measurements Infected, Vaccintaed, No Lesions - 2 measurements Infected, Not Vaccinated, Lesions - 4 measurements Uninfected, Not Vaccinated, No Lesions - 3 measurements

Hi I've been reading the help for by and aggregate but can't get my head round how to do this. I have a data frame - the first three columns are replicate measurements, then the next 3 are replicates etc up to 36 (so 12 variables with 3 replicate measurements each). I want to compute the mean for each of the 12 variables, so that, for each row, I have 12 means. A grouping variable

Hi I can't get R to connect to the internet. I am running R 1.7.1 on Windows XP and whenever I try to download packages etc from within R using the internet, it fails. OK so I am behind a firewall and use a proxy server.... SO, if I go to my MS-DOS prompt and type: RGui.exe --internet2 everything works.... BUT i can't set up a shortcut for this as Windows (I hate Windows) complains

Hi This is a difference between the way matrices and data frames work I guess. I want to replace the NA values in a data frame by 0, and the code works as long as the data frame in question actually includes an NA value. If it doesn't, there is an error: df <- data.frame(c1=c(1,1,1),c2=c(2,2,NA)) df[is.na(df)] <- 0 df df <- data.frame(c1=c(1,1,1),c2=c(2,2,2)) df[is.na(df)] <-

Hi Sorry if this is in the help :-S I've looked at example(dendrogram) and though it gives some indication of what I want, it doesn't do all. OK, so here is what I want to do: draw a tree, and then have an action, on user-click, to either draw a sub tree or a plot of the data. I also want users to be able to search for a particular label and have it highlighted on the tree, say in

I think that title makes sense... I hope it does... I have a data frame, one of the columns of which is a factor. I want the rows of data that correspond to the level in that factor which occurs the most times. I can get a list by doing: by(data,data$pattern,subset) And go through each element of the list counting the rows, to find the maximum.... BUT I can't help thinking there's

Hi Now a more theoretical question. I have two correlation matrices - one of a set of variables under a particular condition, the other of the same set of variables under a different condition. Is there a statistical test I can use to see if these correlation matrices are "different"? Thanks Mick

Hi Windows XP, R 2.0.1. I am drawing a very large barplot using jpeg() - setting the width to 10,000 as there are over 5000 bars. This all works fine and I get exactly what I want - except there is a huge bit of white space between the Y-axis and the first bar - so much in fact that I have to scroll two screens from the Y-axis before I see the first bar. After that the bars are evenly spaced