similar to: PNG files and the png function

platform i586-pc-linux-gnu arch i586 os linux-gnu system i586, linux-gnu status major 1 minor 4.1 year 2002 month 01 day 30 language R I can't seem to get rid of the slightly green background when I use dotplot(). It shows on the screen in the

The result I'm aiming to achieve is a bitmap that can be imported into a PowerPoint file that shows what's behind the lines of the plot. There's a way in PowerPoint that almost works. By choosing a colour to set as transparent, what is behind the graphic is indeed visible, but it's at the expense of losing line and text definition. I notice there have been discussions about

I've been able to do most things I'd like to using the map package, but I can't figure out how to fill a region without drawing a black outline. I can draw the outline in a specified colour, but as soon as I set fill to TRUE, I get a black outline in addition. My map has a lot of activity on the coastline so I don't want a black line obscuring what I do there. A pale colour to

I'm baffled. When I run make check after installing from source, I get a Error 2. From my understanding of how these things work, it would appear to be coming from this (as at the end of base-Ex.Rout.fail: > has.VR <- require(MASS, quietly = TRUE) Attaching package 'MASS': The following object(s) are masked from package:base : confint confint.lm nclass.FD nclass.scott

I have a square matrix wherein a '*' indicates an HSD between the levels indicated by row name and column name. The '.' is simply marking the diagonal. A blank indicates the same group A B C D E F G H I J K L M N A . B . C * * . D * * . E * * * . F * * * .

I've been looking at the kinship package which looks as though it might be appropriate for my purposes. What I can't find is any reference to the data that is used in the example code. A dataframe called d10 with column names, upn, dadid, momid, sex and affect is required. One can get an idea of what sort of values should be in most columns from the description in the pedigree function,

Of course, that's as trivial as it gets on the command line, but I can't work out how to get a column of numbers that are entered as "10^5" from its 'character' format into a numeric one? I feel a bit embarrassed asking such a simple question. Too much Easter.... Thanks -- Patrick Connolly HortResearch Mt Albert Auckland New Zealand Ph: +64-9 815 4200 x 7188

Some years ago when I used S-PLUS, I seem to remember, there was a discussion about a simple method of removing trailing spaces from vector elements. I'd be fairly sure the same would work in R, but I can't find any mention of anything like it in the R archives or with help.search(). There are ways I could do it with substring(), but I seem to remember there was something more elegant.

I seem to remember this was discussed a year or two ago, but I can't find it in the archives. platform i686-pc-linux-gnu arch i686 os linux-gnu system i686, linux-gnu status major 1 minor 7.0 year 2003 month 04 day 16 language R

I've been asked about "Canonical Variance Analysis" (CVA). I don't see any reference to it searching the R site. Does it go by other names? Genstat describes it thus: Canonical variates analysis operates on a within-group sums of squares and products matrix, calculated from a set of variates and factor that specifies the grouping of units. It finds linear combinations of the

V&R describes binomial GLMs with mortality out of 20 budworms. Is it appropriate to use the same approach with mortality out of numbers as low as 3? I feel reticent to do so with data that is not very continuous. There are one continuous and one categorical independent variables. Would it be more appropriate to treat the response as an ordered factor with four levels? If so, what family

Hi, Now suppose I have just one list called FOO, which has 25 objects, e.g.: [[1]] 1 2 3 4 5 [[2]] 6 7 8 9 10 . . . And I want to do something like: FRED <- data.frame(cbind(unlist(FOO[[1]]), unlist(FOO[[2]]), # ... for all 25 subsets )) Is it possible to do this, without doing unlist(FOO[[i]]) 25

I like the fact that the postscript function enables the possbiility of a more useful title than before. However, I'd prefer the default to be the file name. It's very simple for me to make my own postscript function that does just that simply by setting title = file. I always use onefile = TRUE, so it always works (so far). However, I'm a little reluctant to do that in case some

I like the fact that the postscript function enables the possbiility of a more useful title than before. However, I'd prefer the default to be the file name. It's very simple for me to make my own postscript function that does just that simply by setting title = file. I always use onefile = TRUE, so it always works (so far). However, I'm a little reluctant to do that in case some

I investigated why I was getting this error message... Error in Ops.factor(groups[subscripts], vals[1]) : Level sets of factors are different which led me to putting a browser in panel.superpose: Called from: panel.superpose(x, y, subscripts, ...) Browse[1]> vals vals [1] 1 2 3 Levels: 1 2 3 Browse[1]> groups[subscripts] groups[subscripts] [1] 1 2 3 1 2 1 2 3 1 2 3 1 2 3 Levels: 1 2

Say we have a tab delimited file called bug.txt Part Rep Cage Hb pupae 1 1 S 32 1 M 34 L 42 2 S 36 M 28 L 36 read.delim("bug.txt") Part Rep Cage Hb.pupae 1 1 1 S 32 2 NA 1 M 34 3 NA NA L 42 4 NA NA NA 5 NA NA NA 6 NA 2 S 36 7 NA NA M 28 8 NA NA L 36 >

I'm still having trouble getting to grips with time classes. I wish to calculate the difference in days between events. Browse[1]> insp.j$First [1] "2002-02-19 13:00:00 NZDT" Browse[1]> spray.j$Date [1] "2001-11-29 13:00:00 NZDT" Browse[1]> insp.jk - spray.j$Date Time difference of 82 days If I save insp.jk to a vector, I get a nice useful value of 82.

I'm asking on this list because the question will be more easily understood than on a Perl list. Lots of talented people using R also know how to use Perl, so it's easier to ask them. Running a Perl script with the -d switch can do some pretty neat things, but as far as I can tell, every line has to be done individually. Is there a way to specify where to stop in the way browser() does

I'm sure there's a simple way to do this, but I can only think of complicated ones. I have a number of character vectors that look something like this: "12 78 23 9 76 43 2 15 41 81 92 5(92 12) (81 78 5 76 9 41) (23 2 15 43)" I wish to get it into a list of numerical vectors like this: $Group [1] 12 78 23 9 76 43 2 15 41 81 92 5 $Subgroup1 [1] 92 12 $Subgroup2 [1] 81 78 5

Dear R friends My goal is to eliminate this specific group(1) if the # of NAs in this group greater than 50%(specifically say greater than 3). Would you please show me how to do it. I have a sample data as following: Thanks a lot. Kevin Lin y group f1 f2 f3 30 NA 1 1 1 1 27 NA 1 1 2 2 48 NA 1 2 1 2 40 -0.6066416 1 2 2 1 24