similar to: R unable to run on Mac OS 10.4 Tiger

Dear All, I saved by mistake the environment I was working in after typing q(), and now I get the annoying message: [Previously saved workspace restored] I have already deleted all the objects in the environment, saving it as an empty environment, so it's just a matter of nitpicking I suppose. The message does not appear if I start R from any other place in the directory tree. I am

Hi All, I have a list of matrices: > x [,1] [,2] [1,] 1 4 [2,] 2 5 [3,] 3 6 > y [,1] [,2] [,3] [,4] [,5] [,6] [1,] 18 21 24 27 30 33 [2,] 19 22 25 28 31 34 [3,] 20 23 26 29 32 35 > z =list(x,y) I want to create a second list that is has a subset each matrix in the list subsetting so I get the 2nd and 3rd row of each (and

Hi All, I have a vector of data, a vector of bin breakpoints and I want to put my data in the bins and then extract fanciful informations like the mean value of each bin. I know I can write my own function, but I would have thought that R should have somewhere a function that took as arguments something like (data, breaks, what to do with the data in the bins). I surey could not find it

Hi All, a colleague wants to calculate the Fisher information matrix for a model he wrote (not in R). He can easily get the neg-log-likelihood and the best fit parameters at the minimum. He can also get negLLs for other parameter values too. Given these data, is there a way in R to calculate the Fisher information matrix? Best, Federico -- Federico C. F. Calboli Department of Epidemiology

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 All, I have a data frame such as: > test x y p d [1,] 1 0 10 21 0 [2,] 2 3 11 12 0 [3,] 3 4 12 23 0 [4,] 3 5 13 24 0 and I want to perfor some operations on the first two coulums, conditional on the uneqaulity values on the 3rd and 4th columns. For instance: j = 3 test[test[,1] == j, 5] = test[test[,1] == j,2] + test[test[,2] == j,1] gives me the result: test: x y p d

Hi You could try to find an equivalent representation as a string and try to match those. > (A <- cbind(sample(1:2, 10, rep=TRUE), sample(1:2, 10, rep=TRUE))) [,1] [,2] [1,] 1 2 [2,] 1 2 [3,] 1 2 [4,] 2 2 [5,] 1 1 [6,] 1 2 [7,] 1 2 [8,] 1 1 [9,] 1 2 [10,] 1 1 > (B <- unique(A)) [,1] [,2] [1,] 1 2

Hi I am struggling with nested random effects and hope someone can help. I have individuals (ID) who are nested within families (FAM). I want to model an outcome variable, and take account of the intercorrelation of individuals within each family. I think this amounts to two random effects, one nested within the other. How can I model this in R? So far I have tried using the

Hi All, is there an equivalent of the Unix command 'less' (or 'more'), so I can look at what's inside a data.frame or a matrix without having it printed out on console? I am using R on Debian Linux and Mac OS 10.4.5 Cheers, F -- Federico C. F. Calboli Department of Epidemiology and Public Health Imperial College, St Mary's Campus Norfolk Place, London W2 1PG Tel +44

Hi All, I have the following problem: x = c(1,2) x [1] 1 2 as.matrix(x) [,1] [1,] 1 [2,] 2 BUT, if I add: y = c(3,4) as.matrix(rbind(x,y)) [,1] [,2] x 1 2 y 3 4 It does not transpose. Since I will need as.matrix() for a list of data that is in one or more lines, I need as.matrix to behave in a consisten fashions, so I get as.matrix(x, whatever) [,1] [,2] x 1

Hi All, I am using plot(x, type = 'l') for some plotting, but I would like rounded edges rather than jagged edges in the plot (purely for aestetic reasons). How could I achieve that? Cheers, Federico -- Federico C. F. Calboli Department of Epidemiology and Public Health Imperial College, St Mary's Campus Norfolk Place, London W2 1PG Tel +44 (0)20 7594 1602 Fax (+44) 020

Hi All, I am (almost) successfully using apply() to apply a function recursively on a data matrix. The function is question is as.genotype() from the library 'genetics' apply(subset(chr1, names$breed == 'lab'),2,as.genotype,sep ="") Unfortuantely apply puts it's results into a matrix object rather than a data frame, tranforming my factors into numerics and

Hi everyone, I would like to find out when and where exactly I get the following warning in a piece of code I've written: Error in "[<-"(`*tmp*`, iseq, value = numeric(0)) : nothing to replace with The code is a for () loop performing a somewhat trivial calculation, modulated by a number of logical if(){} else(){} conditions, involving the creation of a number of

Hi All, is there a function in R that allows me to work with fractions without transforming them to floats (or whatever) in between? Something that would calculate something like: (1/2 + 1/8) * 1/2 = 5/16 without ever transforming to 0.5 and 0.125? Best, Federico -- Federico C. F. Calboli Department of Epidemiology and Public Health Imperial College, St Mary's Campus Norfolk Place,

Hi everyone, I am confused on how to specify some nesting and interaction terma with lme(). I have a dataset where some flies where selected for accessory gland size, made to mate in presence/absence of another male and the level of some protein measured. Now the complex stuff. The selection has been replicated twice, so that the selection term has got two levels (large and small) with

Hi All, I am looking for references on R as a programming language (apart form the standard R-lang.pdf and the other manuals), reference that would cover _in_depth_ things like loops, code optimisation, debugging tools etc... and is as up-to-date as possible. Can anyone suggest any book or other reference apart from the "green book" and the V&R "S-programming"? Cheers,

Hi All, is there any chance of vectorising the two ifelse() statements in the following code: for(i in gp){ new[i,1] = ifelse(srow[i]>0, new[srow[i],zippo[i]], sample(1:100, 1, prob =Y1, rep = T)) new[i,2] = ifelse(drow[i]>0, new[drow[i]>0,zappo[i]], sample(1:100, 1, prob =Y1, rep = T)) } Where I am forced to check if the value of drow and srow are >0 for each line... in

Hi All, I cannot upgrade r-base and r-recommended because the latest (latest as in it was not listed as 'upgradable' yesterday but today is) r-base-cose is 2.7.1-2hardy0 not hardy1: r-base: Depends: r-base-core (>=2.7.2-1hardy1) but 2.7.1-2hardy0 is to be installed Depends: r-recommended (=2.7.2-1hardy1) but 2.7.1-2hardy0 is to be installed r-base-core: r-recommended:

Hello, I am on the latest Ubuntu 10.04 and I am upgrading R. I get the following error: r-base-core-dbg: Depends: r-base-dev (=2.11.1-2lucid0) but 2.11.1-1lucid0 is to be installed If that helps my CRAN mirror is: http://stat.ethz.ch/CRAN/bin/linux/ubuntu lucid/ I did check on the archives for any email concerning the problem, I hope I am not duplicating a thread. Best, F -- Federico

Dear All, I'm using apply to do some genetic association analysis along a chromosome, with many thousands markers. For each marker the analysis is the same, so I was planning to use apply(chrom, 2, somefunction) In the specific case I do: my.results = apply(chr, 2, function(x){anova(lrm( cpstc.f ~ x + time.cpstc + age + sex + mri))[1,3]}) This is all good and well in theory, but in