similar to: My first package

On 18 Jan 2014, at 14:31, Axel Urbiz <axel.urbiz at gmail.com> wrote: > Hi All, > > > I'm planning to submit my first package to R, and although I read all the > documentation, I'm not very clear on the following 2 items, from which I'd > appreciate your guidance: > > > 1)I understand it is suggested to use the R dev version to build the >

HI All, I have written and succesfully uploaded a new package. The licence it is under is 'GPL' --no version. My assumption is, since all the code is written in R the licence R used for R would affect the code (hence my "GPL" stands for "whatever version of the GPL R is under") I am happy with the licencing I used, but I'd like to ask if there is any transitive

On 20 Jan 2014, at 14:51, G?bor Cs?rdi <csardi.gabor at gmail.com> wrote: > On Mon, Jan 20, 2014 at 4:31 AM, Federico Calboli <f.calboli at imperial.ac.uk> wrote: > [...] > Is it? this is news to me. I have a grand total of 2 packages up and I never ever used R-devel, and never ever had a problem, had a report of a problem or had a note from CRAn about my packages not being

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, 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 am counting the number of occurrences of the terms listed in one vector in another vector. My code runs: for( i in 1:length(vector3)){ vector3[i] = sum(1*is.element(vector2, vector1[i])) } where vector1 = vector containing the terms whose occurrences I want to count vector2 = made up of a number of repetitions of all the elements of vector1 vector3 = a vector of NAs that is

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 would like to output the results of a function into a text file, legible as a such. The function produces a summary quite like: summary(lm(x ~ y + w * z)) [for instance] and I am not clear how to save this summary into a text file 'automagically', because I need to be able to do it in a for() loop. Cheers, Federico -- Federico C. F. Calboli Department of Epidemiology

Dear All, just a quick example: > x = 1:25 > x[12] = NA > x [1] 1 2 3 4 5 6 7 8 9 10 11 NA 13 14 15 16 17 18 19 20 21 22 23 24 25 > y = x[x<10] > y [1] 1 2 3 4 5 6 7 8 9 NA Is there any way of NOT getting NA for y = x[x<10]? Similarly > y = x[x<15] > y [1] 1 2 3 4 5 6 7 8 9 10 11 NA 13 14 How do I get rid of the NA (not post

Dear All, I will soon be working with NIH and possibly FDA. Will I be able to use R or will I be forced to use SAS? 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 75941602 Fax +44 (0)20 75943193 f.calboli [.a.t] imperial.ac.uk f.calboli [.a.t] gmail.com

Hi All, I would like to know, is there a *ballpark* figure for how many parameters the minimisation routines can cope with? I'm asking because I was asked if I knew. 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 75941602 Fax +44 (0)20 75943193 f.calboli [.a.t]

Hi All, I'm having a problem with barplot: mydata [1,] 2 108 0 0 0 1 3 0 0 0 0 0 7 18 3 4 8 20 26 20 19 7 1 1 mycol = c(rep('yellow', 2), rep('white', 3), rep('orange',2), rep('white', 5), rep('orange',3), rep('red',9)) barplot(mydata, col = mycol) gives me an uniformly yellow barplot. How do I solve this? bw Federico

Hello all, a friend has a problem with tiff() which I was unable to help about. I searched the error messages to no avail. When he tries: tiff(filename = "FedeWhyDoesntThisBloodyWork.tif", width = 5, height = 5, units = "cm", bg = "white", res = 1200) Error in tiff(filename = "FedeWhyDoesntThisBloodyWork.tif", width = 5, : unable to start device

Hi All, is there a way of predicting memory usage? I need to build an array of 86000 by 2500 numbers (or I might create a list of 2 by 2500 arrays 43000 long). How much memory should I expect to use/need? Cheers, Fede -- 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 75941602 Fax +44

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 All, I have some data where the dependent variable is a score, low (1:3) or high (8:9), and the independent variables are 21 genotypic markers. I'm fitting a logistic regression on the whole dataset after transforming the score to 0/1 and normal linear regression on the high and low subsets. I all cases I have a numer of cases of data 'duplications', i.e. different

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 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 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, 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