similar to: coxph and completely missing strata/subsetting

Hi everyone, I'm running a cox ph model on a dataset with a number of variables. Each variable has a different number of missing data, so that coxph() drops the individuals who are missing data at one or more variables. Because of this dropping (totally fine btw) I want to know how many events I am left with in the model. Is there a way of extracting them from the coxph() fit? or in any other

Hi, I have a coxph model like coxph(Surv(start, stop, censor) ~ x + y, mydata) I would like to calculate the Schoenfeld residuals for the null, i.e the same model where the beta hat vector (in practical terms, the coeff vector spat out by summary()) is constrained to be all 0s --all lese stays the same. I could calculate it by hand, but I was wondering if there is a way of doing it with

I am writing to get a better handle on a warning I am getting from a coxph analysis I am doing. I am analysing age of onset of dementia *after* the onset of parkinson disease. My data looks like: age.park age.dem age.death censor x1 x2 x3 x4 1 76 87 88 0 16 33 E3 E3 2 75 84 84 0 33 36 E3 E3 3 77 81 81 1 NA NA

Hi All, I am trying to figure out how to get the position of the knots in a pspline used in a cox model. my.model = coxph(Surv(agein, ageout, status) ~ pspline(x), mydata) # x being continuous How do I find out where the knot of the spline are? I would like to know to figure out how many cases are there between each knot. Best, Federico -- Federico C. F. Calboli Department of Epidemiology

Hi All, I have a left truncated, right censored cox model: coxph(Surv(start, stop, censor) ~ x + y, mydata) I would like to know how much of the observed variance (as a number between 0 and 1) is explained by each variable. How could I do that? Adding terms sequentially and then using anova(mod1, mod2) tells me whether I get a significant improvement of the fit, but does not tell me how much

Hi, I am subsetting a matrix thus: test [,1] [,2] [,3] [1,] 1 7 13 [2,] 2 8 14 [3,] 3 9 15 [4,] 4 10 16 [5,] 5 11 17 [6,] 6 12 18 test[cbind(c(1,3,5), c(2,1,3))] [1] 7 3 17 This works fine, and is the equivalent of c(test[1,2], test[3,1], test[5,3]). cbind(c(1,3,5), c(2,1,3)) would obviously look like: [,1] [,2] [1,] 1 2 [2,]

Hello, I'm trying to use the function svyglm in the library survey. I create a data survey object: data_svy<- svydesign(id=~PSU, strata=~sample_domain, weights=~sample_weight, data=data, nest=TRUE) and I try to use svyglm() with little success: R<-svyglm(data_svy[,4]~(data_svy[,iCol]==listModality[[iVar]] [iMod]),design=data_svy, family=binomial(link="logit") Error in

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

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

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, does anyone know how to import binary .bed files generated by Plink (http://pngu.mgh.harvard.edu/~purcell/plink/ ) into R? the Plink FAQ explains how to conver other types of files, not the .bed. 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

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

Hi All, I have the following adjacency matrix for a directed graph: [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [1,] 0 0 0 0 0 0 0 0 [2,] 0 0 0 0 0 0 0 0 [3,] 1 0 0 0 0 0 0 0 [4,] 0 0 1 0 0 0 0 0 [5,] 0 0 1 0 0 0 0 0 [6,] 1 1 0 0 0 0 0 0 [7,] 0 0

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

Hi All, is there a way of using strsplit with a forward slash '/' as the splitting point? For data such as: 1 T/T C/C 16/33 2 T/T C/C 33/36 3 T/T C/C 16/34 4 T/T C/C 16/31 5 C/C C/C 28/29 6 T/T C/C 16/34 strsplit(my.data[1,1], "/") # and any variation thereof Error in strsplit(apoe[1, 1], "/") : non-character

Hi All, is there any general advice about speeding up recursive functions (not mentioning 'don't use them')? Regards, Federico Calboli -- 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]

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 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

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, does anyone ever import old SPSS files in a sl3 format? read.spss('file.sl3') does not seem to work... it's not recognised as a supported SPSS format at all. Best, 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 (0)20 75943193 f.calboli