search for: 337015

...2005 month 12 day 20 svn rev 36812 language R Thanks in advance for any hints, Dimitris ---- Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://www.med.kuleuven.be/biostat/ http://www.student.kuleuven.be/~m0390867/dimitris.htm Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm

Hi all, My question is not really urgent. I can write a loop and solve the problem. But I know that I'll be in a similar situation many more times so it would be useful to find out the answer Is there a fast way to perform linear fit to all the columns of a matrix? (or in the one dimension of a multi-dimensional array.) I'm talking about many single linear fits, not about a multiple fit.

...kage. Can anyone advise me on how to proceed or better point me to examples of how its done. Regards, -- Luwis Diya, Leuven Biostatistics and Statistical Bioinformatics Centre (L-BioStat), Kapucijnenvoer 35 blok d - bus 7001, 3000 Leuven, Belgium Tel: +32 16 336886 or +32 16 336892 Fax: +32 16 337015 [[alternative HTML version deleted]]

...hope it helps. > > Best, > Dimitirs > > ---- > Dimitris Rizopoulos > Ph.D. Student > Biostatistical Centre > School of Public Health > Catholic University of Leuven > > Address: Kapucijnenvoer 35, Leuven, Belgium > Tel: +32/(0)16/336899 > Fax: +32/(0)16/337015 > Web: http://www.med.kuleuven.be/biostat/ > http://www.student.kuleuven.be/~m0390867/dimitris.htm > > > ----- Original Message ----- > From: "Chaouch, Aziz" <achaouch at NRCan.gc.ca> > To: "Dimitris Rizopoulos" <dimitris.rizopoulos at med.ku...

Hi R-users, How do I calculate a number of NA's in a row of every second column in my data frame? As a starting point: dfr <- data.frame(sapply(x, function(x) sample(0:x, 6, replace = TRUE))) dfr[dfr==0] <- NA So, I would like to count the number of NA in row one, two, three etc. of columns X1, X3, X5 etc. Thanks in advance Lauri [[alternative HTML version deleted]]

hi all: xlim and ylim are used to define the interval limits of a plot. I'm interested in the scale of values between this limits. suppose xlim=c(0,10) we can have e.g. 0 5 10 0 2 4 6 8 10 0 1 2 3 4 5 6 7 8 9 10 which is the parameter that allows me to modify this? thanks in advance alexandre

...; Dimitris >> >> ---- >> Dimitris Rizopoulos >> Ph.D. Student >> Biostatistical Centre >> School of Public Health >> Catholic University of Leuven >> >> Address: Kapucijnenvoer 35, Leuven, Belgium >> Tel: +32/16/336899 >> Fax: +32/16/337015 >> Web: http://www.med.kuleuven.ac.be/biostat >> http://www.student.kuleuven.ac.be/~m0390867/dimitris.htm >> >> >> ----- Original Message ----- >> From: "Prasanna Balaprakash" <pbalapra at ulb.ac.be> >> To: <r-help at stat.math.ethz....

...kidney) fit4 fit4$history[[1]]$theta Am I missing something? Thanks in advance for any pointers! Best, Dimitris ---- Dimitris Rizopoulos Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://med.kuleuven.be/biostat/ http://www.student.kuleuven.be/~m0390867/dimitris.htm Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm

....nb, data = quine) Any hints to overcome this are greatly appreciated. Thanks in advance, Dimitris ---- Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://med.kuleuven.be/biostat/ http://www.student.kuleuven.be/~m0390867/dimitris.htm Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm

Dear Rs: outer(1:3, 1:3, function(df1, df2) qf(0.95, df1, df2)) I compare this F distribution results with the table, the answers were perfect. But I need to see for chi-sqaured distribution. When I employed the similar formula outer(1:3, 1:3, function(df1, df2) qchisq(0.95, df1, df2)) , I am getting unexpected results. I need to see the following values: p=0.750 ..... 1 1.323

I have a vector that has 1,974 elements and each element is one of the following (B, F, N, Y). How do I recreate that vector accept in the place of N put 0 and in the place of B, F or Y put a 1? Thanks, Jacob [[alternative HTML version deleted]]

a simple question how does one produce plots on two different graphics devices? / allan

Hi how do I extract those components of a list that satisfy a certain requirement? If jj <- list(list(a=1,b=4:7),list(a=5,b=3:6),list(a=10,b=4:5)) I want just the components of jj that have b[1] ==4 which in this case would be the first and third of jj, viz list (jj[[1]],jj[[3]]). How to do this efficiently? My only idea was to loop through jj, and set unwanted components to NULL,

Hi Everyone, I have some data that predicts both a nominal and ordinal response variable. I was wondering what packages in R would help me analyze the data? I was also curious if anyone could recomend me some textbooks that would help with the analysis of such data? I have the 5th edition of "Applied Multivariate Statistical Analysis" by Richard A. Johnson and Dean W. Wichern

Hello there! My operating system is Windows XP, my version of R is the latest (R-2.5.0). Recently I have downloaded the package "mvtnorm" and a problem with the command "pmvnorm" occured. Trying to enter the lines ... A <- diag(3) A[1,2] <-0.5 A[1,3] <- 0.25 A[2,3] <- 0.5 pvmnorm(lower=c(-Inf,-Inf,-Inf), upper=c(2,2,2),mean = c(0,0,0), corr=A) I got the following

Hello R-ologists, Imagine you have a list "list" like so: >list [[1]] [1] "IPI00776145.1" "IPI00776187.1" [[2]] [1] "Something" "IPI00807764.1" "IPI00807887.1" [[3]] [1] "IPI00807764.1" [[4]] [1] "Somethingelse" What I need to achieve is a filtered list "list2" like so: >list2 [[1]] [1]

...; I hope it helps. > > Best, > Dimitris > > ---- > Dimitris Rizopoulos > Ph.D. Student > Biostatistical Centre > School of Public Health > Catholic University of Leuven > > Address: Kapucijnenvoer 35, Leuven, Belgium > Tel: +32/(0)16/336899 > Fax: +32/(0)16/337015 > Web: http://med.kuleuven.be/biostat/ > http://www.student.kuleuven.be/~m0390867/dimitris.htm > > > ----- Original Message ----- > From: "CG Pettersson" <cg.pettersson at evp.slu.se> > To: <r-help at stat.math.ethz.ch> > Sent: Thursday, November 0...

Hi everyone, I am new to R, but it's really great and helped me a lot! But now I have 2 questions. It would be great, if someone can help me: 1. I want to integrate a normal distribution, given a median and sd. The integrate function works great BUT the first argument has to be a function so I do integrate(dnorm,0,1) and it works with standard m. and sd. But I have the m and sd given.

I am trying to use the LME package to run a multilevel logistic model using the following code: ------------------------------------------------------------------------ ------------------------------------------- Model1 = GLMM(WEAP ~ TSRAT2 , random = ~1 | GROUP , family = binomial, na.action = na.omit ) ------------------------------------------------------------------------

Hi, after reading some R docs, I couldn?t figure out how can I find the solution for the following problem, therefore I would ask this friendly list for an advice. We?re making a least square approximation for an experiment described by the following model: T is the time, Y is some measured value. >From time=0 till time=U: Y = b + p*T >From time=U and on (some effect added): Y = b + p*T +