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Thanks Dimitris!!! That's much clearer now. Still have a lot of work to do this weekend to understand every bit but your code will prove very useful. Cheers, Aziz -----Original Message----- From: Dimitrios Rizopoulos [mailto:Dimitris.Rizopoulos at med.kuleuven.be] Sent: May 12, 2006 4:35 PM To: Chaouch, Aziz Subject: RE: [R] Maximum likelihood estimate of bivariate vonmises-weibulldistribution look at the following code: library(copula) par(mfrow = c(2, 2)) x <- mvdc(normalCopula(sin(0.5 * pi /2)), c(&qu...

...string contains "Ass" but does not contain "Ass.s", so for 'x' that would be positions 1 and 3. I guess this could be programmed around grep() using a suitable regular expression, but I haven't managed to succeed. Thanks in advance. Best, Dimitris -- Dimitris Rizopoulos Assistant Professor Department of Biostatistics Erasmus University Medical Center Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands Tel: +31/(0)10/7043478 Fax: +31/(0)10/7043014 Web: http://www.erasmusmc.nl/biostatistiek/

[My previous message rejected, therefore I am sending same one with some modification] I have 3 vectors with object name : dat1, dat2, dat3 Now I want to create a loop, like : for (i in 1:3) { cat(sd(dati)) } How I can do this in R? Regards,

Dear Group, I have a following dataset: > a A B C D 1 22 3 31 40 2 26 31 36 32 3 3 7 49 16 4 24 40 27 26 5 20 45 47 0 6 34 43 11 18 7 48 48 24 2 8 3 16 39 48 9 20 49 7 21 10 17 36 47 10 > dput(a) structure(list(A = c(22L, 26L, 3L, 24L, 20L, 34L, 48L, 3L, 20L, 17L), B = c(3L, 31L, 7L, 40L, 45L, 43L, 48L, 16L, 49L, 36L), C = c(31L, 36L, 49L, 27L, 47L, 11L, 24L,

...R uses chi-sqaured distribution as below: PVAL <- pchisq(STATISTIC, PARAMETER, lower = FALSE) but still I cant figure out why they are using this pschisq insted of dchisq. Sorry I am wrong!! Thanking you truly Prasanna > >----- Original Message ----- >From: "Dimitris Rizopoulos" <dimitris.rizopoulos at med.kuleuven.ac.be> >To: "Prasanna Balaprakash" <pbalapra at ulb.ac.be> >Cc: <r-help at stat.math.ethz.ch> >Sent: Friday, January 21, 2005 1:05 PM >Subject: Re: [R] chi-Squared distribution > > >> if you check ?qchis...

Hi all. I have a numerical function f(x), with x being a vector of generic size (say k=4), and I wanna take the numerically computed gradient, using deriv or numericDeriv (or something else). My difficulties here are that in deriv and numericDeric the function is passed as an expression, and one have to pass the list of variables involved as a char vector... So, it's a pure R programming

...symbols in the plot. pch = as.character(trial_no) works fine, but truncates the trial number to the first digit. As I have sixteen trials in the series I get into problems.... How do I squeeze in two positions as a symbol in a plot? All the best /CG On Thu, November 9, 2006 9:30 am, Dimitris Rizopoulos said: > try the following: > > x <- runif(100, -4, 4) > y <- 1 + 2 * x + rnorm(100, sd = 2) > fit <- lm(y ~ x) > > plot(x, y) > abline(fit) > legend("topleft", expression(paste(R[adj]^2, " = 0.66"))) > > ## or > > plot(x, y) >...

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 ) ------------------------------------------------------------------------

...the '...' argument in this case? > version _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor 2.1 year 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...

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]]

Dimitris Rizopoulos wrote: > look at ?rowMeans; you can also use "apply(mat, 1, mean)" but > rowMeans() is better. By my reading of the question, this is not what Ezhil wants. He said: ``I have a 992 x 74 matrix. I would like to form a new matrix by averaging each 4 rows from the original o...

Hai, Can anyone help me with some literature (R related) about latent models? Thanx, Wilfred

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

Dear all, I am trying to create a vector name, for example tmax.195012 from tmax., 1950 and 12. Obviously I don't wish to simply type it because the 3 name components are changing in each iteration within a loop. Is there any way of concatenating those 3 components (which are a mixture of numbers and letters)? Thanks for reading, Panos

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.

Dear R helpers Suppose x  <- c(1:3) y  <- matrix(1:12, ncol = 3, nrow = 4) > y      [,1] [,2] [,3] [1,]    1    5    9 [2,]    2    6   10 [3,]    3    7   11 [4,]    4    8   12 I wish to multiply 1st column of y by first element of x i.e. 1, 2nd column of y by 2nd element of x i.e. 2 an so on. Thus the resultant matrix should be like > z      [,1]   [,2]    [,3] [1,]    1   

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

Hello, I'm trying to superimpose a line plot onto a histogram but I'm not having any luck. I've attached the dataset. What I did was: > hist(data,freq=F) Now I'm trying to superimpose the following points with a line connecting them onto the histogram: x y 100 0.535665393824959 200 0.212744329736556 300 0.0844933242968584 400 0.0335572838043417 500

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.

Hi Everybody, I have a matrix of about 45 columns. Some of the rows contain zeros. Using >data1<-data[complete.cases(data),], I can remove the "NA" rows. But I am unable to tackle that of zeros. Can anybody give me an idea of how to remove rows containing zeros in a matrix. Thanks so much Best Ogbos [[alternative HTML version deleted]]