similar to: Geometric mean of rows in matrix

Hi, I tried looking through google search on whether there's a way to computer the median for each row of a nxn matrix and return the medians for each row for further computation. And also if the number of columns in the matrix are even, how could I specify which median to use? Thank you very much! -- Edward Chen [[alternative HTML version deleted]]

Hi everybody, I'm using Windows XP Prof, R 2.5.1 and a Pentium 4 Processor. Now, I want to solve a quadratic optimization program (Portfolio Selection) with the quadprog package I want to minimize (\omega'%*%\Sigma%*%\omega) Subject to (1) \iota' %*% \omega = 1 (full investment) (2) R'%*%\omega = \mu (predefined expectation value) (3) \omega \ge 0 (no short sales). Where

Hi, I would like to generate a correlation matrix with a particular structure. For example, a 3n x 3n matrix : A_(nxn) aI_(nxn) bI_(nxn) aI_(nxn) A_(nxn) cI_(nxn) aI_(nxn) cI_(nxn) A_(nxn) where - A_(nxn) is a *specified* symmetric, positive definite nxn matrix. - I_(nxn) is an identity matrix of order n - a, b, c are (any) real numbers Many attempts have been unsuccessful because a

The code below create an nxn upper triangular matrix of one's. I'm stuck on finding a more efficient vectorized way - Thanks. --Dale n <- 9 data <- matrix(data=NA, nrow=n, ncol=n) data for (i in 1:n) { data[,i] <- c(rep(1,i), rep(0,n-i)) } data

?Dear All I have some data expressed in geometric means and 95% confidence intervals. Can I code them in metafor as: rma(m1i=geometric mean 1, m2i=geometric mean 2, sd1i=geometric mean 1 CI /3.92, sd2i=geometric mean 2 CI/3.92.......etc, measure="MD") All of the studies use geometric means. Thanks! Edward ---------------------------- [[alternative HTML version deleted]]

Dear list, I have got an array with observational values t and I would like to fit a geometric distribution to it. As I understand the geometric distribution, there is only one parameter, the probability p. I estimated it by 1/mean(t). Now I plotted the estimated density function by plot(ecdf(t),do.points=FALSE,col.h="blue"); and I would like to add the geometric distribution. This

Does anyone know how to approach R code for the MLE of a geom. distribution? thanks! [[alternative HTML version deleted]]

Hi all, I need to check for a difference between treatment groups in the parameter of the geometric distribution, but with a cut-off (i.e. right censored). In my experiment I stimulated animals to see whether I got a response, and stopped stimulating if the animal responded OR if I had stimulated 10 times. Since the response could only be to a stimulation, the distribution of response times

---------- Forwarded message --------- From: Sahil Sharma <sahilsharmahimalaya at gmail.com> Date: Tue, Oct 17, 2023 at 12:10?PM Subject: r-stats: Geometric Distribution To: <do-use-Contact-address at r-project.org> Hey I want to raise one issue in *r-stats **geometric distribution * function. I have found the dgeom(x,p) which denotes probability density function of geometric

Sorry to prolong a thread on something that is clearly off topic, but when Michael Meyer wrote >by using the geometric mean all asymptotic results no longer apply. that is flat our wrong. It's true that the geometric mean converges to something different that E[X], but it does indeed have an asymptotic distribution and one that makes sense in some contexts. There is no reason that

Hi, I have a bunch of data which is assumed to be instances of a geometric random variable with outliers. How can I do a robust estimation of the parameter p so that the effect of outliers is minimized? As a part of the estimation process, I also need to know which are the outliers in the data. I found glmrob which does robust estimation of Poisson and binomial random variables but not geometric

Dear useRs, I'd like to announce a new package called geozoo, short for geometric zoo. It's a compilation of functions to produce high-dimensional geometric objects, including hypercubes and hyperspheres, Boy's surface, the hyper torus and a selection of polytopes. For a complete list, as well as images and movies, visit

Hi All, I need to calculate the median for even number of data points.However instead of calculating the arithmetic mean of the two middle values,I need to calculate their geometric mean. Though I can code this in R, possibly in a few lines, but wondering if there is already some built in function. Can somebody give a hint? Thanks in advance [[alternative HTML version deleted]]

better posted on r-sig-ecology? -- or maybe even stack exchange? Cheers, Bert On Mon, Jan 22, 2024 at 7:45?AM Rich Shepard <rshepard at appl-ecosys.com> wrote: > A statistical question, not specific to R. > > I'm asking for a pointer for a source of definitive descriptions of what > types of data are best summarized by the arithmetic, geometric, and > harmonic >

Hello, Can anyone think of a non-iterative way to generate a decreasing geometric sequence in R? For example, for a hypothetical function dg, I would like: > dg(20) [1] 20 10 5 2 1 where I am using integer division by 2 to get each subsequent value in the sequence. There is of course: dg <- function(x) { res <- integer() while(x >= 1) { res <- c(res, x) x

Hi: I need to compute the geometric mean of a population. How i can do it?? thanks. Jorge M. A. Magalhae -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request

Hi, With R version 2.10.0 (2009-10-26) on Windows, I computed the p=1.e-20 quantile of the geometric distribution with parameter prob=0.1. > qgeom(1.e-20,0.1) [1] -1 But this is not possible, since X=0,1,2,... I guess that this might be a bug in the quantile function, which should use the log1p function, instead of the naive formula. Am I correct ? Best regards, Micha?l

By the Strong Law of Large Numbers applied to log(X) the geometric mean of X_1,...,X_n > 0 and IID like X converges toexp(E[log(X)]] which, by Jensen's inequality, is always? <= E[X] and is strictly less than E[X] except in trivial extreme cases. In short: by using the geometric mean all asymptotic results no longer apply. Michael Meyer [[alternative HTML version deleted]]

I want to make a function for geometric seqeunce since testing=function(x){i=1;ans=1;while(true){ans=ans+(1/x)^i ; i=i+1} ;return(ans)} doesn't work... the program is freeze... from my research, i know i should use iterators. I read iterators.pdf at http://cran.r-project.org/web/packages/iterators/iterators.pdf and didnt find it helps solving my problem at all... Is there any sources I

Dear Rusers, Trying to reduce my for loops addiction, could somebody tell me if there are ways to simplify (and perhaps accelerate ?) the following line for (i in 1:N) for (j in 1:N) m[i,j] = b[i]-b[j]; (where m is a NxN matrix and b a vector of length N) Thanks for any hint.