similar to: multiple summation

>>>>> jing hua zhao >>>>> on Mon, 24 Jun 2019 08:51:43 +0000 writes: > Hi All, > Thanks for all your comments which allows me to appreciate more of these in Python and R. > I just came across the matrixStats package, > ## EXAMPLE #1 > lx <- c(1000.01, 1000.02) > y0 <- log(sum(exp(lx))) > print(y0) ## Inf

Hi Hadley, Thanks for the counterpoint. Response below. On Thu, May 16, 2019 at 1:59 PM Hadley Wickham <h.wickham at gmail.com> wrote: > The existing behaviour seems inutitive to me. I would consider these > invariants for n vector x_i's each with size m: > > * nrow(rbind(x_1, x_2, ..., x_n)) equals n > Personally, no I wouldn't. I would consider m==0 a degenerate

Dear All, I am looking for: A software to select the lag length for a lag window spectral estimator. Also, I have a small query in the reprex given below. Background for the above, from the book by Percival and Walden: 1. We are given X_1,...,X_n which is one realization of a stochastic process. 2. We may compute the periodogram using FFT, for example by the function spectrum in R. 3. The

Dear All, Suppose I have a vector X = (x_1, x_2, ...., x_n), X_sort = sort(X) = (x_(1), x_(2), ... , x(n) ), and I would like to know the original position of these ordered x_(i) in X, how can I do it? case 1: all values are unique x <- c( 3, 5, 4, 6) x.sort <- sort(x) # # I would like to obtain a vector (1, 3, 2, 4) which indicates that 3 in x is still the 1st element in x.sort, 5 is at

If I use the Ineq library and the Gini function in this way: >Gini(c(100,0,0,0)) I obtain the result 0.75 instead of 1 (that is the perfect inequality). I think Gini's formula in Ineq is based on a formula as reported here: http://mathworld.wolfram.com/GiniCoefficient.html but in the case of perfect inequality: x_1=.......=x_n-1 =0 x_n>0 these formula are equal to 1 - 1/n, not to

Suppose one wanted to consider random variables X_1,...X_n and from each subtract off the piece which is correlated with the previous variables in the list. i.e. make new variables Z_i so that Z_1=X_1 and Z_i=X_i-cov(X_i,Z_1)Z_1/var(Z_1)-...- cov(X_i,Z__{i-1})Z__{i-1}/var(Z_{i-1}) I have code to do this but I keep getting a "non-conformable array" error in the line with the covariance.

Hi, I have to draw samples from an asymmetric-Laplace-Normal distribution: f(u|y, x, beta, phi, sigma, tau) \propto exp( - sum( ( abs(lo) + (2*tau-1)*lo )/(2*sigma) ) - 0.5/phi*u^2), where lo = (y - x*beta) and y=(y_1, ..., y_n), x=(x_1, ..., x_n) -- sorry for this huge formula -- A WinBUGS Gibbs sampler and the HI package arms sampler were used with the same initial data for all parameters. I

I have a set of n vectors, x_1, ..., x_n, of the same length. I would like to form the vector of dot products -- x_1'x_1, ..., x_n'x_n the fastest way I can think to do this is to put the vectors into a matrix and do diag(crossprod(X)) however, this seems to be very wasteful since this computes n(n+1)/2-n unnecessary dot products. Is there a better way using existing functions in R?

Dear all, The help page on the t distribution says: The most used applications are power calculations for t-tests: Let T= (mX - m0) / (S/sqrt(n)) where mX is the `mean' and S the sample standard deviation (`sd') of X_1,X_2,...,X_n which are i.i.d. N(mu,sigma^2). Then T is distributed as non-centrally t with `df'= n-1 degrees of freedom and non-centrality

Hallo, I would like to add the marginal distributions along the X and the Y axis to a scatter plot. Can anybody help me, please? Thank you, Paolo -- Paolo Bulla Istituto di Metodi Quantitativi Universit?? "L. Bocconi" viale Isonzo 25 20136 Milano paolo.bulla at unibocconi.it

>>>>> William Dunlap via R-devel >>>>> on Sun, 23 Jun 2019 10:34:47 -0700 writes: >>>>> William Dunlap via R-devel >>>>> on Sun, 23 Jun 2019 10:34:47 -0700 writes: > include/Rmath.h declares a set of 'logspace' functions for use at the C > level. I don't think there are core R functions that call

Hi Martin, Thanks for chiming in. Responses inline. On Fri, May 17, 2019 at 12:32 AM Martin Maechler <maechler at stat.math.ethz.ch> wrote: > >>>>> Gabriel Becker > >>>>> on Thu, 16 May 2019 15:47:57 -0700 writes: > > > Hi Hadley, > > Thanks for the counterpoint. Response below. > > > On Thu, May 16, 2019 at 1:59

Hi all, Apologies if this has been asked before (a quick google didn't find it for me),and I know this is a case of behaving as documented but its so unintuitive (to me at least) that I figured I'd bring it up here anyway. I figure its probably going to not be changed, but I'm happy to submit a patch if this is something R-core feels can/should change. So I recently got bitten by

The existing behaviour seems inutitive to me. I would consider these invariants for n vector x_i's each with size m: * nrow(rbind(x_1, x_2, ..., x_n)) equals n * ncol(rbind(x_1, x_2, ..., x_n)) equals m Additionally, wouldn't you expect rbind(x_1[i], x_2[i]) to equal rbind(x_1, x_2)[, i, drop = FALSE] ? Hadley On Thu, May 16, 2019 at 3:26 PM Gabriel Becker <gabembecker at

Hello anyone, I'm trying to install R on Mandriva 2006 distribution via rpm file with the line urpmi R-2.0.0-1mdk.i586.rpm but I got an error message saying that the file is not accessible due to some info problem. I also tried with a source code in R-2.1.1.tar but I think that there is some problem concerning the new version of gcc (4.x), so I downgraded it to gcc 3.4.5 but it does

Hi all, So far I only know one way to get the confidence limit for the Poisson distribution is to use the look-up table given by the 2 parameter (the number of observation x and the confidence level, e.g. 95%) and the table is limit by the maximum number of observations (x <= 50). I know the formula to compute the CI, however, mathematically it is not easy to do it. So, anyone know an R

Dear R users... I made the R-code for this double summation computation http://www.nabble.com/file/p19213599/doublesum.jpg ------------------------------------------------- Here is my code.. sum(sapply(1:m, function(k){sum(sapply(1:m, function(j){x[k]*x[j]*dnorm((mu[j]+mu[k])/sqrt(sig[k]+sig[j]))/sqrt(sig[k]+sig[j])}))})) ------------------------------------------------- In fact, this is

Does this give correct results for special floats (0, infs)? We tried to improve (for single floats) x86 rcp in llvmpipe with newton-raphson, but unfortunately not being able to give correct results for these two cases (without even more additional code) meant it got all disabled in the end (you can still see that code in the driver) since the problems are at least as bad as those due to bad

I'm new R-programming user, I need to use gam function. y<-gam(a~s(b),family=gaussian(link=log),data) y<-gam(loga~s(b), family =gaussian (link=identity),data) why these two command results are different? I guess these two command results are same, but actally these two command results are different, Why? -- View this message in context:

Dear R experts I need to write a function that incorporates double summation, the problem being that the upper limit of the second summation is the index of the first summation, i.e: sum_{j=0}^{x} sum_{i=0}^{j} choose(i+j, i) where x variable or constant, doesn't matter. The following code obviously doesn't work: f=function(x) {j=0:x; i=0:j; sum( choose(i+j,i) ) } Can you help? Thanks