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Hola, Estoy intentando averiguar cómo contar el número de categorías situadas después de ciertos valores. Por ejemplo, en el siguiente vector: x <- c(3, "A", "B", 5, "A", 4, 5, "A", "A", 3) el resultado que quisiera obtener es: Valor -> Resultado 3 -> 1 A y 1 B 4 -> 0 A y 0 B 5 -> 3 A y 0 B ¿Alguien tiene alguna

Hello, I have to compute a multiple summation (not an integration because the independent variables a are discrete) for all the values of a function of several variables f (x_1,...,x_n), that is sum ... sum f(x_1,...,x_n) x_1 x_n have you some suggestion? Is it possible? I know that for multiple integration there is the function adapt, but it has at most n=20. In my case n depends on the dimension of the dataset and, hence, it could be bigger. Thank you, Paolo

...a ) {Yes, this GPL-2 licenced {with Copyright, see file, please keep this one line} ## Copyright (C) 2012 Marius Hofert, Ivan Kojadinovic, Martin Maechler, and Jun Y } ---------------------------------------------------------------------- ##' Properly compute log(x_1 + .. + x_n) for a given (n x d)-matrix of n row ##' vectors log(x_1),..,log(x_n) (each of dimension d) ##' Here, x_i > 0 for all i ##' @title Properly compute the logarithm of a sum ##' @param lx (n,d)-matrix containing the row vectors log(x_1),..,log(x_n) ##' each of dimension...

.... > bld.setPosition(i, true); > - bld.mkOp2(OP_MERGE, TYPE_U64, def, dst[0], dst[1]); > + guess = bld.mkOp2v(OP_MERGE, TYPE_U64, bld.getSSA(8), dst[0], dst[1]); > + > + // 5. Perform 2 Newton-Raphson steps > + if (i->op == OP_RCP) { > + // RCP: x_{n+1} = 2 * x_n - input * x_n^2 > + Value *two = bld.getSSA(8); > + > + bld.mkCvt(OP_CVT, TYPE_F64, two, TYPE_F32, bld.loadImm(NULL, 2.0f)); > + > + guess = bld.mkOp2v(OP_SUB, TYPE_F64, bld.getSSA(8), > + bld.mkOp2v(OP_MUL, TYPE_F64, bld.getSSA(8), two, gues...

....getSSA(8), *neg = bld.getSSA(8), *copy = bld.getSSA(8); + + bld.mkCvt(OP_CVT, TYPE_F64, two, TYPE_F32, bld.loadImm(NULL, 2.0f)); + + /* Square the guess first, since it was for RSQ */ + guess = bld.mkOp2v(OP_MUL, TYPE_F64, bld.getSSA(8), guess, guess); + + // RCP: x_{n+1} = 2 * x_n - input * x_n^2 + guess = bld.mkOp2v(OP_SUB, TYPE_F64, bld.getSSA(8), + bld.mkOp2v(OP_MUL, TYPE_F64, bld.getSSA(8), two, guess), + bld.mkOp2v(OP_MUL, TYPE_F64, bld.getSSA(8), abs, + bld.mkOp2v(OP_MUL, TYPE_F64,...

...back into the original destination. bld.setPosition(i, true); - bld.mkOp2(OP_MERGE, TYPE_U64, def, dst[0], dst[1]); + guess = bld.mkOp2v(OP_MERGE, TYPE_U64, bld.getSSA(8), dst[0], dst[1]); + + // 5. Perform 2 Newton-Raphson steps + if (i->op == OP_RCP) { + // RCP: x_{n+1} = 2 * x_n - input * x_n^2 + Value *two = bld.getSSA(8); + + bld.mkCvt(OP_CVT, TYPE_F64, two, TYPE_F32, bld.loadImm(NULL, 2.0f)); + + guess = bld.mkOp2v(OP_SUB, TYPE_F64, bld.getSSA(8), + bld.mkOp2v(OP_MUL, TYPE_F64, bld.getSSA(8), two, guess), +...

...hanks 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 case, where there is no data, but I personally find matrices (or data.frames) with rows but no columns a very strange concept. The converse is not true, I understand the utility of columns but no rows, particularly i...

...on(i, true); >> - bld.mkOp2(OP_MERGE, TYPE_U64, def, dst[0], dst[1]); >> + guess = bld.mkOp2v(OP_MERGE, TYPE_U64, bld.getSSA(8), dst[0], dst[1]); >> + >> + // 5. Perform 2 Newton-Raphson steps >> + if (i->op == OP_RCP) { >> + // RCP: x_{n+1} = 2 * x_n - input * x_n^2 >> + Value *two = bld.getSSA(8); >> + >> + bld.mkCvt(OP_CVT, TYPE_F64, two, TYPE_F32, bld.loadImm(NULL, 2.0f)); >> + >> + guess = bld.mkOp2v(OP_SUB, TYPE_F64, bld.getSSA(8), >> + bld.mkOp2v(OP_MUL, TYPE_F64,...

Hi, This is most likely only a minor technicality, but I saw the following: On page 6 of the 'parallel' vignette (http://stat.ethz.ch/R-manual/R-devel/library/parallel/doc/parallel.pdf), the random-number generator "L'Ecuyer-CMRG" is said to have seed "(x_n, x_{n-1}, x_{n-2}, y_n, y_{n-1}, y_{n-2})". However, in L'Ecuyer et al. (2002), the seed is given with 'increasing' indices, so should rather be "(x_{n-2}, x_{n-1}, x_n, y_{n-2}, y_{n-1}, y_n)" (or, even more intuitively, "(x_{n-3}, x_{n-2}, x_{n-1}, y_{n-3}, y_{n-2}...

...i 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 1. ....I don't know where I'm wrong.... -- Michele De Meo http://micheledemeo.blogspot.com/ [[alternative HTML version deleted]]

Hi, I have two groups of data of different size: group A: x1, x2, ...., x_n; group B: y1, y2, ...., y_m; (m is not equal to n) The two groups are independent but observations within each group are not independent, i.e., x1, x2, ..., x_n are not independent; but x's are independent from y's I wonder if randomization test is still applicable to this case. Does...

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 s...

Hello, Say I want to make a multiple regression model with the following expression: lm(y~x1 + x2 + x3 + ... + x_n,data=mydata) It gets boring to type in the whole independent variables, in this case x_i. Is there any simple way to do the metaprogramming for this? (There are different cases where the names of the independent variables might sometimes have apparent patterns or not)

>>>>> 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, 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 compared the mean from both the Gibbs sample and the arms sample for several y and x. Surprisingly, both means always differed by the same constant....

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

...n't get it to work, and wonder if I am not using the right function. What I want is a Kalman filter that accepts exogenous inputs where the input is found using the algebraic Ricatti equation solution to a penalty function. If K is the gain matrix then the exogenous input would be u_t = -Kx_n, where x_n is the Kalman filter state estimate. These inputs would be entered as such x_t = Ax_t-1 + Bu_t-1 + Ge_t. Is l.SS in the dse1 package the correct parametrization of the Kalman filter? Thank you very much, Todd Remund

...t; This is most likely only a minor technicality, but I saw the > following: On page 6 of the 'parallel' vignette > (http://stat.ethz.ch/R-manual/R-devel/library/parallel/doc/parallel.pdf), > the random-number generator "L'Ecuyer-CMRG" is said to have seed > "(x_n, x_{n-1}, x_{n-2}, y_n, y_{n-1}, y_{n-2})". However, in L'Ecuyer > et al. (2002), the seed is given with 'increasing' indices, so should > rather be "(x_{n-2}, x_{n-1}, x_n, y_{n-2}, y_{n-1}, y_n)" (or, even > more intuitively, "(x_{n-3}, x_{n-2}, x_{n-1}, y...

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&quo...

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 produ...