similar to: non-numeric argument to binary operator

Hello, suppose I have a list with matrices: a=list(x1=matrix(rnorm(10),5,2),x2=matrix(rnorm(10),5,2),x3=matrix(rnorm(10),5,2)) I want to compute for all combination of xi and xj (x1,x2 x1,x3 and x2,x3) a value. This value is given for the pair x1,x2 by trace(x1%*%t(x1)%*%x2%*%t(x2)) / trace(x1%*%t(x1))*trace(x2%*%t(x2)) I know that product matrices t(xi)%*%xi can be obtained by:

Dear R-ians, I'm looking for a computational simplified formula to calculate a measure for heterogeneity (let's say H ): H = sqrt [ (Si (Sj (Xi - Xj)?? ) ) /n ] where: sqrt = square root Si = summation over i (= 0 to n) Sj = summation over j (= 0 to n) Xi = element of X with index i Xj = element of X with index j I can simplify the formula to: H = sqrt [ ( 2 * n * Si (Xi) - 2 Si (Sj

Hi, I am tring to write a loop to compute this, ========================== x1=c( rep(-1,4), rep(1,4) ) x2=c( rep(c(-1,-1,1,1),2) ) x3=c( rep(c(-1,1),4) ) x1*x2 x1*x3 x2*x3 ======================== suppose i have x1,x2,x3 i want to compute their ' two factor interactions', x1x2,x1x3 and x2x3, I wrote ======================== for(i in 1:2){ for( j in i+1:3){ xij=c()

Dear R Core, pairwise.wilcox.test does not handle "paired = TRUE" correctly; e.g. set.seed(13) x <- rnorm(20) g <- c(rep(1, 10), rep(2, 10)) wilcox.test(x ~ g)$p.value # 0.075 pairwise.wilcox.test(x, g)$p.value # 0.075, o.k wilcox.test(x ~ g, paired = TRUE)$p.value # 0.105 pairwise.wilcox.test(x, g, paired = TRUE)$p.value # 0.075, wrong The line wilcox.test(xi, xj,

Dear Friends: I like so much to work with R program. Congratulations for your work. I need R for work with multivariate data. My question is: With the pairs(X) command my output is a pairwise scatterplot symmetric matrix. Like: | X1 |X1 vs X2|X1 vs X3|X1 vs X4| |X2 vs X1| X2 |X2 vs X3|X2 vs X4| |X3 vs X1|X3 vs X2| X3 |X3 vs X4| |X4 vs X1|X4 vs X2|X4 vs X3| X4 | It is

Conversion of a data frame to a matrix using as.matrix() when a column of the data frame is POSIXt and all other columns are numeric has changed in R 1.9.0 from R 1.8.1. The new behavior issues a warning message and converts to a character matrix. In R 1.8.1, such an object was converted to a numeric matrix. Here is an example. #### R 1.9.0 #### > foo <- data.frame(

Dear list-reader, by running the following script: library(zoo) sessionInfo() search() packageDescription("zoo") data(EuStockMarkets) dax <- as.zoo(EuStockMarkets[1:10, "DAX"]) daxr <- diff(log(dax)) identical(as.vector(qnorm(daxr)), qnorm(coredata(daxr))) qqnorm(coredata(daxr)) qqnorm(daxr) qqnorm() produces an error: > qqnorm(daxr) Fehler in if (xi == xj) 0L

I have found that order() fails in a rather arcane circumstance, as in this example: > foo <- I( c('x','\265g') ) > order(foo) Error in if (xi > xj) 1L else -1L : missing value where TRUE/FALSE needed > foo <-c('x','\265g') > order(foo) [1] 1 2 > sessionInfo() R version 3.1.1 (2014-07-10) Platform: x86_64-apple-darwin13.1.0 (64-bit)

Hi all and thanks for your time in advance, I can't figure out why summary.lmrob complains when lmrob is used on a zooreg object. If the zooreg object is converted to vector before calling lmrob, no problems appear. Let me clarify this with an example: >library(robustbase) >library(zoo) >dad<-c(801.4625,527.2062,545.2250,608.2313,633.8875,575.9500,797.0500,706.4188,

I have not used POSIX classes previously and now have a need to use them. I have sports data with times of some athletes after different events. I need to perform some simple analyses using the times. I think I've figured out how to do this. I just want to confirm with others who have more experience that this is indeed the correct approach. If not, please suggest a more appropriate way.

Hi, I need this plot: given: x,y - numerical vectors of length N plot xi vs mean(yj such that |xj - xi|<epsilon) (running mean?) alternatively, discretize X as if for histogram plotting and plot mean y over the center of the histogram group. is there a simple way? thanks! -- Sam Steingold (http://sds.podval.org/) on CentOS release 5.6 (Final) X 11.0.60900031 http://thereligionofpeace.com

Dear, I am looking for the simplification of a formula to improve the calculation speed of my program. Therefore I want to simplify the following formula: H = sum{i=0..n-1 , [ sum {j=0..m-1 , sqrt ( (Ai - Bj)^2 + (Ci - Dj)^2) } ] } where: A, C = two vectors (with numerical data) of length n B, D = two vectors (with numerical data) of length m sqrt = square root Ai = element of A with index

Is there a way to set up a regression in R that forces two coefficients to be equal but opposite in sign? I'm trying to setup a model where a subject appears in a pair of environments where a measurement X is made. There are a total of 5 environments, one of which is a baseline. But each observation is for a subject in only two of them, and not all subjects will appear in each

Hi, I have the following two measurements stored in mat: > print(mat) [,1] [,2] [1,] -14.80976 -265.786 [2,] -14.92417 -54.724 [3,] -13.92087 -58.912 [4,] -9.11503 -115.580 [5,] -17.05970 -278.749 [6,] -25.23313 -219.513 [7,] -19.62465 -497.873 [8,] -13.92087 -659.486 [9,] -14.24629 -131.680 [10,] -20.81758 -604.961 [11,] -15.32194 -18.735 To calculate the ranking

I have a math (and maybe an R) question. I want to find equilibrium population density values for a system of linear population growth equations (e.g., Pimm and Lawton 1977). The eventual goal is to perform stability analysis. I can find the partial differential equations for the Jacobian matrix, but I get stuck trying finding the equilibriuym pop densities (X*). Here is the (incorrect?) R code

Hello, I would like to use mapply to avoid using a loop but for some reason, I can't seem to get it to work. I've included copies of my code below. The first set of code uses a loop (and it works fine), and the second set of code attempts to use mapply but I get a "subscript out of bounds" error. Any guidance would be greatly appreciated. Xj, Yj, and Wj are also lists, and s2,

Dear all, I am trying to find a suitable R-function for parametric proportional hazard regressions. The package survival contains the coxph() function which performs a Cox regression which leaves the base hazard unspecified, i.e. it is a semi-parametric method. The package Design contains the function pphsm() which is good for parametric proportional hazard regressions when the underlying base

I'm trying to figure out how to write a r function that will calculate the extreme spread of a group of points given their (x,y) coordinates. Extreme Spread is the maximal Euclidean distance between two points in a group ex.spread = max{ sqrt [ (xi-xj)^2 - (yi-yj)^2 ] } for i not equal to j I have 60 levels to apply this to. There is the combination function in the dprep package but

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Below is a version of [.data.frame that is faster for subscripting rows of large data frames; it avoids calling duplicated(rows) if there is no need to check for duplicate row names, when: i is logical attr(x, "dup.row.names") is not NULL (S+ compatibility) i is numeric and negative i is strictly increasing "[.data.frame" <- function (x, i, j,