search for: covs

Hello, I have a vector {a, b1, b2, b3, b4}. How can I calculate the following matrix: var(a) cov(a, b1) cov(a, b2) cov(a, b3) cov(a, b4) cov(a, b1) var(b1) cov(a, b2) cov(a, b3) cov(a, b4) ... ... cov(a, b1) cov(a, b2) cov(a, b3) cov(a, b4) var(b4) I would very appreciate your inputs. Thank you very much. Sincerely, Jialin Huang [[alternative HTML version deleted]]

rkotler at mipsswbrd006-le:~/caviumllvm/build/test$ make Making LLVM 'lit.site.cfg' file... Making LLVM unittest 'lit.site.cfg' file... ( ulimit -t 600 ; ulimit -d 512000 ; ulimit -m 512000 ; ulimit -s 8192 ; \ /usr/bin/python /home/rkotler/workspace/llvm/utils/lit/lit.py -s -v . ) XPASS: LLVM :: tools/llvm-cov/llvm-cov.test (8916 of 9784) ******************** TEST

Dear R community Have just recently got back into R after a long break and have been amazed at how much it has grown, and how active the list is! Thank you so much to all those who contribute to this amazing project. My question: I am trying to calculate Mahalanobis distances for a matrix called "fgmatrix" >dim(fgmatrix) [1] 76 15 >fg.cov <- cov.wt(fgmatrix)

...art, : Variablenl?ngen sind unterschiedlich (gefunden f?r '(id)') --> failure, length of variables are different. #----------------------------------------------------------------- fold <- function(data, time, event, cov, cov.names=paste('covariate', '.', 1:ncovs, sep=""), suffix='.time', cov.times=0:ncov, common.times=TRUE, lag=0){ vlag <- function(x, lag) c(rep(NA, lag), x[1:(length(x)-lag)]) xlag <- function(x, lag) apply(as.matrix(x), 2, vlag, lag=lag) all.cov <- unlist(cov) if (!is.list(cov)) cov <...

Hi, I have the following problem: I have a data.frame with 36 sample sites (colums) for which I have covariates in 3 categories: Area, Month and River. Each Area consists of 3 rivers, which were sampled over 3 month. Now I want to fuse River 1-3 for one area in one month. To get a data.frame with 12 colums. I am trying to do a "for loop" (which may be a complicated solution, but I

...data.frame(onset=c(46, 32, 53, 76, 64, 43), case=c(1,1,1,0,0,0), ooph=c(NA, 30, 38, 50, NA, NA), ocp=c(1,1,0,0,1,0), parity=c(2,0,1,3,3,2), age.preg=c(28,NA,27,20,22,23)) make.dep <- function(onset, case, time.dep, covs=NULL) { if (is.null(n.time.dep <- ncol(time.dep))) { if (!is.null(time.dep)) { n.time.dep <- 1 time.dep <- as.matrix(time.dep) }else{ n.time.dep <- 0 warning("No time dependent covariates") } } if (is.null(n.covs <- ncol(covs))) {...

Dear all I am trying to use the cov function in the stats library. I have no problem using this function from the console. However, in my R script I received a "function not found" message. Then I called stats::cov(...) and received an error message that the function was not exported. Then I tried stats:::cov (three colons) and received the error Error in get(name, envir =

Hi all, I want to get results for cox proportional hazards on SNP data. I'm trying to get HRs, CI's, & p-values for each individual SNP - this is represented by cov[,i]. When I run my code, I get the following error: subscript out of bounds. I don't know why I am getting this error. I have looked through the R logs, but nothing that has been previously suggested has helped so

I am running a 3-parameter nonlinear fit using the default Gauss-Newton method of nls. initialValues.L = list(b=4,d=0.04,t=180); fit.nls.L = nls( myModel.nlm , fData.L, start = initialValues.L, control = nls.control(warnOnly = TRUE), trace=T ); summary.nls.L = summary(fit.nls.L); I run the same analysis in SAS proc NLIN. proc nlin data=apples outest=a; parms b=4 d=.04 t=180; model Y =

Hi, Excuse me for asking this silly question. But I really couldn't understand why cov() and ccov() don't work for my calculation of covariance matrix. a <- matrix(1:8, 2, 4) a [,1] [,2] [,3] [,4] [1,] 1 3 5 7 [2,] 2 4 6 8 > ccov(a) Error in solve.default(cov, ...) : Lapack routine dgesv: system is exactly singular I also tried colume bind, but it

Dear list, what is the fastest way to compute a multivariate mean and cov-matrix? I presume that the mean is computed in cov, so it may be a waste of time to compute the mean first and then a second time inside of cov. Is it faster to use cov.wt, which gives cov-matrix and center? And: If mean and cov should be computed on a part of the data, is it faster to use cov.wt with some weights zero, or

dear list I am student M.S. statistics in department statistics . I am working in the function "nls" in the [R 2.3.1] with 246 data and want to fit the "exp" model to vectors( v and u ) but I have a problem to use it u 5.000000e-13 2.179057e+03 6.537171e+03 1.089529e+04 1.525340e+04 1.961151e+04 2.396963e+04 2.832774e+04 3.268586e+04 3.704397e+04 4.140209e+04

Hi all, Why cov(y, y) only gives one value, and cov(t(y), t(y)) gives 3x3 NA matrix? Here my y is listed below and it is a 3x1 matrix. I am expecting that if I have a random vector y=[y1 y2 y3]', here " ' " denotes a transposition so that y is a column vector, where y1, y2, y3 are independent random variables... then cov(y, y) should be E[ y * y' ] - E[y] * E[y] ',

> On Sep 21, 2016, at 9:36 AM, Kostya Serebryany via llvm-dev <llvm-dev at lists.llvm.org> wrote: > > Exciting! > > (btw, I'd prefer libfuzzer at googlegroups.com <mailto:libfuzzer at googlegroups.com> for such discussions, please start new topics there) You mean a LLVM library has a separate mailing-list? Why? — Mehdi > > I can reproduce this too, but

There appears to be a problem with name matching in list assignment: Version 0.63.0 (November 14, 1998) ... > r <- list() > r$forecast.cov.trend <- 1:12 > r$forecast.cov.zero <- 1:12 > r$forecast.cov <- 1:2 > length(r$forecast.cov) [1] 0 #should be 2 > But note that this works correctly: > r <- list() >

Hi all, I'm trying to support llvm-cov for a new architecture and I have successfully built compiler-rt for my arch. Following steps shown in https://clang.llvm.org/docs/SourceBasedCodeCoverage.html , I encountered an error for the last step(step of llvm-cov show). The command line was (supposed my arch is XXXX) "llvm-cov show -arch=XXXX ./foo -instr-profile=foo.profdata" and the

At the risk of ridicule for my deficient linear algebra skills, I ask for help using the solve.QP function to do portfolio optimization. I am trying to following a textbook example and need help converting the problem into the format required by solve.QP. Below is my sample code if anyone is willing to go through it. This problem will not solve because it is not set up properly. I hope I

Hi vedant, The program didn't pass the checking "OF->getArch() != Triple(Arch).getArch()" loadBinaryFormat in CoverageMappingReader.cpp and returned an error. It's because "OF->getArch()" returned null and "Triple(Arch).getArch()" returned XXXX(name of my arch). The returned value of " OF->getArch()" is decided by "

I recently started using R 2.7.2, and noticed a surprising change in the behavior of var() on NA data: R 2.6.1 (Patched), 2007-11-26, svn.rev 43541, x86_64-unknown-linux-gnu: > stdev(rep(NA,3), na.rm=F) [1] NA > stdev(rep(NA,3), na.rm=T) [1] NA > var(rep(NA,3), na.rm=T, use="complete.obs") [1] NA R 2.7.2 (Patched), 2008-09-02, svn.rev 46491,

Hi, I am attempting to fit a ARMA/GARCH regression model without success. ### ARIMA-GARCH model with regressor ### ### Time series data: A multivariate data set. cov.ts.dq = cov.ts[1:4,"dq1"][!is.na(cov.ts[,"dq1"])] cov.ts.day = ts.intersect(dq = diff(q.ts), day = lag(q.ts, -1)) ### The following R scripts work: (summary(no.day.fitr <- garchFit(dq ~ arma(0,3) +