search for: interprated

Bonjour, je voulais savoir si wxruby2 est intégré à l''interpréteur ruby? Que veut dire $0 ? merci Sebastien _______________________________________________ wxruby-users mailing list wxruby-users@rubyforge.org http://rubyforge.org/mailman/listinfo/wxruby-users

Hello, [really hoping not to ask something the 1001th time] I need single backslashes printed for generating TeX-code. That works: > print("\begin{tabular}") [1] "\begin{tabular}" But \h seems to be interprated in some way: > print('\hline') [1] "hline" Hm, trying to mask the backslash: > print('\\hline') [1] "\\hline" But I need "\hline"... Is there a way without generating strange code in a temporary file for "sed"ing it? It wo...

Hi, I am wondering how I should interpreate the output of optFederov() in AlgDesign. Specially I want to know what is $D, $A, $Ge and $Dea, which one I can use as an efficiency to say how good the optimal design is. I only know when a orthogonal design comes, $D = 1. I red the pdf document -- vignette("AlgDesign") [Just type: vignette("AlgDesign") in R, you will get

If do a scattrplot of data ( x and y ) and there are two clouds of points. One cloud is in the left bottom corner of the plot and the other cloud is in the upper right. If I fit a regression line to this data ( or equivalently , calculate a correlation ), then obviously, it is going to seem like x and y are related because a line has to be connected between the 2 clouds. But, there must be a

hello, I need help because I don't understand the syntaxe "else" how can I write it for example I writed a script to cut missings values and I have errors > if(na==length(C)){ + pos=match(0,match(donGeno[[na-1]],donGeno[[na]],nomatch=0)) + for(k in 1:(na-1)) { + if(pos==1) {donGeno[[k]] <-

Dear all, I have an implementation of streams in R. The current implementation of delay() and force() is inspired from the LISP implementation found in Part VI "Languages for AI problem solving" of "Artificial Intelligence" by G. Luger. I have tested it with the Fibonacci example in the same book (see examples below). It works but I do run into a problem when I try to

I think median() behaves as designed: As long as the argument can be ordered, the "middle observation" makes sense, except when the middle falls between two categories, and you can't define and average of the two candidates for a median. The "sick man" would seem to be var(). Notice that it is also inconsistent with cov(): >

For the first question x'(y-bx)<>0 or Cov(Ui,Xi)<>0 you have heterogeinity problem ! OLS estimator in this case in biased. The bias is equal to (Exx')^{-1}Exu You obtain obtain correct estimator by use instrumental variable or 2SLS ... Justin BEM Elève Ingénieur Statisticien Economiste BP 294 Yaoundé. Tél (00237)9597295. ----- Message d'origine ---- De

Hi, In stlmethods(stats) says: range.bars logical indicating if each plot should have a bar at its right side which are of equal heights in user coordinates. I don't understand the meaning of: "which are of equal heights in user coordinates" I'm working with monthly time series, with a clear seasonal cycle, but a not so clear trend, is just what I'm looking for. How

I stand corrected. The part "If set to 'NULL', it implies 'useNA="always"'." is even in the documentation in R 2.8.0. It was my fault not to check carefully. I wonder, why "always" was chosen for 'useNA' for exclude = NULL. Why exclude = NULL is so special? What about another 'exclude' of length zero, like character(0) (not c(),

Dear all, I'm facing a problem in estimation of glm model with weibull distribution. I run this : eqn0<-formula(fdh~cup1+cup2+cup3+cup4+fin1+vd1+cm2+cm4+milieu+cpro1+cpro2+cpro3a+cpro3b+schef+log(y)) regWeib0<-vglm(eqn0,family=weibull,subset(br, fdh<1)) I have en estimation but there is a message saying that regularity conditions are violated : Message d'avis : In eval(expr,

Hello, Is there a reason for the following behaviour? > mean(c("1","2","3")) [1] NA Warning message: In mean.default(c("1", "2", "3")) : l'argument n'est ni num?rique, ni logique : renvoi de NA But: > var(c("1","2","3")) [1] 1 And also: >

On Mon, Apr 27, 2015 at 1:02 PM, Joerg Schilling <Joerg.Schilling at fokus.fraunhofer.de> wrote: > > > The GPL makes claims that are in conflict with the law because these claims are > not amongst what the list in the law permits and that are thus void. The GPL is all that gives you permission to distribute. If it is void then you have no permission at all to distribute any

Jean-Luc, Please keep the communications on the list, for the benefit of others, now and in the future, via the list archive. I am adding r-devel back here. I can't speak to the rationale in some of these cases. As I noted, it may be (is likely) due to differing authors over time, and there may have been relevant use cases at the time that the code was written, resulting in the various

Les Mikesell <lesmikesell at gmail.com> wrote: > On Mon, Apr 27, 2015 at 1:02 PM, Joerg Schilling > <Joerg.Schilling at fokus.fraunhofer.de> wrote: > > > > > The GPL makes claims that are in conflict with the law because these claims are > > not amongst what the list in the law permits and that are thus void. > > The GPL is all that gives you

I have been experimenting with different possibilities for an "introduction" page for my packages. That is, a good place to tell users about the most important things in a package, and where to start. Recently there was a discussion about this, and a suggestion to use <foo>-package.Rd, and also a function that generates a skeleton document. My problem with this suggestion is

On Mon, Apr 27, 2015 at 2:13 PM, Joerg Schilling <Joerg.Schilling at fokus.fraunhofer.de> wrote: > >> The GPL is all that gives you permission to distribute. If it is >> void then you have no permission at all to distribute any covered >> code. > > Fortunately judges know better than you.... > > If you read the reasoning from judgements, you would know that

>>>>> Suharto Anggono Suharto Anggono via R-devel <r-devel at r-project.org> >>>>> on Thu, 11 Aug 2016 16:19:49 +0000 writes: > I stand corrected. The part "If set to 'NULL', it implies > 'useNA="always"'." is even in the documentation in R > 2.8.0. It was my fault not to check carefully. I wonder,

My impression from reading http://llvm.org/docs/DeveloperPolicy.html#clp is that it's intended to be possible to compile programs using llvm and distribute the resulting binaries freely. This does not seem to be the case. I'm assuming no portion of LLVM is included in the compiled binaries, only the runtime library components, so that the compiled binaries are not derived from LLVM.

useNA <- if (missing(useNA) && !missing(exclude) && !(NA %in% exclude)) "ifany" An example where it change 'table' result for non-factor input, from https://stat.ethz.ch/pipermail/r-help/2005-April/069053.html : x <- c(1,2,3,3,NA) table(as.integer(x), exclude=NaN) I bring the example up, in case that the change in result is not intended.