similar to: Compilation of R (linux) package on windows

Hello, I'm a newbie on this list. I have a R code but its execution take a very long time. Is it possible to compile it (in C for example) to decrease the execution time ? -- ---------------- Alexandre DEPIRE INRETS / GARIG

Hello, I'm a newbie on this list. I have a R code but its execution take a very long time. Is it possible to compile it (in C for example) to decrease the execution time ? -- ---------------- Alexandre DEPIRE INRETS / GARIG

Hi, list, I am using 'sgeostat' package by Albrecht Gebhardt and I am trying to put a correlation coefficient of some kind on the lagplots. Is there a possiblity to do so? Thanks Mahdi -- ----------------------------------- Mahdi Osman (PhD) E-mail: m_osm at gmx.net ----------------------------------- GMX im TV ... Die Gedanken sind frei ... Schon gesehen? Jetzt Spot online ansehen:

Hello there! When I do freq=F on hist, I get on the left a small number, what exactly does that mean? Thanks, Martin

Hello, I would like to know if it is possible to insert LaTeX typesetting in R output. I want to obtain a graph with LaTeX label in order to incorporate it as postscript or pdf, x<-seq(0,1,length=100) y<-x*x plot(x,y,xlab="$X$",ylab="$X^2$")

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Hi. Here are my patches and makefiles for compiling ogg, vorbis and vorbis-tools on DOS with DJGPP. They can build them from cvs snapshots and need minimal maintainance (even extract source lists from `makefile.am's). Everything seems to work, except one FP exception on encode at middle bitrates (lowest two and highest bitrates don't cause it) - I don't think it's my

I have made available the work of the last weekend that I mentioned in my previous messages on ftp://sirio.stat.unipd.it/pub/R. Many thanks to Peter D. and Ross I. for their suggestions. If you think that the work is of some interest, I can upload it to CRAN. guido m. This is the README enclosed in the distribution. -------------------------------------------------------------------------

Hello all, How do I actually use the output of predict.lm(..., type="terms") to predict new term values from new response values? I'm a chromatographer trying to use R (2.15.1) for one of the most common calculations in that business: - Given several chromatographic peak areas measured for control samples containing a molecule at known (increasing) concentrations, first

I'm a week-old R user, and have become stuck trying to create usable CSV outputs for post-processing. I am using the package Rioja, which provides small datasets of results. I am running several analyses in a loop and iteratively adding the results to a *list* ("combined"). Within each iteration I use the following: > combined[[i]] <- performance(fit) With two iterations I

Dear All, I'm trying to use waldtest to test poolability (parameter stability) between two logistic regressions. Because I need to use robust standard errors (using sandwich), I cannot use anova. anova has no problems running the test, but waldtest does, indipendently of specifying vcov or not. waldtest does not appear to see that my models are nested. H0 in my case is the the vector of

Dear Spencer and Prof. Fox, Thank you for your replies. I'll very appreciate, if you have any ideas concerning the problem described below. First, I'd like to describe the model in brief. In general I consider a model with three equations. First one is for annual GRP growth - in general it looks like: 1) GRP growth per capita = G(investment, migration, initial GRP per

Dear R users this may be a simple question - but i would appreciate any thoughts does anyone know how you would get one lower and one upper confidence interval for a set of data that consists of proportions. i.e. taking a usual confidence interval for normal data would result in the lower confidence interval being negative - which is not possible given the data (which is constrained between

Caching computation in rails? Simple example: factorial modulus a large number input: integer x output: factorial( x ) % 12345678901234567 I want it so that if it computes factorial of N once, it will not have to compute for N again. code: class SiteController < ApplicationController caches_action :factorial, :inv def examine @inv = @params[''inv''] @outv =

[Hope this is the right list where to send...] An attempt to update package 'mnormt' involves the addition of a small new function called 'pd.solve'. When I come to the package checking stage, an error occurs in parsing pd.solve.Rd. The full transcript of the outcome is copied below (it includes details on my installation) but the critical point is where the \examples{} section

Hi, I want to do a meta-analysis with count data for treatement/control cases. Mi problem is that I need to use zero values (an informative zero value) for the mean and standard deviation for one of the treatement, but R has a problem: "Studies with zero values for sd.e or sd.c get no weight in meta-analysis". I can agroup the case by Family (byvar=Family). ¿Can you help me? Thanks!

Hi All, I *think* it's ddply because the function recognizes vr1, etc, in other parts of the function. Here's some code: # create dataset PROV.PM.FBCTS <- c(0.00 ,0.00, 33205.19, 25994.56, 23351.37, 26959.56 ,27632.58, 26076.24, 0.00, 0.00 , 6741.42, 18665.09 ,18129.59 ,21468.39 ,21294.60 ,22764.82, 26076.73) FBCTS.INV.TOT <- c(0 , 0, 958612, 487990, 413344, 573347,

I'm new to R and i'm having some trouble with a bubble chart. Basically I have 3 series (a,b,c), but the third one is a binnary variable (assumes only 0 or 1 to the entire data). How can I use these binnary information to make 2 different colours in a bubble chart?. I.e., I'm using this code: symbols(inv$a, inv$b, circles=radius, inches=0.35, fg="white", bg="red",

I submit the following matrix to both MATLAB and R x= 0.133 0.254 -0.214 0.116 0.254 0.623 -0.674 0.139 -0.214 -0.674 0.910 0.011 0.116 0.139 0.011 0.180 MATLAB's inv(x) provides the following 137.21 -50.68 -4.70 -46.42 -120.71 27.28 -8.94 62.19 -58.15 6.93 -7.89 36.94 8.35 11.17 10.42 -14.82 R's solve(x) provides: 261.94 116.22 150.92 -267.78 116.22 344.30 286.68