On 2/11/10 6:30 PM, John A Schroeder wrote: > > Sorry, I have been unable to install and use rpy version 1. I have made > several attempts. The Windows binary distribution does not work with the > latest version of R, and I can't apply the patch for the version problem > because rpy won't build on my platforms. Appears to be an issue with a > missing header file, and I am too inexperienced with python to get past > the problems. I have experimented briefly with the rpy classic import > from rpy2 but it does not appear to be a complete representation of rpy.
This does not have to remain this way. Earlier calls on this list for getting this improved did not trigger much reaction, but specific reports of problems with rpy_classic have all been addressed so far ( http://bitbucket.org/lgautier/rpy2/issues/?q=rpy_classic ). The offer is still on the table. I am not using rpy any longer, neither do use rpy_classic much... initially I just roughly implemented the rpy interface with the new low-level rpy2.rinterface layer to possible. Laurent > At least not sufficient to run the faithful.py program on the RPy web site. > > John A. Schroeder > Idaho National Laboratory > Battelle Energy Alliance, LLC > P.O. Box 1625 > Idaho Falls, ID 83415-3850 > > Ph: 208-526-8755 > FAX: 208-526-2930 > > > *zahra sheikhbahaee <sheikhbah...@gmail.com>* > > 02/11/2010 10:11 AM > Please respond to > "RPy help, support and design discussion list" > <rpy-list@lists.sourceforge.net> > > > > To > "RPy help, support and design discussion list" > <rpy-list@lists.sourceforge.net> > cc > > Subject > Re: [Rpy] Rpy problem > > > > > > > > > Hi John, > > Actually, I used rpy classic in my program. I wonder whether it is > possible to introduce the results of ks package for rpy2 and then I > distinguish my result or not?? Since rpy classic has problem with the > index of lists and '$' , therefore I think it would be practical but I > do not know how I could do this exchange. I do not know rpy2 enough to > solve my problem and I am not sure that it is plausible for this problem > or not?? > > On Thu, Feb 11, 2010 at 3:18 PM, John A Schroeder > <_john.schroe...@inl.gov_ <mailto:john.schroe...@inl.gov>> wrote: > > Hi Zahra, > > I have been struggling with the same issues. How you do it apparently > depends on the version of RPy. I am currently using 2.0.8 on my windows > machine and the latest 2.1alpha on my linux machine. Of course they are > very different , and very different from the v1 code shown in the most > helpful examples on the RPy web site. > > The following is some code I am working on that illustrates some > possibilities. It is derived from the "faithful" demo on the RPy web > site. This works in 2.0.8 . To see the structure of the object I am > trying to access I typically have to print it first (diagnostic prints > are commented out). Then I try to work out the access details. The > answer to your problem is most likely in the shapiro_test or ks_test > output below. I am still looking for the best way myself, so I am > posting this code with the hope that more experienced users might offer > improvements. I do have this same code working for version 2.1alpha, but > can't easily reconstruct it from memory. > > # Use RPy for processing simulation results > > # Arrange access to some of the required R objects/functions > r = robjects.r > dev_off = robjects.r('dev.off') > shapiro_test = robjects.r('shapiro.test') > ks_test = robjects.r('ks.test') > > # Read the unavailability (UA) data into an R vector > > uaData = r.scan("uaData.txt") > > # Print mean and std. dev. of UA data. > print "RESULTS SUMMARY -- USING R\n" > print " uaData observatons = %7.3E" % (r.length(uaData)[0]) > print " uaData mean = %7.3E" % (r.mean(uaData)[0]) > print " uaData std dev = %7.3E" % (_r.sd_ <http://r.sd/>(uaData)[0]) > print " " > > # Print some distribution statistics provided by R summary() > sumr = r.summary(uaData) > print " Output from R summary() " > #print sumr > print " Min = %7.3E" %(sumr[0]) > print " 1st Qu = %7.3E" %(sumr[1]) > print " Median = %7.3E" %(sumr[2]) > print " Mean = %7.3E" %(sumr[3]) > print " 3rd Qu = %7.3E" %(sumr[4]) > print " Max = %7.3E" %(sumr[5]) > print " " > > # Print a stem-and-leaf plot provided by R stem() > #print "Stem-and-leaf plot of unavailability data" > #print r.stem(uaData) > > # Plot histogram of the UA data. Show normal curve with same mean, std. > dev. > r.X11() > #r.png('uaHistogram.png',width=733,height=550) > r.par(ann=0) # disables automatic label annotations > r.hist(uaData,breaks=25, prob=1, > main="Unavailability Data Histogram",xlab="Unavailability") > #r.plot(h.counts, verticals=1, log="x", > # main="Unavailability data") > x = r.seq(0.00001, .1, length=100) > r.lines(x, r.dnorm(x, mean=r.mean(uaData), sd=r.sqrt(r.var(uaData))), > lty=3, col="red") > #dev_off() > > # Plot cumulative distribution of UA data. Show normal curve with same > # mean, std. dev. > #r.X11() > r.png('uaECDF.png',width=733,height=550) > r.library('stats') > r.plot(r.ecdf(uaData), verticals=1, > main="Empirical cumulative distribution function of unavailability data") > x = r.seq(.001,.1,length=100) > r.lines(x,r.pnorm(x,mean=r.mean(uaData), > sd=r.sqrt(r.var(uaData))), lty=3, lwd=2, col="red") > dev_off() > > # Plot Q-Q curve > #r.X11() > r.png('ua_Q-Q.png',width=733,height=550) > r.par(pty="s") > r.qqnorm(uaData,col="blue") > r.qqline(uaData,col="red") > dev_off() > > # Run the Shapiro-Wilks normality test. > sw = shapiro_test(uaData) > print " Shapiro-Wilks normality test of the UA data" > #print sw > print " W = %.4f" % sw.r["statistic"][0][0] > print " p-value = %.4E" % sw.r["p.value"][0][0] > print " " > > # Run the Kolmogorov-Smirnov test. > ks = ks_test(uaData,"pnorm", mean=r.mean(uaData), sd=r.sqrt(r.var(uaData))) > print " One-sample Kolmogorov-Smirnov test of the UA data" > #print ks > print " D = %.4f" % ks.r['statistic'][0][0] > print " p-value = %.4f" % ks.r['p.value'][0][0] > print " Alternative hypothesis: %s" % ks.r['alternative'][0][0] > print " " > > > > John A. Schroeder > Idaho National Laboratory > Battelle Energy Alliance, LLC > P.O. Box 1625 > Idaho Falls, ID 83415-3850 > > Ph: 208-526-8755 > FAX: 208-526-2930 > > *zahra sheikhbahaee <**_sheikhbah...@gmail.com_* > <mailto:sheikhbah...@gmail.com>*>* > > 02/11/2010 04:58 AM > > > Please respond to > "RPy help, support and design discussion list" > <_rpy-l...@lists.sourceforge.net_ <mailto:rpy-list@lists.sourceforge.net>> > > > > To > _rpy-l...@lists.sourceforge.net_ <mailto:rpy-list@lists.sourceforge.net> > cc > > Subject > Re: [Rpy] Rpy problem > > > > > > > > > > > Hi all, > > I tried the method that one of you proposed for attributing components > on a list but the problem is that when I want to devote eval.points > which is a list and contains two components. It could only devote the > first component and actually it could not recognize the second one. > >>>list_names = [name for name in r.names(j)] > > >>> value= j[list_names.index('eval.points')] > >>> r.print_(value) > $eval.points > $eval.points[[1]] > [1] -2.89805909 -2.77157575 -2.64509241 -2.51860907 -2.39212573 -2.26564239 > [7] -2.13915905 -2.01267571 -1.88619237 -1.75970903 -1.63322569 -1.50674235 > [13] -1.38025901 -1.25377567 -1.12729233 -1.00080899 -0.87432565 -0.74784231 > [19] -0.62135897 -0.49487563 -0.36839229 -0.24190895 -0.11542561 0.01105773 > [25] 0.13754107 0.26402441 0.39050775 0.51699108 0.64347442 0.76995776 > [31] 0.89644110 1.02292444 1.14940778 1.27589112 1.40237446 1.52885780 > [37] 1.65534114 1.78182448 1.90830782 2.03479116 2.16127450 2.28775784 > [43] 2.41424118 2.54072452 2.66720786 2.79369120 2.92017454 3.04665788 > [49] 3.17314122 3.29962456 > > > {'eval.points': {'': [-2.8980590912448885, -2.771575751386762, > -2.6450924115286356, -2.5186090716705092, -2.3921257318123832, > -2.2656423919542568, -2.1391590520961303, -2.0126757122380039, > -1.8861923723798777, -1.7597090325217513, -1.6332256926636251, > -1.5067423528054986, -1.3802590129473722, -1.253775673089246, > -1.1272923332311195, -1.0008089933729933, -0.87432565351486691, > -0.74784231365674048, -0.62135897379861404, -0.49487563394048806, > -0.36839229408236163, -0.2419089542242352, -0.11542561436610876, > 0.011057725492017667, 0.1375410653501441, 0.26402440520827009, > 0.39050774506639652, 0.51699108492452295, 0.64347442478264938, > 0.76995776464077581, 0.8964411044989018, 1.0229244443570282, > 1.1494077842151547, 1.2758911240732806, 1.4023744639314075, > 1.5288578037895335, 1.6553411436476604, 1.7818244835057864, > 1.9083078233639124, 2.0347911632220392, 2.1612745030801652, > 2.2877578429382921, 2.4142411827964181, 2.5407245226545441, > 2.6672078625126709, 2.7936912023707969, 2.9201745422289238, > 3.0466578820870498, 3.1731412219451767, 3.2996245618033027]}} > >>> r.mode(value) > 'list' > >>> r.length(value) > 1 > > It is very urgent. Could someone help me? > > Zahra. > > On Wed, Feb 10, 2010 at 11:58 AM, zahra sheikhbahaee > <_sheikhbah...@gmail.com_ <mailto:sheikhbah...@gmail.com>> wrote: > Hi, > > I have written a program in python and then I used the ks(kernel density > estimator)package of R for smoothing my dataset. Now, I need to extract > the information which has been calculated by the package. The result is > somthing like that: > r.str(j) > List of 4 > $ x : num [1:3105, 1:2] 0.952 0.891 0.902 0.889 0.864 ... > $ eval.points:List of 2 > ..$ : num [1:50] -2.84 -2.72 -2.59 -2.47 -2.35 ... > ..$ : num [1:50] -2.9 -2.77 -2.65 -2.52 -2.39 ... > $ estimate : num [1:50, 1:50] 0 0 0 0 0 ... > $ H : num [1:2, 1:2] 0.00751 0.00452 0.00452 0.00802 > - attr(*, "class")= chr "kde" > For analysing the results I need eval.points, which give me the points > in two dimension and they have list format but I do not know how I could > attribute these two components to two python components. > > How could I tackle with this problem? > > Cheers, > Zahra. > ------------------------------------------------------------------------------ > SOLARIS 10 is the OS for Data Centers - provides features such as DTrace, > Predictive Self Healing and Award Winning ZFS. 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