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>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(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>*
>
> 02/11/2010 04:58 AM
> Please respond to
> "RPy help, support and design discussion list" <
> rpy-list@lists.sourceforge.net>
>
> To
> 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* <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.
>
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