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.
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
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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>
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
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rpy-list@lists.sourceforge.net>
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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> 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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