I've runned some algorithms and wanted to make some statistics analysis with the results. I have two vectors with the averages of the error rate.
With R, using the line below I would get everything.
t.test(methodresults1,methodresults2,var.equal=FALSE,paired=FALSE,alternative="less")
Since I'm using Python, I wanted to use Rpy2 project.
I tried that:
import rpy2.robjects as R
# methodresults1 and methodresults2 are numpy arrays.
# kolmogorov test
normality_res = R.r['ks.test'](R.FloatVector(methodresults1.tolist()),'pnorm',mean=R.FloatVector(methodresults1.mean().tolist()),sd=R.FloatVector(methodresults1.std().tolist())))
# t-test
res = R.r['t.test'](R.FloatVector(methodresults1.tolist()),R.FloatVector(methodresults2.tolist()),alternative='two.sided',var.equal=FALSE,paired=FALSE)
res.rx('p.value')[0][0]
res.rx('statistic')[0][0]
res.rx('parameter')[0][0]
I wasn't able to perform both tests.
I found also that the problem with the t-test is with the var.equal statement and it gives me an * SyntaxError: keyword can't be an expression (, line 1).
Extra question: Is there a better way to work with numpy and Rpy2?
to perform ks test with python, in case of a two-sample test, you can
where
x
,y
are twonupmy.array
:first value is the test statistics, and second value is the p-value. if the p-value is less than 95 (for a level of significance of 5%), this means that you cannot reject the Null-Hypothese that the two sample distributions are identical.
for one sample ks test, see for example here: http://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.kstest.html#scipy.stats.kstest
this test lets you test the goodness of fit of your empirical distribution to a given probability distribution.
As it says: "SyntaxError: keyword can't be an expression (, line 1)."
In Python, symbols cannot contain the character ".".
Check the rpy2 documentation about functions for more details.