[R] ecdf manipulation

Martin Maechler maechler at stat.math.ethz.ch
Sat Aug 16 17:50:32 CEST 2008


>>>>>   <harryandlaura at talktalk.net>
>>>>>     on Fri, 15 Aug 2008 00:51:56 +0100 writes:

    > I’m using the edcf function to look at a number of
    > empirical distributions graphically for run-time analyses
    > of stochastic optimization algorithms.  When dealing with
    > problems where the optimal solution for these problems is
    > always found everything is fine and the graphs are very
    > useful for comparative observations. These distributions
    > have a vertical axis height of one i.e. a probability of
    > one. However, I’ve hit a problem when the optimal
    > solution is not always obtained during the allotted
    > run-time. In the cases I’m looking these graphs are
    > only concerned with the behaviour those runs that find the
    > optimal solution.
 
    > e.g. say we have two algorithms one solves a given problem
    > 1000 times out of 1000 runs and the second solves the same
    > problem 800 times out of 1000 runs then the first plot
    > rises from 0 to 1 where as the second should only rise to
    > 0.8
 
    > One idea is that the ecdf R code relies upon the number of
    > samples n (1000 in this case) is it possible to manipulate
    > this R code and pass an extra argument to have n defined
    > when the function is called, opposed to the value of n
    > being set to the size of the vector being passed in as
    > appears to be the current case, whilst maintaining its
    > graphical capability?

Yes, this is possible 
 { install.packages("fortunes"); fortunes::fortune("Yoda") }
 
    > If so how and where do I get hold of the ecdf R code to
    > manipulate?

  https::/svn.r-project.org/R/trunk/src/library/stats/R/ecdf.R

always has the (R-devel version of the) ecdf code;
you may also want to study the  stepfun code in stepfun.R (same
directory)  which is made use of by ecdf and its methods.

Martin Maechler,
ETH Zurich


    > If not then does anyone have any suggestions?
 
    > Thanks
    > Harry Venables



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