Path: utzoo!mnetor!uunet!lll-winken!lll-lcc!ptsfa!pacbell!att-ih!ihnp4!inuxc!iuvax!pur-ee!uiucdcs!uxc.cso.uiuc.edu!uicsrd.csrd.uiuc.edu!jaxon
From: jaxon@uicsrd.csrd.uiuc.edu
Newsgroups: comp.lang.apl
Subject: Re: MODE
Message-ID: <49700001@uicsrd.csrd.uiuc.edu>
Date: 29 Feb 88 17:20:00 GMT
References: <1428@water.waterloo.edu>
Lines: 28
Nf-ID: #R:water.waterloo.edu:1428:uicsrd.csrd.uiuc.edu:49700001:000:810
Nf-From: uicsrd.csrd.uiuc.edu!jaxon    Feb 29 11:20:00 1988



For a large array, the outer product in the above solution
can be slow.  A solution that makes fewer comparisons is:

$ MODE is MODE V;#CT;FREQ;NUB;T
[1]  #CT is 0   note Need transitivity of =
[2]  V is V[gradeup V]
[3]  NUB is 1,(1 drop V) neq (^1 drop V)
[4]  FREQ is T - ^1 drop 0,T is NUB/iota rho T
[5]  MODE is (FREQ=max/FREQ)/NUB/V
[6]  MODE is MODE[floor .5 times ''rho rho MODE]
$

After obtaining the FREQuency counts in line 4
you might wish to select the MODE a different way,
Line[5]  selects all of them.

Iverson's Window Reduction operator could be used
on lines [3] and [4].

Since gradeup is  o(n log n) this algorithm is
faster in the limit cases, but probably won't
beat Prof. Dickey over short vectors.


P.S.  $ = del,  ^ = high minus,  # = quad.

Greg Jaxon   (jaxon@uicsrd.uiuc.edu)