source: python/aubio/median.py @ 0ce9acc3

"""Copyright (C) 2004 Paul Brossier <piem@altern.org>
print aubio.__LICENSE__ for the terms of use
"""

__LICENSE__ = """\
         Copyright (C) 2004 Paul Brossier <piem@altern.org> 
                
         This program is free software; you can redistribute it and/or modify
         it under the terms of the GNU General Public License as published by
         the Free Software Foundation; either version 2 of the License, or
         (at your option) any later version.

         This program is distributed in the hope that it will be useful,
         but WITHOUT ANY WARRANTY; without even the implied warranty of
         MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
         GNU General Public License for more details.

         You should have received a copy of the GNU General Public License
         along with this program; if not, write to the Free Software
         Foundation, Inc., 675 Mass Ave, Cambridge, MA 02139, USA.
"""            

""" 
original author Tim Peters
modified by Paul Brossier <piem@altern.org>
inspired from http://www.ics.uci.edu/~eppstein/161/python/peters-selection.py
"""

def short_find(a, rank):
    a.sort()
    return a[rank - 1]

# Find the rank'th-smallest value in a, in worst-case linear time.
def percental(a, rank):
    n = len(a)
    assert 1 <= rank <= n
    if n <= 7:
        return short_find(a, rank)

    ## Find median of median-of-7's.
    ##medians = [short_find(a[i : i+7], 4) for i in xrange(0, n-6, 7)]
    #median = find(medians, (len(medians) + 1) // 2)
    
    # modified to Find median
    median = short_find([a[0], a[-1], a[n//2]], 2)

    # Partition around the median.
    # a[:i]   <= median
    # a[j+1:] >= median
    i, j = 0, n-1
    while i <= j:
        while a[i] < median:
            i += 1
        while a[j] > median:
            j -= 1
        if i <= j:
            a[i], a[j] = a[j], a[i]
            i += 1
            j -= 1

    if rank <= i:
        return percental(a[:i], rank)
    else:
        return percental(a[i:], rank - i)