2004-12-21-1741Z


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I found some code on the web (gumuz'), but you know me, I had to code my own:

#!/usr/pkg/bin/python
"""testing stuff for huffman compression, based on gumuz' and others

this is public domain!"""

import sys, types

def huffwalk(*args): """we're using lists of lists... offset [-1] will either be:

a list of length 3, which is a number and two branches, or a list of length 2, just a number and letter, or just a number, which means a list or a letter is at offset -2""" try: code, branch, codes = args except: return huffwalk('', huffmantree('test'), ()) if len(branch) == 0: return codes elif type(branch[-1]) == types.IntType: if type(branch[-2]) == types.StringType: return codes + tuple(branch) + (code,) else: # walk the left and right branches recursively return huffwalk(code + '0', branch[-2], codes) + \ huffwalk(code + '1', branch[-3], codes) else: # we should only get here on the first iteration if at all if len(code) > 0: raise Exception, 'lost! %s' % repr((branch, code, codes)) return huffwalk('', branch[-1], codes)

def huffmantree(*args): if args[0] == 'test': # example at http://www.maths.abdn.ac.uk/~igc/tch/mx4002/notes/node59.html nodes = [['i', 6], ['a', 3], ['b', 3], ['d', 2], ['m', 4], ['e', 5], ['o', 3], ['c', 1], ['f', 1], ['g', 2], ['s', 4], ['t', 3], ['l', 2], ['r', 2], ['n', 5], ['p', 1], ['u', 2], [' ', 11]] else: nodes = wordfrequencies(args) while len(nodes) > 1: combined = [nodes.pop(0), nodes.pop(0)] # get the 2 lowest-weight nodes combined.append(combined[0][-1] + combined[1][-1]) # add the weights #DebugPrint('combined lowest nodes: ', combined) nodes.append(combined) # plop new node into the existing list nodes.sort(lambda a, b: cmp(a[-1], b[-1])) # re-sort for lowest weights return nodes

if __name__ == "__main__": sys.stderr.write('running huffman test\n') print repr(huffwalk('test')) else: sys.stderr.write('loaded huffman module\n')

There may be ways to simplify it even further, but this is good enough for now. Note that my result is not the same as on the referenced webpage; Huffman trees are not unique; as far as I know there's no rule that says the highest weight must go to the same side of the node. It was just easier to code it that way.

I haven't tried it on any real data yet, I may find this to be totally broken. What the hell, I'll figure it out eventually.

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