dive-into-python/soundex.py at master · ypeels/dive-into-python · GitHub

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"""Soundex algorithm

This program is part of "Dive Into Python", a free Python book for
experienced programmers.  Visit http://diveintopython.org/ for the
latest version.
"""

__author__ = "Mark Pilgrim (mark@diveintopython.org)"
__version__ = "$Revision: 1.5 $"
__date__ = "$Date: 2004/05/11 19:11:21 $"
__copyright__ = "Copyright (c) 2004 Mark Pilgrim"
__license__ = "Python"

import string

allChar = string.uppercase + string.lowercase
charToSoundex = string.maketrans(allChar, "91239129922455912623919292" * 2)     # 18.4: changed from { "A": 9, ...}
                                                                                # the resulting translation matrix (specialized data structure) is faster (and less general) than a dictionary
def soundex(source):
    "convert string to Soundex equivalent"

    # Soundex requirements:
    # source string must be at least 1 character
    # and must consist entirely of letters
    if (not source) or (not source.isalpha()):                                  # 18.3: changed to "obscure string method" from (compiled) regex
        return "0000"

    # Soundex algorithm:
    # 1. make first character uppercase
    # 2. translate all other characters to Soundex digits
    digits = source[0].upper() + source[1:].translate(charToSoundex)

    # 3. remove consecutive duplicates                                          # from 8.4/BaseHTMLProcessor.py: shouldn't this be done with a list of chars instead? (saves STORAGE but not time?) i guess string APPENDING is fast??
    digits2 = digits[0]                                                         # 18.5: unchanged from original. "Sometimes it doesn't pay to be clever."
    for d in digits[1:]:
        if digits2[-1] != d:
            digits2 += d

    # 4. remove all "9"s
    # 5. pad end with "0"s to 4 characters
    return (digits2.replace('9', '') + '000')[:4]                               # 18.6: changed from (slow) regex and second step to append '000'

    # my list-based version, significantly slower                               # hey, soundex3b, 3c.py try list-based solutions too!
    #digitList = [digits[0]]
    #for d in digits[1:]:
    #    if digitList[-1] != d:
    #        digitList.append(d)
    #return ( "".join([d for d in digitList if d is not '9']) + '0'*3 )[:4]

if __name__ == '__main__':
    import sys
    if sys.argv[1:]:
        print soundex(sys.argv[1])
    else:
        from timeit import Timer                                                # 18.2: the only class in module timeit
        names = ('Woo', 'Pilgrim', 'Flingjingwaller')
        for name in names:
            statement = "soundex('%s')" % name
            t = Timer(statement, "from __main__ import soundex")
            print name.ljust(15), soundex(name), min(t.repeat())                # Timer.repeat(numtests=3, callspertest=1000000)
                                                                                # see also Timer.timeit(), which calls statement 1 million times. ALSO see module hotshot.
                                                                                # 18.2 has a pretty unbaked justification for taking min() instead of average()