18.5. Optimizing List Operations

The third step in the Soundex algorithm is eliminating consecutive duplicate digits. What's the best way to do this?

Here's the code we have so far, in soundex/stage2/soundex2c.py:

    digits2 = digits[0]
    for d in digits[1:]:
        if digits2[-1] != d:
            digits2 += d

Here are the performance results for soundex2c.py:

C:\samples\soundex\stage2>python soundex2c.py
Woo             W000 12.6070768771
Pilgrim         P426 14.4033353401
Flingjingwaller F452 19.7774882003

The first thing to consider is whether it's efficient to check digits[-1] each time through the loop. Are list indexes expensive? Would we be better off maintaining the last digit in a separate variable, and checking that instead?

To answer this question, here is soundex/stage3/soundex3a.py:

    digits2 = ''
    last_digit = ''
    for d in digits:
        if d != last_digit:
            digits2 += d
            last_digit = d

soundex3a.py does not run any faster than soundex2c.py, and may even be slightly slower (although it's not enough of a difference to say for sure):

C:\samples\soundex\stage3>python soundex3a.py
Woo             W000 11.5346048171
Pilgrim         P426 13.3950636184
Flingjingwaller F452 18.6108927252

Why isn't soundex3a.py faster? It turns out that list indexes in Python are extremely efficient. Repeatedly accessing digits2[-1] is no problem at all. On the other hand, manually maintaining the last seen digit in a separate variable means we have two variable assignments for each digit we're storing, which wipes out any small gains we might have gotten from eliminating the list lookup.

Let's try something radically different. If it's possible to treat a string as a list of characters, it should be possible to use a list comprehension to iterate through the list. The problem is, the code needs access to the previous character in the list, and that's not easy to do with a straightforward list comprehension.

However, it is possible to create a list of index numbers using the built-in range() function, and use those index numbers to progressively search through the list and pull out each character that is different from the previous character. That will give you a list of characters, and you can use the string method join() to reconstruct a string from that.

Here is soundex/stage3/soundex3b.py:

    digits2 = "".join([digits[i] for i in range(len(digits))
                       if i == 0 or digits[i-1] != digits[i]])

Is this faster? In a word, no.

C:\samples\soundex\stage3>python soundex3b.py
Woo             W000 14.2245271396
Pilgrim         P426 17.8337165757
Flingjingwaller F452 25.9954005327

It's possible that the techniques so far as have been “string-centric”. Python can convert a string into a list of characters with a single command: list('abc') returns ['a', 'b', 'c']. Furthermore, lists can be modified in place very quickly. Instead of incrementally building a new list (or string) out of the source string, why not move elements around within a single list?

Here is soundex/stage3/soundex3c.py, which modifies a list in place to remove consecutive duplicate elements:

    digits = list(source[0].upper() + source[1:].translate(charToSoundex))
    for item in digits:
        if item==digits[i]: continue
    del digits[i+1:]
    digits2 = "".join(digits)

Is this faster than soundex3a.py or soundex3b.py? No, in fact it's the slowest method yet:

C:\samples\soundex\stage3>python soundex3c.py
Woo             W000 14.1662554878
Pilgrim         P426 16.0397885765
Flingjingwaller F452 22.1789341942

We haven't made any progress here at all, except to try and rule out several “clever” techniques. The fastest code we've seen so far was the original, most straightforward method (soundex2c.py). Sometimes it doesn't pay to be clever.

Example 18.5. Best Result So Far: soundex/stage2/soundex2c.py

import string, re
allChar = string.uppercase + string.lowercase
charToSoundex = string.maketrans(allChar, "91239129922455912623919292" * 2)
isOnlyChars = re.compile('^[A-Za-z]+$').search
def soundex(source):
    if not isOnlyChars(source):
        return "0000"
    digits = source[0].upper() + source[1:].translate(charToSoundex)
    digits2 = digits[0]
    for d in digits[1:]:
        if digits2[-1] != d:
            digits2 += d
    digits3 = re.sub('9', '', digits2)
    while len(digits3) < 4:
        digits3 += "0"
    return digits3[:4]
if __name__ == '__main__':
    from timeit import Timer
    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())