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Performance tuning is a many-splendored thing. Just because Python is an interpreted language doesn't mean you shouldn't worry about code optimization. But don't worry about it too much.
There are so many pitfalls involved in optimizing your code, it's hard to know where to start.
Let's start here: are you sure you need to do it at all? Is your code really so bad? Is it worth the time to tune it? Over the lifetime of your application, how much time is going to be spent running that code, compared to the time spent waiting for a remote database server, or waiting for user input?
Second, are you sure you're done coding? Premature optimization is like spreading frosting on a half-baked cake. You spend hours or days (or more) optimizing your code for performance, only to discover it doesn't do what you need it to do. That's time down the drain.
This is not to say that code optimization is worthless, but you need to look at the whole system and decide whether it's the best use of your time. Every minute you spend optimizing code is a minute you're not spending adding new features, or writing documentation, or playing with your kids, or writing unit tests.
Oh yes, unit tests. It should go without saying that you need a complete set of unit tests before you begin performance tuning. The last thing you need is to introduce new bugs while fiddling with your algorithms.
With these caveats in place, let's look at some techniques for optimizing Python code. The code in question is an implementation of the Soundex algorithm. Soundex was a method used in the early 20th century for categorizing surnames in the United States census. It grouped similar-sounding names together, so even if a name was misspelled, researchers had a chance of finding it. Soundex is still used today for much the same reason, although of course we use computerized database servers now. Most database servers include a Soundex function.
There are several subtle variations of the Soundex algorithm. This is the one used in this chapter:
For example, my name, Pilgrim, becomes P942695. That has no consecutive duplicates, so nothing to do there. Then you remove the 9s, leaving P4265. That's too long, so you discard the excess character, leaving P426.
Another example: Woo becomes W99, which becomes W9, which becomes W, which gets padded with zeros to become W000.
Here's a first attempt at a Soundex function:
もしまだダウンロードしていないのなら, 良かったらこの本で使われているこの例や他の例をダウンロードしてみてください.
import string, re charToSoundex = {"A": "9", "B": "1", "C": "2", "D": "3", "E": "9", "F": "1", "G": "2", "H": "9", "I": "9", "J": "2", "K": "2", "L": "4", "M": "5", "N": "5", "O": "9", "P": "1", "Q": "2", "R": "6", "S": "2", "T": "3", "U": "9", "V": "1", "W": "9", "X": "2", "Y": "9", "Z": "2"} def soundex(source): "convert string to Soundex equivalent" # Soundex requirements: # source string must be at least 1 character # and must consist entirely of letters allChars = string.uppercase + string.lowercase if not re.search('^[%s]+$' % allChars, source): return "0000" # Soundex algorithm: # 1. make first character uppercase source = source[0].upper() + source[1:] # 2. translate all other characters to Soundex digits digits = source[0] for s in source[1:]: s = s.upper() digits += charToSoundex[s] # 3. remove consecutive duplicates digits2 = digits[0] for d in digits[1:]: if digits2[-1] != d: digits2 += d # 4. remove all "9"s digits3 = re.sub('9', '', digits2) # 5. pad end with "0"s to 4 characters while len(digits3) < 4: digits3 += "0" # 6. return first 4 characters 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())
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