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18.6. Optimizing String Manipulation

## 18.6. Optimizing String Manipulation

The final step of the Soundex algorithm is padding short results with zeros, and truncating long results. What is the best way to do this?

This is what we have so far, taken from soundex/stage2/soundex2c.py:

```    digits3 = re.sub('9', '', digits2)
while len(digits3) < 4:
digits3 += "0"
return digits3[:4]
```

These are the 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 replacing that regular expression with a loop. This code is from soundex/stage4/soundex4a.py:

```    digits3 = ''
for d in digits2:
if d != '9':
digits3 += d
```

Is soundex4a.py faster? Yes it is:

```C:\samples\soundex\stage4>python soundex4a.py
Woo             W000 6.62865531792
Pilgrim         P426 9.02247576158
Flingjingwaller F452 13.6328416042
```

But wait a minute. A loop to remove characters from a string? We can use a simple string method for that. Here's soundex/stage4/soundex4b.py:

```    digits3 = digits2.replace('9', '')
```

Is soundex4b.py faster? That's an interesting question. It depends on the input:

```C:\samples\soundex\stage4>python soundex4b.py
Woo             W000 6.75477414029
Pilgrim         P426 7.56652144337
Flingjingwaller F452 10.8727729362
```

The string method in soundex4b.py is faster than the loop for most names, but it's actually slightly slower than soundex4a.py in the trivial case (of a very short name). Performance optimizations aren't always uniform; tuning that makes one case faster can sometimes make other cases slower. In this case, the majority of cases will benefit from the change, so let's leave it at that, but the principle is an important one to remember.

Last but not least, let's examine the final two steps of the algorithm: padding short results with zeros, and truncating long results to four characters. The code you see in soundex4b.py does just that, but it's horribly inefficient. Take a look at soundex/stage4/soundex4c.py to see why:

```    digits3 += '000'
return digits3[:4]
```

Why do we need a while loop to pad out the result? We know in advance that we're going to truncate the result to four characters, and we know that we already have at least one character (the initial letter, which is passed unchanged from the original source variable). That means we can simply add three zeros to the output, then truncate it. Don't get stuck in a rut over the exact wording of the problem; looking at the problem slightly differently can lead to a simpler solution.

How much speed do we gain in soundex4c.py by dropping the while loop? It's significant:

```C:\samples\soundex\stage4>python soundex4c.py
Woo             W000 4.89129791636
Pilgrim         P426 7.30642134685
Flingjingwaller F452 10.689832367
```

Finally, there is still one more thing you can do to these three lines of code to make them faster: you can combine them into one line. Take a look at soundex/stage4/soundex4d.py:

```    return (digits2.replace('9', '') + '000')[:4]
```

Putting all this code on one line in soundex4d.py is barely faster than soundex4c.py:

```C:\samples\soundex\stage4>python soundex4d.py
Woo             W000 4.93624105857
Pilgrim         P426 7.19747593619
Flingjingwaller F452 10.5490700634
```

It is also significantly less readable, and for not much performance gain. Is that worth it? I hope you have good comments. Performance isn't everything. Your optimization efforts must always be balanced against threats to your program's readability and maintainability.

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