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      <title>18.6.&nbsp;优化字符串操作</title>
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                  <h2 class="title"><a name="soundex.stage4"></a>18.6.&nbsp;优化字符串操作
                  </h2>
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            <div></div>
         </div>
         <div class="abstract">
            <p>Soundex 算法的最后一步是对短结果补零和截短长结果。最佳的做法是什么？</p>
         </div>
         <p>这是目前在 <tt class="filename">soundex/stage2/soundex2c.py</tt> 中的做法：
         </p>
         <div class="informalexample"><pre class="programlisting">
    digits3 = re.sub(<span class='pystring'>'9'</span>, <span class='pystring'>''</span>, digits2)
    <span class='pykeyword'>while</span> len(digits3) &lt; 4:
        digits3 += <span class='pystring'>"0"</span>
    <span class='pykeyword'>return</span> digits3[:4]
</pre></div>
         <p>这里是 <tt class="filename">soundex2c.py</tt> 的表现：
         </p>
         <div class="informalexample"><pre class="screen">
<tt class="prompt">C:\samples\soundex\stage2&gt;</tt><span class="userinput">python soundex2c.py</span>
<span class="computeroutput">Woo             W000 12.6070768771
Pilgrim         P426 14.4033353401
Flingjingwaller F452 19.7774882003</span>
</pre></div>
         <p>思考的第一件事是以循环取代正则表达式。这里的代码来自 <tt class="filename">soundex/stage4/soundex4a.py</tt>：
         </p>
         <div class="informalexample"><pre class="programlisting">
    digits3 = <span class='pystring'>''</span>
    <span class='pykeyword'>for</span> d <span class='pykeyword'>in</span> digits2:
        <span class='pykeyword'>if</span> d != <span class='pystring'>'9'</span>:
            digits3 += d
</pre></div>
         <p><tt class="filename">soundex4a.py</tt> 快了吗？是的：
         </p>
         <div class="informalexample"><pre class="screen">
<tt class="prompt">C:\samples\soundex\stage4&gt;</tt><span class="userinput">python soundex4a.py</span>
<span class="computeroutput">Woo             W000 6.62865531792
Pilgrim         P426 9.02247576158
Flingjingwaller F452 13.6328416042</span>
</pre></div>
         <p>但是，等一下。一个从字符串去除字符的循环？我们可以用一个简单的字符串方法做到。这便是  <tt class="filename">soundex/stage4/soundex4b.py</tt>：
         </p>
         <div class="informalexample"><pre class="programlisting">
    digits3 = digits2.replace(<span class='pystring'>'9'</span>, <span class='pystring'>''</span>)
</pre></div>
         <p><tt class="filename">soundex4b.py</tt> 快了吗？这是个有趣的问题，它取决输入值：
         </p>
         <div class="informalexample"><pre class="screen">
<tt class="prompt">C:\samples\soundex\stage4&gt;</tt><span class="userinput">python soundex4b.py</span>
<span class="computeroutput">Woo             W000 6.75477414029
Pilgrim         P426 7.56652144337
Flingjingwaller F452 10.8727729362</span>
</pre></div>
         <p><tt class="filename">soundex4b.py</tt> 中的字符串方法对于大多数名字比循环快，但是对于短小的情况 (很短的名字) 却比 <tt class="filename">soundex4a.py</tt> 略微慢些。性能优化并不总是一致的，对于一个情况快些，却可能对另外一些情况慢些。就此而言，大多数情况将会从改变中获益，所以就改吧，但是别忘了原则。
         </p>
         <p>最后仍很重要的是，让我们检测算法的最后两步：以零补齐短结果和截短超过四字符的长结果。你在  <tt class="filename">soundex4b.py</tt> 中看到的代码就是做这个工作的，但是太没效率了。看一下 <tt class="filename">soundex/stage4/soundex4c.py</tt> 找出原因：
         </p>
         <div class="informalexample"><pre class="programlisting">
    digits3 += <span class='pystring'>'000'</span>
    <span class='pykeyword'>return</span> digits3[:4]
</pre></div>
         <p>我们为什么需要一个 <tt class="literal">while</tt> 循环来补齐结果？我们早就知道我们需要把结果截成四字符，并且我们知道我们已经有了至少一个字符 (直接从 <tt class="varname">source</tt> 中拿过来的起始字符)。这意味着我们可以仅仅在输出的结尾添加三个零，然后截断它。不要害怕重新理解问题，从不太一样的角度看问题可以获得简单的解决方案。
         </p>
         <p>我们丢弃 <tt class="literal">while</tt> 循环后从 <tt class="filename">soundex4c.py</tt> 中获得怎样的速度？太明显了：
         </p>
         <div class="informalexample"><pre class="screen">
<tt class="prompt">C:\samples\soundex\stage4&gt;</tt><span class="userinput">python soundex4c.py</span>
<span class="computeroutput">Woo             W000 4.89129791636
Pilgrim         P426 7.30642134685
Flingjingwaller F452 10.689832367</span>
</pre></div>
         <p>最后，还有一件事可以令这三行运行得更快：你可以把它们合并为一行。看一眼 <tt class="filename">soundex/stage4/soundex4d.py</tt>：
         </p>
         <div class="informalexample"><pre class="programlisting">
    <span class='pykeyword'>return</span> (digits2.replace(<span class='pystring'>'9'</span>, <span class='pystring'>''</span>) + <span class='pystring'>'000'</span>)[:4]
</pre></div>
         <p>在 <tt class="filename">soundex4d.py</tt> 中把所有代码放在一行可以比 <tt class="filename">soundex4c.py</tt> 稍微快那么一点：
         </p>
         <div class="informalexample"><pre class="screen">
<tt class="prompt">C:\samples\soundex\stage4&gt;</tt><span class="userinput">python soundex4d.py</span>
<span class="computeroutput">Woo             W000 4.93624105857
Pilgrim         P426 7.19747593619
Flingjingwaller F452 10.5490700634</span>
</pre></div>
         <p>它非常难懂，而且优化也不明显。这值得吗？我希望你有很好的见解。性能并不是一切。你在优化方面的努力应该与程序的可读性和可维护性相平衡。</p>
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