<p>一般方法:</p>
<pre><code>def checkEqual1(iterator):
iterator = iter(iterator)
try:
first = next(iterator)
except StopIteration:
return True
return all(first == rest for rest in iterator)
</code></pre>
<p>一行:</p>
<pre><code>def checkEqual2(iterator):
return len(set(iterator)) <= 1
</code></pre>
<p>还有一行:</p>
<pre><code>def checkEqual3(lst):
return lst[1:] == lst[:-1]
</code></pre>
<p>三个版本的区别在于:</p>
<ol>
<li>在<code>checkEqual2</code>中,内容必须是可散列的。</li>
<li><code>checkEqual1</code>和<code>checkEqual2</code>可以使用任何迭代器,但是<code>checkEqual3</code>必须接受序列输入,通常是列表或元组之类的具体容器。</li>
<li><code>checkEqual1</code>发现差异后立即停止。</li>
<li>由于<code>checkEqual1</code>包含更多的Python代码,因此当许多项在开始时相等时,效率会降低。</li>
<li>由于<code>checkEqual2</code>和<code>checkEqual3</code>总是执行O(N)复制操作,因此如果大多数输入返回False,它们将花费更长的时间。</li>
<li>对于<code>checkEqual2</code>和<code>checkEqual3</code>来说,很难适应从<code>a == b</code>到<code>a is b</code>的比较。</li>
</ol>
<hr/>
<p><code>timeit</code>结果,对于Python 2.7和(只有s1、s4、s7、s9应该返回True)</p>
<pre><code>s1 = [1] * 5000
s2 = [1] * 4999 + [2]
s3 = [2] + [1]*4999
s4 = [set([9])] * 5000
s5 = [set([9])] * 4999 + [set([10])]
s6 = [set([10])] + [set([9])] * 4999
s7 = [1,1]
s8 = [1,2]
s9 = []
</code></pre>
<p>我们得到</p>
<pre><code> | checkEqual1 | checkEqual2 | checkEqual3 | checkEqualIvo | checkEqual6502 |
|-----|-------------|-------------|--------------|---------------|----------------|
| s1 | 1.19 msec | 348 usec | 183 usec | 51.6 usec | 121 usec |
| s2 | 1.17 msec | 376 usec | 185 usec | 50.9 usec | 118 usec |
| s3 | 4.17 usec | 348 usec | 120 usec | 264 usec | 61.3 usec |
| | | | | | |
| s4 | 1.73 msec | | 182 usec | 50.5 usec | 121 usec |
| s5 | 1.71 msec | | 181 usec | 50.6 usec | 125 usec |
| s6 | 4.29 usec | | 122 usec | 423 usec | 61.1 usec |
| | | | | | |
| s7 | 3.1 usec | 1.4 usec | 1.24 usec | 0.932 usec | 1.92 usec |
| s8 | 4.07 usec | 1.54 usec | 1.28 usec | 0.997 usec | 1.79 usec |
| s9 | 5.91 usec | 1.25 usec | 0.749 usec | 0.407 usec | 0.386 usec |
</code></pre>
<p>注:</p>
<pre><code># http://stackoverflow.com/q/3844948/
def checkEqualIvo(lst):
return not lst or lst.count(lst[0]) == len(lst)
# http://stackoverflow.com/q/3844931/
def checkEqual6502(lst):
return not lst or [lst[0]]*len(lst) == lst
</code></pre>