<p>只是把这个<code>map</code>版本扔进去。列表理解更像是python,但这里有一个替代方法。你知道吗</p>
<pre><code>>>> list1 = ["aa", "bb", "cc"]
>>> list2 = ["dd", "ee", "ffg"]
>>> map("".join, zip(list1, list2))
['aadd', 'bbee', 'ccffg']
</code></pre>
<p>你甚至可以用这个来创建一个生成器。你知道吗</p>
<pre><code>>>> from itertools import izip, imap
>>> final = imap("".join, izip(list1, list2))
>>> next(final)
'aadd'
>>> next(final)
'bbee'
</code></pre>
<p>下面是三种不同解决方案的一些计时结果。(发电机一号打败了所有人)</p>
<pre><code>>>> timeit('[a+b for a, b in zip(list1, list2)]', 'list1 = ["aa", "bb", "cc"]*100; list2 = ["dd", "ee", "ffg"]*100', number=10000)
0.4470004917966719
>>> timeit('map("".join, zip(list1, list2))', 'list1 = ["aa", "bb", "cc"]*100; list2 = ["dd", "ee", "ffg"]*100', number=10000)
0.43502864982517053
>>> timeit('imap("".join, izip(list1, list2))', 'from itertools import imap, izip; list1 = ["aa", "bb", "cc"]*100; list2 = ["dd", "ee", "ffg"]*100', number=10000)
0.011020268755800089
>>> timeit('[a+b for a, b in izip(list1, list2)]', 'from itertools import izip; list1 = ["aa", "bb", "cc"]*100; list2 = ["dd", "ee", "ffg"]*100', number=10000)
0.32172862839627214
>>> timeit('map(lambda x,y: x + y, list1, list2)', 'list1 = ["aa", "bb", "cc"]*100; list2 = ["dd", "ee", "ffg"]*100', number=10000)
0.5423113458890612
</code></pre>