擅长:python、mysql、java
<p>更多<code>itertools</code>糖:</p>
<pre><code>>>> parts = s.split(designatedstring)
>>> num = len(parts) - 1
>>> replacements = itertools.product([designatedstring, replacerstring], repeat=num)
>>> replacements = list(replacements)
>>> replacements.remove((designatedstring,) * num)
>>> for r in replacements:
... print ''.join(itertools.chain(*zip(parts, r + ('',))))
...
thrupought
throughput
throughpought
</code></pre>
<p>如果你能忍受结果中的原始字符串,你可以省略4行和5行的难看的转换。在</p>