擅长:python、mysql、java
<p>这是第二次尝试理解您的问题:</p>
<pre><code>def replaceit(s, replacefrom, replaceto, n):
new_s, count = '', 0
for letter in s:
if letter == replacefrom:
count += 1
if count == n:
new_s += replaceto
continue
new_s += letter
return new_s
</code></pre>
<p>这与您的示例相匹配:</p>
^{pr2}$
<p>如果这不是你想要的,请解释清楚。在</p>
<p>也可以使用正则表达式实现相同的效果:</p>
<pre><code>def replaceit(s, replacefrom, replaceto, n):
import re
if n <= 0:
return s
return re.sub('(.*?%s)%s' % (('%s.*?' % replacefrom) * (n-1), replacefrom), r'\1%s' % replaceto, s)
</code></pre>