擅长:python、mysql、java
<p>您可以使用regex:</p>
<pre><code>import re
s = ['10.10.10.0/255.255.255.0', '10.10.20.0/255.255.255.192']
final_ips = ['{}/{}'.format(a, sum([bin(int(x)).count('1') for x in b.split('.')])) for a, b in map(lambda x:re.findall('[\d\.]+', x), s)]
</code></pre>
<p>输出:</p>
^{pr2}$