擅长:python、mysql、java
<p>使用<code>re.findall()</code>函数的短解:</p>
<pre><code>s = '2014 2026 202 20 1000 1949 194 195092 20111a a2011a a2011 keep this text n0t th1s th0ugh 1 0 2015 2025 2026'
result = ''.join(re.findall(r'\b(19[5-9][0-9]|20[01][0-9]|202[0-5]|[a-z]+|[^0-9a-z]+)\b', s, re.I))
print(result)
</code></pre>
<p>输出:</p>
^{pr2}$