这一串代码的xpath怎么写

尝试提取class=title,也就是 勿忘2020,前行2021

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    <a class="title" href="/p/f9d3626adf38">勿忘2020,前行2021</a>
    <p class="abstract">2020年像是患了拖延症,连个人总结都拖到了大年三十。是的,罗罗攀又陪伴了大家一年的时光,当然更新的频率不高,大几个月都处于断更状态,感谢还没有取关的你们。 是的,还是每年都...</p>

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import re
a = re.findall(r'<a class="title".*</a>',input())
print(a)

不如用re来解决