擅长:python、mysql、java
<p>请你试试:</p>
<pre><code>df.replace(regex=True, inplace=True, to_replace=r'^\(?(?:[ivxlcdm]+|[a-zA-Z]+|[0-9]+)[).]', value='')
</code></pre>
<p>输入:</p>
<pre><code>(i) The cow has four legs.
(ii) The cow eats grass.
(iii) Cow gives us milk.
a.The cow has four legs.
b.The cow eats grass.
c.Cow gives us milk.
1.The cow has four legs.
2.The cow eats grass.
3.Cow gives us milk.
a)The cow has four legs.
b)The cow eats grass.
c)Cow gives us milk.
</code></pre>
<p>输出:</p>
<pre><code>The cow eats grass.
Cow gives us milk.
The cow has four legs.
The cow eats grass.
Cow gives us milk.
The cow has four legs.
The cow eats grass.
Cow gives us milk.
The cow has four legs.
The cow eats grass.
Cow gives us milk.
</code></pre>
<p>regex <code>^\(?(?:[ivxlcdm]+|[a-zA-Z]+|[0-9]+)[).]</code>的解释:</p>
<ul>
<li><code>^</code>表示字符串的开头</李>
<li><code>\(?</code>匹配零或一个左括号</李>
<li><code>(?:[ivxlcdm]+|[a-zA-Z]+|[0-9]+)</code>可以分解为以下任一种:
<ul>
<li><code>[ivxlcdm]+</code>匹配罗马数字的</李>
<li><code>[a-zA-Z]+</code>匹配字母表的</李>
<li><code>[0-9]+</code>匹配数字</李>
</ul>
</li>
<li><code>[).]</code>匹配右括号或点</李>
</ul>