擅长:python、mysql、java
<p>你可以简单地写下:</p>
<pre><code>df['difference'] = df.groupby('player')['date'].diff().fillna(0)
</code></pre>
<p>这将为新的timedelta列提供正确的值:</p>
<pre><code> player date difference
0 A 2010-01-01 0 days
1 A 2010-01-09 8 days
2 A 2010-01-11 2 days
3 A 2010-01-15 4 days
4 B 2010-02-01 0 days
5 B 2010-02-10 9 days
6 B 2010-02-21 11 days
7 B 2010-02-23 2 days
</code></pre>
<p>(我使用了名称“difference”而不是“diff”来区分名称和方法<code>diff</code>)</p>