擅长:python、mysql、java
<p>可能不是最有效的解决方案,但使用<code>itertools.groupby</code>和{<cd2>}:</p>
<pre><code>>>> df['player'] = list(itertools.chain.from_iterable([key] + [float('nan')]*(len(list(val))-1)
for key, val in itertools.groupby(df['player'].tolist())))
>>> df
player points
0 LBJ 25
1 NaN 32
2 NaN 26
3 Kyrie 21
4 NaN 29
5 LBJ 21
6 NaN 35
</code></pre>
<p>更具体地说,这说明了它是如何工作的:</p>
^{pr2}$
<p>给予:</p>
<pre><code>['LBJ', nan, nan]
['Kyrie', nan]
['LBJ', nan]
</code></pre>
<p>然后被“锁”在一起。在</p>