python - Pandas 识别组中第一行的列值

Question

我有一个包含三列的数据框:

    ID       Date    Status
0    1   1/1/2000  Complete
1    1   1/4/2000  ReOpened
2    1  1/10/2000  ReOpened
3    1  1/11/2000    Closed
4    1  1/15/2000  ReOpened
5    2   1/2/2000  ReOpened
6    2   1/4/2000  ReOpened
7    2  1/10/2000    Closed
8    3  1/20/2000    Closed
9    3  1/22/2000    Closed
10   4  1/25/2000  ReOpened

对于每个 ID，如果有“重新打开”状态，我需要根据日期获取显示第一次“重新打开”的行。所以我的输出看起来像:

   ID ProductionDate    Status
0   1       1/4/2000  ReOpened
1   2       1/2/2000  ReOpened
2   4      1/25/2000  ReOpened

我尝试过: df = pd.np.where(df.Status.str.contains("ReOpened"), df.groupby(['ID']).first(),0) 但这并不工作。

Answer 1

在蒙版上使用 groupby 和 cumsum 执行此操作:

df[df['Status'].eq('ReOpened').groupby(df['ID']).cumsum() == 1] 

    ID       Date    Status
1    1   1/4/2000  ReOpened
5    2   1/2/2000  ReOpened
10   4  1/25/2000  ReOpened 

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您还可以在过滤后使用 groupby 和 first 来仅获取第一行:

df[df['Status'].eq('ReOpened')].groupby('ID', as_index=False).first()  

   ID       Date    Status
0   1   1/4/2000  ReOpened
1   2   1/2/2000  ReOpened
2   4  1/25/2000  ReOpened

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如果性能很重要，您可以使用 eq 和 duplicated 将上述内容简化为单个 bool 索引操作:

df[df['Status'].eq('ReOpened') & ~df.duplicated(['ID', 'Status'])] 

    ID       Date    Status
1    1   1/4/2000  ReOpened
5    2   1/2/2000  ReOpened
10   4  1/25/2000  ReOpened