Question

我想对数据集进行分组并返回最大和最小时间戳。这是我的数据

id  timestamp
1   2017-09-17 10:09:01
2   2017-10-02 01:13:15
1   2017-09-17 10:53:07
1   2017-09-17 10:52:18
2   2017-09-12 21:59:40

这是我想要的输出

id    max                   min
1     2017-09-17 10:53:07   2017-09-17 10:09:01
2     2017-10-02 01:13:15   2017-09-12 21:59:40

在这里，我做了什么，代码效率似乎不高，我希望在熊猫上做更好的方法

data1 = df.sort_values('timestamp').drop_duplicates(['customer_id'], keep='last')
data2 = df.sort_values('timestamp').drop_duplicates(['customer_id'], keep='first')
data1['max'] = data1['timestamp']
data2['min'] = data2['timestamp']
data = data1.merge(data2, on = 'customer_id', how='left')
data = data.drop(['timestamp_x','timestamp_y'], axis=1)

似乎熊猫有这种类型的支点

Answer 1

我认为需要agg：

df = df.groupby('id')['timestamp'].agg(['min','max']).reset_index()
print (df)
   id                 min                 max
0   1 2017-09-17 10:09:01 2017-09-17 10:53:07
1   2 2017-09-12 21:59:40 2017-10-02 01:13:15

或者稍微修改一下你的解决方案（应该更快）：

data = df.sort_values('timestamp')
data1 = data.drop_duplicates(['id'], keep='last').set_index('id')
data2 = data.drop_duplicates(['id'], keep='first').set_index('id')

df = pd.concat([data1['timestamp'], data2['timestamp']],keys=('max','min'), axis=1)

print (df)
                   max                 min
id                                        
1  2017-09-17 10:53:07 2017-09-17 10:09:01
2  2017-10-02 01:13:15 2017-09-12 21:59:40

如何对pandas dataframe

1 个答案: