[英]Dataframe groupby - list of values
我有以下數據幀:
driver_id status dttm
9f8f9bf3ee8f4874873288c246bd2d05 free 2018-02-04 00:19
9f8f9bf3ee8f4874873288c246bd2d05 busy 2018-02-04 01:03
8f174ffd446c456eaf3cca0915d0368d free 2018-02-03 15:43
8f174ffd446c456eaf3cca0915d0368d enroute 2018-02-03 17:02
3列:driver_id,status,dttm
我需要做的是按驅動程序ID分組,並將所有狀態列表及其各自的dttm值放入名為'driver_info'
新列中:
driver_id driver_info
9f8f9bf3ee8f4874873288c246bd2d05 [("free", 2018-02-04 00:19), ("busy", 2018-02-04 01:03)]
8f174ffd446c456eaf3cca0915d0368d [("free", 2018-02-03 15:43), ("enroute", 2018-02-03 17:02) ...]
我如何在python 3中做到這一點?
我試過了
dfg = df.groupby("driver_id").apply(lambda x: pd.concat((x["status"], x["dttm"])))
但結果與我的預期不同......
嘗試:使用zip和apply(列表)
df['driver_info'] = list(zip(df['status'], df['dttm']))
df = df.groupby('driver_id')['driver_info'].apply(list)
將GroupBy.apply
與list
和zip
一起使用以獲取元組列表:
df1 = (df.groupby('driver_id')
.apply(lambda x: list(zip(x['status'], x['dttm'])))
.reset_index(name='driver_info'))
print (df1)
driver_id \
0 8f174ffd446c456eaf3cca0915d0368d
1 9f8f9bf3ee8f4874873288c246bd2d05
driver_info
0 [(free, 2018-02-03 15:43), (enroute, 2018-02-0...
1 [(free, 2018-02-04 00:19), (busy, 2018-02-04 0...
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