CODEWITHSUNDEEP
pythonpandaspandas-groupby
Egor Maksimov
Egor Maksimov

Reputation: 62

Dataframe groupby - list of values

I have a following dataframe:

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 columns : driver_id, status, dttm

What I need to do is to group by driver id and make list of all statuses with their respective dttm values into new column called 'driver_info':

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) ...]

How do I do that in python 3?

I tried 

dfg = df.groupby("driver_id").apply(lambda x: pd.concat((x["status"], x["dttm"])))

but the result differs from what I expect it to be...

Upvotes: 3

Views: 125

Answers (2)

jezrael
jezrael

Reputation: 862641

Use GroupBy.apply with list and zip for list of tuples:

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...

Upvotes: 2

Frenchy
Frenchy

Reputation: 17007

try: using zip and apply(list)

df['driver_info'] = list(zip(df['status'], df['dttm']))
df = df.groupby('driver_id')['driver_info'].apply(list)

Upvotes: 2

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