Pandas: Drop duplicates in col[A] keeping row based on condition on col[B]

问题

Given the dataframe:

df = pd.DataFrame({'col1': ['A', 'A', 'A','B','B'], 'col2': ['type1', 'type2', 'type1', 'type2', 'type1'] , 'hour': ['18:03:30','18:00:48', '18:13:46', '18:11:29', '18:06:31']  })

col1 col2   hour
A   type1   18:03:30 # Drop this row as (A type1) already present
A   type2   18:00:48
A   type1   18:13:46 # keep this row as (A type1) already present.
B   type2   18:11:29
B   type1   18:06:31

I want to drop duplicates based on col1,col2.

eg.(row(0): A type1, row(2): A type1)

keeping only the row that has the latest hour eg.(18:13:46).

I tried using groupby to return subset based on col1, and drop_duplicates to drop the duplicate in col2. I need to find a way to pass the condition (latest hour)

example code:

for key, grp in df.groupby('col1'):
  grp.drop_duplicates(subset='col2', keep="LATEST OF HOUR") 

Expected outcome:

col1 col2   hour
A   type1   18:03:30
A   type2   18:00:48
B   type2   18:11:29
B   type1   18:06:31

EDIT adding context

my original dataframe is larger, the solution needs to work for also:

col1 col2   other  hour
A   type1   h  18:03:30 # Drop this row as (A type1) already present
A   type2   ss 18:00:48
A   type1   ll 18:13:46 # keep this row as (A type1) already present
B   type2   mm 18:11:29
B   type1   jj 18:06:31

it would still need to drop the column based on the hour

回答1:

df.drop_duplicates(['col1','col2'] , keep = 'last')

回答2:

Following anky_91's comment I solved it like this:

df.sort_values('hour').drop_duplicates(['col1','col2'] , keep = 'last')

This sorts based on the column 'hour' so that you are sure that keep='last' gets the last element

来源：https://stackoverflow.com/questions/57807305/pandas-drop-duplicates-in-cola-keeping-row-based-on-condition-on-colb