熊猫基于日期的分组以返回条件中的行

Question

I have a dataframe that has data like

1. id       date                   remarks
2. 1       12-01-2015 12:00:15     Good
3. 2       12-01-2015 1:00:14      OK
4. 1       13-01-2015 12:00:15     Not Ok
5. 2       14-01-2015 1:00:15      Bad
6. 3       15-01-2015 1:00:15      Good

I need the output in such a way that for each id the highest date and remarks is returned, so for id 2 it would return 14-01-2015 1:00:15 and remark as bad

Answer 1

You need sort_values + groupby + GroupBy.last :

df['date'] = pd.to_datetime(df['date'], dayfirst=True)

df1 = df.sort_values('date').groupby('id', as_index=False).last()
print (df1)
   id                date remarks
0   1 2015-01-13 12:00:15  Not Ok
1   2 2015-01-14 01:00:15     Bad
2   3 2015-01-15 01:00:15    Good

Answer 2

I hope your date column is in dayfirst format if thats so,you need groupby on id with idxmax on date and reuse then from loc lookup. If its not in dayfirst format then idxmin() will help

df.loc[df.groupby('id')['date'].idxmax()]

Output:

id                date remarks
2   1 2015-01-13 12:00:15  Not Ok
3   2 2015-01-14 01:00:15     Bad
4   3 2015-01-15 01:00:15    Good

If you dont want the index and intend to create a new dataframe with new index then (Thanks @Zero)

df.loc[df.groupby('id')['date'].idxmax()].reset_index(drop=T‌​rue)