Simple DataFrame:
df = pd.DataFrame({'A': [1,1,2,2], 'B': [0,1,2,3], 'C': ['a','b','c','d']})
df
A B C
0 1 0 a
1 1 1 b
2 2 2 c
3 2 3 d
I wish for every value (groupby) of column A, to get the value of column C, for which column B is maximum. For example for group 1 of column A, the maximum of column B is 1, so I want the value "b" of column C:
A C
0 1 b
1 2 d
No need to assume column B is sorted, performance is of top priority, then elegance.
Check with sort_values +drop_duplicates
df.sort_values('B').drop_duplicates(['A'],keep='last')
Out[127]:
A B C
1 1 1 b
3 2 3 d
df.groupby('A').apply(lambda x: x.loc[x['B'].idxmax(), 'C'])
# A
#1 b
#2 d
Use idxmax to find the index where B is maximal, then select column C within that group (using a lambda-function
Here's a little fun with groupby and nlargest:
(df.set_index('C')
.groupby('A')['B']
.nlargest(1)
.index
.to_frame()
.reset_index(drop=True))
A C
0 1 b
1 2 d
Or, sort_values, groupby, and last:
df.sort_values('B').groupby('A')['C'].last().reset_index()
A C
0 1 b
1 2 d
Similar solution to @Jondiedoop, but avoids the apply:
u = df.groupby('A')['B'].idxmax()
df.loc[u, ['A', 'C']].reset_index(drop=1)
A C
0 1 b
1 2 d
