Remove duplicates from dataframe, based on two columns A,B, keeping row with max value in another column C
Question:
I have a pandas dataframe which contains duplicates values according to two columns (A and B):
A B C
1 2 1
1 2 4
2 7 1
3 4 0
3 4 8
I want to remove duplicates keeping the row with max value in column C. This would lead to:
A B C
1 2 4
2 7 1
3 4 8
I cannot figure out how to do that. Should I use drop_duplicates(), something else?
Answers:
I think groupby should work.
df.groupby(['A', 'B']).max()['C']
If you need a dataframe back you can chain the reset index call.
df.groupby(['A', 'B']).max()['C'].reset_index()
You can do it using group by:
c_maxes = df.groupby(['A', 'B']).C.transform(max)
df = df.loc[df.C == c_maxes]
c_maxes is a Series of the maximum values of C in each group but which is of the same length and with the same index as df. If you haven’t used .transform then printing c_maxes might be a good idea to see how it works.
Another approach using drop_duplicates would be
df.sort('C').drop_duplicates(subset=['A', 'B'], take_last=True)
Not sure which is more efficient but I guess the first approach as it doesn’t involve sorting.
EDIT:
From pandas 0.18 up the second solution would be
df.sort_values('C').drop_duplicates(subset=['A', 'B'], keep='last')
or, alternatively,
df.sort_values('C', ascending=False).drop_duplicates(subset=['A', 'B'])
In any case, the groupby solution seems to be significantly more performing:
%timeit -n 10 df.loc[df.groupby(['A', 'B']).C.max == df.C]
10 loops, best of 3: 25.7 ms per loop
%timeit -n 10 df.sort_values('C').drop_duplicates(subset=['A', 'B'], keep='last')
10 loops, best of 3: 101 ms per loop
You can do it with drop_duplicates as you wanted
# initialisation
d = pd.DataFrame({'A' : [1,1,2,3,3], 'B' : [2,2,7,4,4], 'C' : [1,4,1,0,8]})
d = d.sort_values("C", ascending=False)
d = d.drop_duplicates(["A","B"])
If it’s important to get the same order
d = d.sort_index()
You can do this simply by using pandas drop duplicates function
df.drop_duplicates(['A','B'],keep= 'last')
I have a pandas dataframe which contains duplicates values according to two columns (A and B):
A B C
1 2 1
1 2 4
2 7 1
3 4 0
3 4 8
I want to remove duplicates keeping the row with max value in column C. This would lead to:
A B C
1 2 4
2 7 1
3 4 8
I cannot figure out how to do that. Should I use drop_duplicates(), something else?
I think groupby should work.
df.groupby(['A', 'B']).max()['C']
If you need a dataframe back you can chain the reset index call.
df.groupby(['A', 'B']).max()['C'].reset_index()
You can do it using group by:
c_maxes = df.groupby(['A', 'B']).C.transform(max)
df = df.loc[df.C == c_maxes]
c_maxes is a Series of the maximum values of C in each group but which is of the same length and with the same index as df. If you haven’t used .transform then printing c_maxes might be a good idea to see how it works.
Another approach using drop_duplicates would be
df.sort('C').drop_duplicates(subset=['A', 'B'], take_last=True)
Not sure which is more efficient but I guess the first approach as it doesn’t involve sorting.
EDIT:
From pandas 0.18 up the second solution would be
df.sort_values('C').drop_duplicates(subset=['A', 'B'], keep='last')
or, alternatively,
df.sort_values('C', ascending=False).drop_duplicates(subset=['A', 'B'])
In any case, the groupby solution seems to be significantly more performing:
%timeit -n 10 df.loc[df.groupby(['A', 'B']).C.max == df.C]
10 loops, best of 3: 25.7 ms per loop
%timeit -n 10 df.sort_values('C').drop_duplicates(subset=['A', 'B'], keep='last')
10 loops, best of 3: 101 ms per loop
You can do it with drop_duplicates as you wanted
# initialisation
d = pd.DataFrame({'A' : [1,1,2,3,3], 'B' : [2,2,7,4,4], 'C' : [1,4,1,0,8]})
d = d.sort_values("C", ascending=False)
d = d.drop_duplicates(["A","B"])
If it’s important to get the same order
d = d.sort_index()
You can do this simply by using pandas drop duplicates function
df.drop_duplicates(['A','B'],keep= 'last')