Reputation: 344
Say I have the pandas DataFrame below:
A B C D
1 foo one 0 0
2 foo one 2 4
3 foo two 4 8
4 cat one 8 4
5 bar four 6 12
6 bar three 7 14
7 bar four 7 14
I would like to select all the rows that have equal values in A but differing values in B. So I would like the output of my code to be:
A B C D
1 foo one 0 0
3 foo two 4 8
5 bar three 7 14
6 bar four 7 14
What's the most efficient way to do this? I have approximately 11,000 rows with a lot of variation in the column values, but this situation comes up a lot. In my dataset, if elements in column A are equal then the corresponding column B value should also be equal, however due to mislabeling this is not the case and I would like to fix this, it would be impractical for me to do this one by one.
Upvotes: 11
Views: 29878
Reputation: 8816
You can try groupby()
+ filter
+ drop_duplicates()
:
>>> df.groupby('A').filter(lambda g: len(g) > 1).drop_duplicates(subset=['A', 'B'], keep="first")
A B C D
0 foo one 0 0
2 foo two 4 8
4 bar four 6 12
5 bar three 7 14
OR, in case you want to drop duplicates between the subset of columns A
& B
then can use below but that will have the row having cat
as well.
>>> df.drop_duplicates(subset=['A', 'B'], keep="first")
A B C D
0 foo one 0 0
2 foo two 4 8
3 cat one 8 4
4 bar four 6 12
5 bar three 7 14
Upvotes: 12
Reputation: 502
The current answers are correct and may be more sophisticated too. If you have complex criteria, filter function will be very useful. If you are like me and want to keep things simple, i feel following is more beginner friendly way
>>> df = pd.DataFrame({
'A': ['foo', 'foo', 'foo', 'cat', 'bar', 'bar', 'bar'],
'B': ['one', 'one', 'two', 'one', 'four', 'three', 'four'],
'C': [0,2,4,8,6,7,7],
'D': [0,4,8,4,12,14,14]
}, index=[1,2,3,4,5,6,7])
>>> df = df.drop_duplicates(['A', 'B'], keep='last')
A B C D
2 foo one 2 4
3 foo two 4 8
4 cat one 8 4
6 bar three 7 14
7 bar four 7 14
>>> df = df[df.duplicated(['A'], keep=False)]
A B C D
2 foo one 2 4
3 foo two 4 8
6 bar three 7 14
7 bar four 7 14
keep='last'
is optional here
Upvotes: 1
Reputation: 61910
result = df.groupby('A').filter(lambda g: len(g) > 1).groupby(['A', 'B']).head(1)
print(result)
Output
A B C D
0 foo one 0 0
2 foo two 4 8
4 bar four 6 12
5 bar three 7 14
The first group-by and filter will remove the rows with no duplicated A
values (i.e. cat
), the second will create groups with same A, B
and for each of those get the first element.
Upvotes: 4