How to drop unique rows in a pandas dataframe?

Solutions for select all duplicated rows:

You can use duplicated with subset and parameter keep=False for select all duplicates:

df = df[df.duplicated(subset=['A','B'], keep=False)]
print (df)
     A  B  C
1  foo  1  A
2  foo  1  B

Solution with transform:

df = df[df.groupby(['A', 'B'])['A'].transform('size') > 1]
print (df)
     A  B  C
1  foo  1  A
2  foo  1  B

A bit modified solutions for select all unique rows:

#invert boolean mask by ~
df = df[~df.duplicated(subset=['A','B'], keep=False)]
print (df)
     A  B  C
0  foo  0  A
3  bar  1  A

df = df[df.groupby(['A', 'B'])['A'].transform('size') == 1]
print (df)
     A  B  C
0  foo  0  A
3  bar  1  A

I came up with a solution using groupby:

groupped = df.groupby(['A', 'B']).size().reset_index().rename(columns={0: 'count'})
uniques = groupped[groupped['count'] == 1]
duplicates = df[~df.index.isin(uniques.index)]

Duplicates now has the proper result:

    A       B   C
2   foo     1   B
3   bar     1   A

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Also, my original attempt in the question can be fixed by simply adding keep=False in the drop_duplicates method:

# Load Dataframe
df = pd.DataFrame({"A":["foo", "foo", "foo", "bar"], "B":[0,1,1,1], "C":["A","A","B","A"]})

uniques = df[['A', 'B']].drop_duplicates(keep=False)
duplicates = df[~df.index.isin(uniques.index)]

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Please @jezrael answer, I think it is safest(?), as I am using pandas indexes here.