The currently selected solution produces incorrect results. To correctly solve this problem, we can perform a left-join from df1
to df2
, making sure to first get just the unique rows for df2
.
First, we need to modify the original DataFrame to add the row with data [3, 10].
df1 = pd.DataFrame(data = {'col1' : [1, 2, 3, 4, 5, 3],
'col2' : [10, 11, 12, 13, 14, 10]})
df2 = pd.DataFrame(data = {'col1' : [1, 2, 3],
'col2' : [10, 11, 12]})
df1
col1 col2
0 1 10
1 2 11
2 3 12
3 4 13
4 5 14
5 3 10
df2
col1 col2
0 1 10
1 2 11
2 3 12
Perform a left-join, eliminating duplicates in df2
so that each row of df1
joins with exactly 1 row of df2
. Use the parameter indicator
to return an extra column indicating which table the row was from.
df_all = df1.merge(df2.drop_duplicates(), on=['col1','col2'],
how='left', indicator=True)
df_all
col1 col2 _merge
0 1 10 both
1 2 11 both
2 3 12 both
3 4 13 left_only
4 5 14 left_only
5 3 10 left_only
Create a boolean condition:
df_all['_merge'] == 'left_only'
0 False
1 False
2 False
3 True
4 True
5 True
Name: _merge, dtype: bool
A few solutions make the same mistake - they only check that each value is independently in each column, not together in the same row. Adding the last row, which is unique but has the values from both columns from df2
exposes the mistake:
common = df1.merge(df2,on=['col1','col2'])
(~df1.col1.isin(common.col1))&(~df1.col2.isin(common.col2))
0 False
1 False
2 False
3 True
4 True
5 False
dtype: bool
This solution gets the same wrong result:
df1.isin(df2.to_dict('l')).all(1)