Pandas: Find rows which don't exist in another DataFrame by multiple columns

Question

same as this python pandas: how to find rows in one dataframe but not in another? but with multiple columns

This is the setup:

import pandas as pd

df = pd.DataFrame(dict(
    col1=[0,1,1,2],
    col2=['a','b','c','b'],
    extra_col=['this','is','just','something']
))

other = pd.DataFrame(dict(
    col1=[1,2],
    col2=['b','c']
))

Now, I want to select the rows from df which don't exist in other. I want to do the selection by col1 and col2

In SQL I would do:

select * from df 
where not exists (
    select * from other o 
    where df.col1 = o.col1 and 
    df.col2 = o.col2
)

And in Pandas I can do something like this but it feels very ugly. Part of the ugliness could be avoided if df had id-column but it's not always available.

key_col = ['col1','col2']
df_with_idx = df.reset_index()
common = pd.merge(df_with_idx,other,on=key_col)['index']
mask = df_with_idx['index'].isin(common)

desired_result =  df_with_idx[~mask].drop('index',axis=1)

So maybe there is some more elegant way?

Solution 1

Since 0.17.0 there is a new indicator param you can pass to merge which will tell you whether the rows are only present in left, right or both:

In [5]:
merged = df.merge(other, how='left', indicator=True)
merged

Out[5]:
   col1 col2  extra_col     _merge
0     0    a       this  left_only
1     1    b         is       both
2     1    c       just  left_only
3     2    b  something  left_only

In [6]:    
merged[merged['_merge']=='left_only']

Out[6]:
   col1 col2  extra_col     _merge
0     0    a       this  left_only
2     1    c       just  left_only
3     2    b  something  left_only

So you can now filter the merged df by selecting only 'left_only' rows

Solution 2

Interesting

cols = ['col1','col2']
#get copies where the indeces are the columns of interest
df2 = df.set_index(cols)
other2 = other.set_index(cols)
#Look for index overlap, ~
df[~df2.index.isin(other2.index)]

Returns:

    col1 col2  extra_col
0     0    a       this
2     1    c       just
3     2    b  something

Seems a little bit more elegant...