python - pandas - filter dataframe by another dataframe by row elements

Question

I have a dataframe df1 which looks like:

   c  k  l
0  A  1  a
1  A  2  b
2  B  2  a
3  C  2  a
4  C  2  d

and another called df2 like:

   c  l
0  A  b
1  C  a

I would like to filter df1 keeping only the values that ARE NOT in df2. Values to filter are expected to be as (A,b) and (C,a) tuples. So far I tried to apply the isin method:

d = df[~(df['l'].isin(dfc['l']) & df['c'].isin(dfc['c']))]

That seems to me too complicated, it returns:

   c  k  l
2  B  2  a
4  C  2  d

but I'm expecting:

   c  k  l
0  A  1  a
2  B  2  a
4  C  2  d

Answer 1

You can do this efficiently using isin on a multiindex constructed from the desired columns:

df1 = pd.DataFrame({'c': ['A', 'A', 'B', 'C', 'C'],
                    'k': [1, 2, 2, 2, 2],
                    'l': ['a', 'b', 'a', 'a', 'd']})
df2 = pd.DataFrame({'c': ['A', 'C'],
                    'l': ['b', 'a']})
keys = list(df2.columns.values)
i1 = df1.set_index(keys).index
i2 = df2.set_index(keys).index
df1[~i1.isin(i2)]

I think this improves on @IanS's similar solution because it doesn't assume any column type (i.e. it will work with numbers as well as strings).

---

(Above answer is an edit. Following was my initial answer)

Interesting! This is something I haven't come across before... I would probably solve it by merging the two arrays, then dropping rows where df2 is defined. Here is an example, which makes use of a temporary array:

df1 = pd.DataFrame({'c': ['A', 'A', 'B', 'C', 'C'],
                    'k': [1, 2, 2, 2, 2],
                    'l': ['a', 'b', 'a', 'a', 'd']})
df2 = pd.DataFrame({'c': ['A', 'C'],
                    'l': ['b', 'a']})

# create a column marking df2 values
df2['marker'] = 1

# join the two, keeping all of df1's indices
joined = pd.merge(df1, df2, on=['c', 'l'], how='left')
joined

# extract desired columns where marker is NaN
joined[pd.isnull(joined['marker'])][df1.columns]

There may be a way to do this without using the temporary array, but I can't think of one. As long as your data isn't huge the above method should be a fast and sufficient answer.