CODEWITHSUNDEEP
pythonpandasdataframe
Dave
Dave

Reputation: 129

Join two pandas dataframes based on lists columns

I have 2 dataframes containing columns of lists.
I would like to join them based on 2+ shared values on the lists. Example:

ColumnA ColumnB        | ColumnA ColumnB        
id1     ['a','b','c']  | id3     ['a','b','c','x','y', 'z']
id2     ['a','d,'e']   |

In this case we can see that id1 matches id3 because there are 2+ shared values on the lists. So the output will be (columns name are not important and just for example):

    ColumnA1 ColumnB1     ColumnA2   ColumnB2        
    id1      ['a','b','c']  id3     ['a','b','c','x','y', 'z']

How can I achieve this result? I've tried to iterate each row in dataframe #1 but it doesn't seem a good idea.
Thank you!

Upvotes: 2

Views: 94

Answers (2)

SeaBean
SeaBean

Reputation: 23217

If you are using pandas 1.2.0 or newer (released on December 26, 2020), cartesian product (cross joint) can be simplified as follows:

    df = df1.merge(df2, how='cross')         # simplified cross joint for pandas >= 1.2.0

Also, if system performance (execution time) is a concern to you, it is advisable to use list(map... instead of the slower apply(... axis=1)

Using apply(... axis=1):

%%timeit
df['overlap'] = df.apply(lambda x: 
                         len(set(x['ColumnB1']).intersection(
                             set(x['ColumnB2']))), axis=1)


800 µs ± 59.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

while using list(map(...:

%%timeit
df['overlap'] = list(map(lambda x, y: len(set(x).intersection(set(y))), df['ColumnB1'], df['ColumnB2']))

217 µs ± 25.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Notice that using list(map... is 3x times faster!

Whole set of codes for your reference:

    data = {'ColumnA1': ['id1', 'id2'], 'ColumnB1': [['a', 'b', 'c'], ['a', 'd', 'e']]}
    df1 = pd.DataFrame(data)

    data = {'ColumnA2': ['id3', 'id4'], 'ColumnB2': [['a','b','c','x','y', 'z'], ['d','e','f','p','q', 'r']]}
    df2 = pd.DataFrame(data)

    df = df1.merge(df2, how='cross')             # for pandas version >= 1.2.0

    df['overlap'] = list(map(lambda x, y: len(set(x).intersection(set(y))), df['ColumnB1'], df['ColumnB2']))

    df = df[df['overlap'] >= 2]
    print (df)

Upvotes: 0

mujjiga
mujjiga

Reputation: 16876

Using cartesian product of rows and checking each row

Code is documented in-line

df1 = pd.DataFrame(
    {
        'ColumnA': ['id1', 'id2'],
        'ColumnB': [['a','b','c'], ['a','d','e']],
    }
)

df2 = pd.DataFrame(
    {
        'ColumnA': ['id3'],
        'ColumnB': [['a','b','c','x','y', 'z']],
    }
)

# Take cartesian product of both dataframes
df1['k'] = 0
df2['k'] = 0
df = pd.merge(df1, df2, on='k').drop('k',1)
# Check the overlap of the lists and find the overlap length
df['overlap'] = df.apply(lambda x: len(set(x['ColumnB_x']).intersection(
                                   set(x['ColumnB_y']))), axis=1)
# Select whoes overlap length > 2
df = df[df['overlap'] > 2]
print (df)

Output:

  ColumnA_x  ColumnB_x ColumnA_y           ColumnB_y  overlap
0       id1  [a, b, c]       id3  [a, b, c, x, y, z]        3

Upvotes: 2

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