pandas: how to aggregate two list columns when joining data frames

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I have the following two data frames:

    id  websites
    --   ---
0   1   [cnn.com, bbc.com]
1   2   [ebay.com, facebook.com]

________________

    id  websites
    --   ---
0   2   [google.com, facebook.com]
1   3   [amazon.com, youtube.com]

I want to outer join them on the id column by aggregating unique websites for matched rows. The output should be as below:

    id  websites    
    --   ---
0   1   [cnn.com, bbc.com]  
1   2   [ebay.com, facebook.com, google.com]
2   3   [amazon.com, youtube.com]

I have tried the following so far:

import pandas as pd

df_a = pd.DataFrame({'id':[1,2],'websites':[['cnn.com','bbc.com'],['ebay.com','facebook.com']]})
df_b = pd.DataFrame({'id':[2,3],'websites':[['google.com','facebook.com'],['amazon.com','youtube.com']]})
df_a.merge(df_b, on='id', how='outer')

which is giving me the following output:

    id  websites_x                 websites_y
    --   ---                        ---
0   1   [cnn.com, bbc.com]         NaN
1   2   [ebay.com, facebook.com]   [google.com, facebook.com]
2   3   NaN                        [amazon.com, youtube.com]
pandas outer-join dataframe list python
1 Answers

You can concat them and then groupby on id column:

df_a = pd.DataFrame({'id':[1,2],'websites':[['cnn.com','bbc.com'],
                    ['ebay.com','facebook.com']]})
df_b = pd.DataFrame({'id':[2,3],'websites':[['google.com','facebook.com'],
                    ['amazon.com','youtube.com']]})

Solution:

Method1:

a = df_a.explode('websites') #requires pandas version 0.25+
b = df_b.explode('websites') #requires pandas version 0.25+
out = pd.concat((a,b)).groupby('id')['websites'].apply(pd.unique).reset_index()
#or out = pd.concat((a,b)).groupby('id')['websites'].agg(set).reset_index()
print(out)

Method2:

Another solution using itertools.chain.from_iterable which doesnot need the exploded dataframes:

from itertools import chain
out = (pd.concat((df_a,df_b)).groupby('id')['websites']
     .apply(lambda x : dict.fromkeys(chain.from_iterable(x)).keys()).reset_index())
print (out)

   id                              websites
0   1                    [cnn.com, bbc.com]
1   2  [ebay.com, facebook.com, google.com]
2   3             [amazon.com, youtube.com]
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