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Consider the following DataFrame:

import pandas as pd



df = pd.DataFrame({'id': [1, 2, 3],

               'json_col': [ [{'aa' : 1, 'ab' : 1}, {'aa' : 3, 'ab' : 2, 'ac': 6}],

                             [{'aa' : 1, 'ab' : 2, 'ac': 1}, {'aa' : 5}],

                             [{'aa': 3, 'ac': 2}] ]})

df

Out[134]: 

   id                                           json_col

0   1  [{'aa': 1, 'ab': 1}, {'aa': 3, 'ab': 2, 'ac': 6}]

1   2           [{'aa': 1, 'ab': 2, 'ac': 1}, {'aa': 5}]

2   3                               [{'aa': 3, 'ac': 2}]

We can see that we have a list of jsons for each id.

I'd like, for each 'id' and for each corresponding json in its list, to have a 'row' in the DataFrame. So the following DataFrame will look like this:

   id  aa   ab   ac

0   1   1  1.0  NaN

1   1   3  2.0  6.0

2   2   1  2.0  1.0

3   2   5  NaN  NaN

4   3   3  NaN  2.0

We can see, id '1' had 2 corresponding jsons in it's list and therefor it gets 2 rows in the new DataFrame

Is there a pythonic way to do so using panda, numpy or json functionality?

Adding the run times of the solutions

setup = """

import pandas as pd

df = pd.DataFrame({'id': [1, 2, 3],

               'json_col': [ [{'aa' : 1, 'ab' : 1}, {'aa' : 3, 'ab' : 2, 'ac': 6}],

                             [{'aa' : 1, 'ab' : 2, 'ac': 1}, {'aa' : 5}],

                             [{'aa': 3, 'ac': 2}] ]})

"""



s1 = """

df = pd.concat(

       [pd.DataFrame(j, index=[i]*len(j)) for i, j in enumerate(df['json_col'], 1)],

       sort=False

     )                             

"""



s2 = """

recs = df.apply(lambda x: [{**{'id': x.id}, **d} for d in x.json_col], axis=1).sum()

df2 = pd.DataFrame.from_records(recs)

"""



%timeit(s1, setup)

52.3 ns ± 2.6 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)

%timeit(s2, setup)

50.6 ns ± 3.28 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)

Here is one quick way by converting all the json_col's lists of dictionaries to DataFrame and concatenating them together plus some tweaks to create the id column:

In [51]: df = pd.concat(

           [pd.DataFrame(j, index=[i]*len(j)) for i, j in enumerate(json_col, 1)],

           sort=False

         )



In [52]: df.index.name = 'id'



In [53]: df.reset_index()

Out[53]: 

   id  aa   ab   ac

0   1   1  1.0  NaN

1   1   3  2.0  6.0

2   2   1  2.0  1.0

3   2   5  NaN  NaN

4   3   3  NaN  2.0

a short way to accomplish this would be the following, although I don't personally consider it very pythonic as the code is a little hard to read, and not terribly performant, but for small data wrangling this should do the trick:

recs = df.apply(lambda x: [{**{'id': x.id}, **d} for d in x.json_col], axis=1).sum()

df2 = pd.DataFrame.from_records(recs)

# outputs:

   aa   ab   ac  id

0   1  1.0  NaN   1

1   3  2.0  6.0   1

2   1  2.0  1.0   2

3   5  NaN  NaN   2

4   3  NaN  2.0   3

How it works:

1. The applied lambda creates a new dictionary by merging the contents of {id: x.id} to each dictionary in the list of dictionaries in x.json_col (where x is a row).



2. 
   

   This is then summed. Since summing a lists of list of elements unites them into a big list of elements, recs has the following form

   
   
   

   [{'id': 1, 'aa': 1, 'ab': 1},
   
    {'id': 1, 'aa': 3, 'ab': 2, 'ac': 6},
   
    {'id': 2, 'aa': 1, 'ab': 2, 'ac': 1},
   
    {'id': 2, 'aa': 5},
   
    {'id': 3, 'aa': 3, 'ac': 2}]
   
   

   
   



3. A new data frame is then simply constructed from the records.