Pandas Column Dict Split To New Column And Rows

I have a dict in pandas dataframe column, the input is, import pandas as pd df = pd.DataFrame([{'A': {'k1': 10}}, {'A': {'k2': 20, 'k3': 30}}, {'A': {'k4': 15}}]) df

Solution 1:

Use list comprehension with flatenning for tuples and then DataFrame contructor:

L = [(k1, v1) for k, v indf['A'].to_dict().items() for k1, v1 in v.items()]

df = pd.DataFrame(L, columns = ['keys','values'])
print (df)
  keys  values
0   k1      10
1   k2      20
2   k3      30
3   k4      15

Or create DataFrame and stack:

df = (pd.DataFrame(df['A'].values.tolist())
       .stack().reset_index(level=0, drop=True)
       .reset_index())
df.columns = ['keys','values']
print (df)
  keys  values
0   k1    10.0
1   k2    20.0
2   k3    30.0
3   k4    15.0

Solution 2:

Option 1 (If you have all unique keys in sub-dictionaries) dict with collections.ChainMap

from collections import ChainMap   
dct = dict(ChainMap(*[i['A'] for i in d]))
pd.DataFrame(list(dct.items()), columns=['key', 'value'])

  key  value
0  k1     101  k4     152  k2     203  k3     30

Option 2 (If you might have duplicate keys) itertools.chain.from_iterable

dct = list(itertools.chain.from_iterable([i['A'].items() for i in d]))
df = pd.DataFrame(dct, columns=['key', 'value'])

  key  value
0  k1     10
1  k2     20
2  k3     30
3  k4     15