Interacting through dicts, grabbing their values and transitioning to a panda df

Question

I have a list of dicts

[{a:'jeffrey',b:'pineapple',c:'apple'},{a:'epstein',c:'banana'},{a:'didnt kill'},{a:'himself',b:'jebus'}]

What I want to do is transition those values to a pandas df. But as you can see a few dicts are lacking a few keys and therefore lacking values. So I took a glance at defaultdict object so I could transform the list object to something that pandas actually is able to interpret. And transform it into a dataframe.

dd = defaultdict(list)

for d in l:
    for k in d.keys():
        dd[k]

for d in l:
    for k in dd.keys():
        try: 
            dd[k].append(d[k])
        except KeyError:
            dd[k].append(0)
# Dict auto adaptavél

The code works, and folows the order given of those events meaning with the key is empty return a 0. But I was wondering if there better alternative or a code which has a better o(n) complexity

Wanted result:

defaultdict(<class 'list'>, {'a': ['jeffrey', 'epstein', 'didnt kill', 'himself'], 'b': ['pineapple', 0, 0, 'jebus'], 'c': ['apple', 'banana', 0, 0]})

Answer 1

You can use DataFrame constructor and fill missing values with 0 then use to_dict method to export the dataframe as a dict of lists:

>>> pd.DataFrame(l).fillna(0).to_dict('list')
{'a': ['jeffrey', 'epstein', 'didnt kill', 'himself'],
 'b': ['pineapple', 0, 0, 'jebus'],
 'c': ['apple', 'banana', 0, 0]}

Intermediate result:

>>> pd.DataFrame(l)
            a          b       c
0     jeffrey  pineapple   apple
1     epstein        NaN  banana
2  didnt kill        NaN     NaN
3     himself      jebus     NaN