List of dictionaries:
data = [{
\'a\':{\'l\':\'Apple\',
\'b\':\'Milk\',
\'d\':\'Meatball\'},
\'b\':{\'favou
You can do the following, using itertools.chain:
>>> from itertools import chain
# timeit: ~3.40
>>> [dict(chain(*map(dict.items, d.values()))) for d in data]
[{'l': 'Apple',
'b': 'Milk',
'd': 'Meatball',
'favourite': 'coke',
'dislike': 'juice'},
{'l': 'Apple1',
'b': 'Milk1',
'dislike': 'juice3',
'favourite': 'coke2',
'd': 'Meatball2'}]
The usage of chain, map, * make this expression a shorthand for the following doubly nested comprehension which actually performs better on my system (Python 3.5.2) and isn't that much longer:
# timeit: ~2.04
[{k: v for x in d.values() for k, v in x.items()} for d in data]
# Or, not using items, but lookup by key
# timeit: ~1.67
[{k: x[k] for x in d.values() for k in x} for d in data]
RoadRunner's loop-and-update approach outperforms both these one-liners at timeit: ~1.37