If you must use a list comprehension, then you must also rebuild each dictionary:
matchesData = [{k: str(v) if k == 'match_date' else v for k, v in d.items()} for d in matchesData]
or use
matchesData = [dict(d, match_date=str(d['match_date'])) for d in matchesData]
The first example uses a dictionary comprehension, that simply creates a new dictionary from the old by looping over each key-value pair. For the 'match_date' key the values is passed through the str() function. This means the match_date key is entirely optional.
The second version requires 'match_date' to exist, and creates a copy of the original dictionary with the dict() function, adding in an extra key (which will replace the original 'match_date' key-value pair from d).
Demo:
>>> from pprint import pprint >>> matchesData = [ ... {'match_id': 1L, 'player_b_id': 2L, 'round_id': 1L, 'match_winner_id': 2L, 'match_date': datetime.date(2016, 3, 9), 'player_a_id': 1L, 'tournament_id': 1L}, ... {'match_id': 2L, 'player_b_id': 4L, 'round_id': 1L, 'match_winner_id': 4L, 'match_date': datetime.date(2016, 3, 10), 'player_a_id': 3L, 'tournament_id': 1L} ... ] >>> pprint([{k: str(v) if k == 'match_date' else v for k, v in d.items()} for d in matchesData]) [{'match_date': '2016-03-09', 'match_id': 1L, 'match_winner_id': 2L, 'player_a_id': 1L, 'player_b_id': 2L, 'round_id': 1L, 'tournament_id': 1L}, {'match_date': '2016-03-10', 'match_id': 2L, 'match_winner_id': 4L, 'player_a_id': 3L, 'player_b_id': 4L, 'round_id': 1L, 'tournament_id': 1L}] >>> pprint([dict(d, match_date=str(d['match_date'])) for d in matchesData]) [{'match_date': '2016-03-09', 'match_id': 1L, 'match_winner_id': 2L, 'player_a_id': 1L, 'player_b_id': 2L, 'round_id': 1L, 'tournament_id': 1L}, {'match_date': '2016-03-10', 'match_id': 2L, 'match_winner_id': 4L, 'player_a_id': 3L, 'player_b_id': 4L, 'round_id': 1L, 'tournament_id': 1L}]
