Summarize a list of dictionaries based on common key values

Question

I have a list of dictionaries like so:

dictlist = [{\'day\': 0, \'start\': \'8:00am\', \'end\': \'5:00pm\'},
            {\'day\': 1, \'start\': \'10:00am\', \'e         


        

Answer 1

If you don't need the exact format that you provide you could use defaultdict

dictlist = [{'day': 0, 'start': '8:00am', 'end': '5:00pm'},
            {'day': 1, 'start': '10:00am', 'end': '7:00pm'},
            {'day': 2, 'start': '8:00am', 'end': '5:00pm'},
            {'day': 3, 'start': '10:00am', 'end': '7:00pm'},
            {'day': 4, 'start': '8:00am', 'end': '5:00pm'},
            {'day': 5, 'start': '11:00am', 'end': '1:00pm'}]

from collections import defaultdict

dd = defaultdict(list)

for d in dictlist:
    dd[(d['start'],d['end'])].append(d['day'])

Result:

>>> dd
defaultdict(, {('11:00am', '1:00pm'): [5], ('10:00am', '7:00pm'): [1, 3], ('8:00am', '5:00pm'): [0, 2, 4]})

And if format is important to you could do:

>>> my_list = [(v, k[0], k[1]) for k,v in dd.iteritems()]
>>> my_list
[([5], '11:00am', '1:00pm'), ([1, 3], '10:00am', '7:00pm'), ([0, 2, 4], '8:00am', '5:00pm')]
>>> # If you need the output sorted:  
>>> sorted_my_list = sorted(my_list, key = lambda k : len(k[0]), reverse=True)
>>> sorted_my_list
[([0, 2, 4], '8:00am', '5:00pm'), ([1, 3], '10:00am', '7:00pm'), ([5], '11:00am', '1:00pm')]