Consolidating a dictionary from nested dictionary

Question

Currently learning from "LPTHW - Zed Shaw" and "Automating the boring stuff".

There is a nested dictionary of the type {'string1': {inner dict 1}, 'string2':{inner dict 2},...}.

I want a non nested consolidated dictionary of all the keys and values from all of the {inner dicts}. I have written the code below, but I am not convinced it is the most efficient way. I am looking for suggestions on how the code can be simplified by utilising logic within a single loop or by making use of some other modules and methods.

Is that new method efficient from a processing point of view?

allGuests = {'Alice': {'apples': 5, 'pretzels': 12},
           'Bob': {'ham sandwiches': 3, 'apples': 2},
           'Carol': {'cups': 3, 'apple pies': 1}}

item_list = []
count_list = []
consolidated = {}
all_items_list = list(allGuests.values()) #This is a list of sub dictionaries

#Loop for the number of guests who brought items
for guest in range(0, len(allGuests)):

#Creating 2 separate lists for 'items' and 'their count'

    item_list = item_list + list(all_items_list[guest].keys())
    count_list = count_list + list(all_items_list[guest].values())

#Creating a consolidated dictionary    
for index in range(0, len(item_list)):

    if not consolidated.get(item_list[index]):
        consolidated.setdefault(item_list[index], count_list[index])

    else:
        consolidated[item_list[index]] = consolidated[item_list[index]] + count_list[index]

    pprint.pprint(consolidated)

Expected output:

{'apple pies': 1, 'apples': 7, 'cups': 3, 'ham sandwiches': 3, 'pretzels': 12}

Answer 1

You can use .get method with default parameter.

consolidated = {}
for guest in allGuests.values():
    for key, value in guest.items():
        consolidated[key] = consolidated.get(key, 0) + value

Output: {'apples': 7, 'pretzels': 12, 'ham sandwiches': 3, 'cups': 3, 'apple pies': 1}

Answer 2

If your dictionary always has the structure {key: {key: int}}, i.e. the values of the inner dictionaries are all integers, this can be easily achieved using collections.Counter.update:

from collections import Counter

def consolidate(d):
    consolidated = Counter()
    for sub_dict in d.values():
        consolidated.update(sub_dict)
    return dict(consolidated)


if __name__ == "__main__":
    all_guests = {'Alice': {'apples': 5, 'pretzels': 12},
                  'Bob': {'ham sandwiches': 3, 'apples': 2},
                  'Carol': {'cups': 3, 'apple pies': 1}}
    print(consolidate(all_guests))
    # {'apple pies': 1, 'apples': 7, 'cups': 3, 'ham sandwiches': 3, 'pretzels': 12}

Note that Python has an official style-guide, PEP8, that recommends using lower_case both for functions and for variables. I also put this code into a function, so it is reusable, and only called it with the test case under a if __name__ == "__main__": guard to allow importing from this module without executing the test case.

Answer 3

Below code worked as per one suggestion on stackoverflow

  consolidated = {}
  for guest in allGuests.values():
      for key, value in guest.items():
          if not consolidated.get(key):
              consolidated.setdefault(key, value)

          else:
              consolidated[key] = consolidated[key] + value

          pprint.pprint(consolidated)