Update nested dict with missing key value pairs

Question

I have a dict with a unique id and nested dicts of key-value pairs corresponding to cardinal positions:

{'925328413719459770': {'N': 14,
  'NNW': 116,
  'NW': 38,
  'S': 1,
  'SE': 2,
  'SSE': 1,
  'W': 16,
  'WNW': 146,
  'WSW': 16}}

However, not all nested dicts will contain each of the 16 possible cardinal positions:

cp = ["N", "NNE", "NE", "ENE", "E", "ESE", "SE", "SSE",
        "S", "SSW", "SW", "WSW", "W", "WNW", "NW", "NNW"]

I'm able to get the difference between the set of cardinal positions in the dict compared to the full list cp

for leg in legdict.values():
    print(list(set(cp) - set(leg.keys())))
    
['ESE', 'SSW', 'NNE', 'NE', 'ENE', 'SW', 'E']

How would one then go about updating the nested dict with the missing positions and setting their values to 0? The output would be something like this:

  {'925328413719459770': {'N': 14,
      'NNW': 116,
      'NW': 38,
      'S': 1,
      'SE': 2,
      'SSE': 1,
      'W': 16,
      'WNW': 146,
      'WSW': 16,
      'ESE': 0, 
      'SSW': 0, 
      'NNE': 0, 
      'NE': 0, 
      'ENE':0, 
      'SW': 0, 
      'E': 0 }}

Is there an efficient way of doing this in python using the update() method, or can a list be used to update a dict while setting all elements of that list to 0? The actual legdict is very large so any suggestions about an efficient method would be appreciated.

Answer 1

What about this:

cardinal_positions = [
    "N", "NNE", "NE", "ENE", "E", "ESE", "SE", "SSE",
    "S", "SSW", "SW", "WSW", "W", "WNW", "NW", "NNW"
]

def fill_missing_keys(the_huge_dict):
    """Fills in missing cardinal positions with 0"""

    for inner_dict in the_huge_dict.values():
        for key in cardinal_positions:
            if key not in inner_dict:
                inner_dict[key] = 0

Should be reasonably efficient in that it doesn't make any copies, it just iterates over everything and fills in missing keys where necessary.

As a general rule, you should always just go with the simplest solution first. If it turns out to be too slow, optimize later.