Python：组合复杂字典的最有效方法

Question

Let's say I have two dicts consisting of multi-leveled sets:

houses:
  - house4:
      - level1:
          - unitA
          - unitB
      - level8:
          - unitG
  - house6:
      - level5:
          - unitK

and

houses:
  - house6:
      - level4:
          - unitT
      - level5:
          - unitK
  - house3:
      - level8:
          - unitG

What is the most efficient way to combine them without duplicates?

I can write a three level loop and generate a new dict by recombining each subset with set.union but that kind of doesn't look nice.

Answer 1

a ={"house4": {
        "level1": ["unitA", "unitB"],
        "level8": ["unitG"],
   },
    "house6": {
    "level6": ["unitK"]
    }
}
b = {"house6": {
        "level4": ["unitT"],
        "level5": ["unitK"]
    },
    "house3": {
        "level8": ["unitG"]
    }
}
a.update(b)

keys in "a" are replaced by "b" ie "a" once updated will have the value of "house 6" from dictionary "b". If you need to resolve duplicates at a lower level in your dictionary, make sure to post a question that demos exactly what you need--it's unclear.

Answer 2

Assuming that your last level of these dictionaries are sets of strings, you could use nested loops:

for k1,v1 in b.items():                         # e.g. 'houses': {...
    d1 = a.setdefault(k1,dict())
    for k2,v2 in v1.items():                    # e.g. 'house6': {...
        d2 = d1.setdefault(k2,dict())
        for k3,v3 in v2.items():                # e.g. 'level4': {'unitT'}
            d2.setdefault(k3,list()).update(v3)

Note that the sub-dictionaries added to a are new instances (not references to the ones in b . So if the content of the b structure is changed afterward, you won't get undesirable side effect in the a structure.

If the number of levels in the dictionaries is unknown, you can define a recursive function that does the same thing for any number of levels:

def merge(a,b):
    if isinstance(a,dict):
        for k,v in b.items(): merge(a.setdefault(k,type(v)()),v)
    elif isinstance(a,set):   a.update(b)
    elif isinstance(a,list):  a.extend(b)

Answer 3

Assuming the units are sets, you can do it with 2 for-loops.

Iterate over b and check if a house in it exists in a , if not just add it to a . If it is, then iterate over the levels in that house and check if a level in it exists in the same house in a . If there is, then add these new units to the existing level in a , if not add this level to the same house in a :

for house, levels in b.items():
    if house in a:
        for level, units in levels.items():
            if level in a[house]:
                a[house][level].update(units)
            else:
                a[house][level] = units
    else:
        a[house] = levels

The same can be constructed using dict.setdefault as well:

for house, levels in b.items():
    a.setdefault(house, dict())
    for level, units in levels.items():
        a[house].setdefault(level, set()).update(units)

print(a)

Output:

{'house4': {'level1': {'unitA', 'unitB'}, 'level8': ['unitG']},
 'house6': {'level6': {'unitK'}, 'level4': {'unitT'}, 'level5': {'unitK'}},
 'house3': {'level8': {'unitG'}}}