Python 3：显示字典键值对之间的差异

Question

I'd like to find the differences between two dictionaries that have multiple values for a key. All the examples that I find, the dictionaries have a key which contains only one value. Say you have keys that have multiple values as in the following example:

pizza_1 = {"toppings": ["cheese", "pepperoni", "mushroom"],
           "crust": ["deep dish", "hand tossed", "thin"],
           "size": ["large", "medium", "small"],
           "price": ["$12.99", "$9.99", "$7.99"]}

pizza_2 = {"toppings": ["cheese", "pepperoni", "olive"],
           "crust": ["deep dish", "traditional", "thin"],
           "size": ["large", "medium", "small"],
           "brand": ["Domino's", "Pizza Hut", "Little Caesars"]}

I want to return only the differences in the two dictionaries including both the key and the values. Either the difference in pizza_1 or pizza_2, it does not matter which dictionary. Example of what I'm looking for below:

print(differences)
"toppings": ["mushroom"]
"crust": ["hand tossed"]
"price": ["$12.99", "$9.99", "$7.99"]

I'm not sure exactly how it will output, but wanted to give an example of what I am sort of looking for. Thanks in advance for taking your time to help!

Answer 1

Loop over the union of keys, treat the values as sets and print the set difference :

for key in pizza_1.keys() | pizza_2:  # union of the dict views
    difference = set(pizza_1.get(key, [])).difference(pizza_2.get(key, []))
    if difference:
        print(key, list(difference))

I'm using the dict.keys() dictionary view here to provide the union of dictionary keys. The if test filters out empty results.

If you wanted this as a dictionary, you can produce one with a generator expression plus a dict comprehension to avoid producing the sets more than once:

differences = ((key, list(set(pizza_1.get(key, [])).difference(pizza_2.get(key, []))))
              for key in pizza_1.keys() | pizza_2)
differences = {k: v for k, v in differences if v}

Demo:

>>> for key in pizza_1.keys() | pizza_2:  # union of the dict views
...     difference = set(pizza_1.get(key, [])).difference(pizza_2.get(key, []))
...     if difference:
...         print(key, list(difference))
...
crust ['hand tossed']
toppings ['mushroom']
price ['$9.99', '$7.99', '$12.99']
>>> differences = ((key, list(set(pizza_1.get(key, [])).difference(pizza_2.get(key, []))))
...               for key in pizza_1.keys() | pizza_2)
>>> {k: v for k, v in differences if v}
{'crust': ['hand tossed'], 'toppings': ['mushroom'], 'price': ['$9.99', '$7.99', '$12.99']}

Answer 2

I propose a dictionary with set-values as your output-datastructure.

>>> {k: set(v).difference(pizza_2.get(k, {})) for k, v in pizza_1.items()}
{'price': {'$9.99', '$7.99', '$12.99'}, 'size': set(), 'toppings': {'mushroom'}, 'crust': {'hand tossed'}}

The result will hold an empty set for a key where there's no difference (see size).

Answer 3

What I did was loop through keys and use list comprehensions to get the difference (specifically what's in pizza_1 that's not in pizza_2 as your example output shows).

def getDiff(dict1, dict2):
    diff = {}

    for key in dict1:
        if key not in dict2:
            diff[key] = dict1[key]
        elif dict1[key] != dict2[key]:
            diff[key] = [e for e in dict1[key] if e not in dict2[key]]

    return diff