将Python字典与列表作为值进行比较

Question

我正在为蒸汽编写一个解析器机器人，它将跟踪从蒸汽用户的库存中进出的物品。我编写了一个代码，它获取用户的所有项，并以字典的形式返回，其中KEY =用户名、VALUE = ITEM名称及其数量，现在需要比较Data1和Data2 (更新的数据)。

Data1 = {
            'user1': [('AWP', 1), ('DEAGLE', 2), ('AK-47', 3)], 
            'user2': [('MP-7', 1), ('KNIFE', 1), ('GLOVES', 1)]

        }

Data2 = {
              
            'user1': [('AWP', 1), ('DEAGLE', 2), ('AK-47', 3), ('M4A4', 1)], 
            'user2': [('MP-7', 1), ('KNIFE', 1), ('GLOVES', 1), ('GLOCK-18', 1)] 

        }

例如，Data2中出现了一个新项，我需要将它写到第三个字典中，保留键。如果项目离开Data2，也必须这样做(这意味着这些项目不再存在于用户的库存中)。

Result = {
              
            'user1': [('M4A4', 1)], 
            'user2': [('GLOCK-18', 1)] 

        }

Answer 1

使用sets

result_addition = {d:list(set(Data2[d]) - set(Data1[d])) for d in Data2.keys()}
print(result_addition)

>> {'user1': [('M4A4', 1)], 'user2': [('GLOCK-18', 1)]}


result_missing = {d:list(set(Data1[d]) - set(Data2[d])) for d in Data2.keys()}
print(result_missing )

>> {'user1': [], 'user2': []}

Answer 2

def dict_diff(dict_1, dict_2):
    keys = set(dict_1.keys()).union(set(dict_2.keys()))
    result = {}
    for key in keys:
        if key not in Data1:
            result[key] = Data2[key]
            continue
        if key not in Data2:
            result[key] = Data1[key]
            continue
        
        diff1 = set(Data2[key]).difference(set(Data1[key]))
        diff2 = set(Data1[key]).difference(set(Data2[key]))
        
        
        result[key] = list(diff1.union(diff2))

    return result
                
    
dict_diff(Data1, Data2)