将 Python 字典与列表作为值进行比较

Question

我正在为 Steam 编写一个解析器机器人，它将跟踪哪些物品从 Steam 用户的库存中进出。我写了一段代码，以带有嵌套列表的字典形式获取所有用户的项目和 returns，其中 KEY = USER NAME，VALUE = ITEM NAME AND ITS QUANTITY. 现在我需要比较 Data1 和 Data2（更新后的数据）。

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)] 

        }

比如Data2中出现了一个新条目，我需要将它写入第三个字典，保留key。如果项目离开 Data2（这意味着这些项目不再在用户的库存中），则必须执行相同的操作。

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)

将 Python 字典与列表作为值进行比较

Compare Python dictionary with list as value

python

list

dictionary

steam