python - 如何获取两个列表之间普遍存在的键值？

Question

我正在尝试获取 first_list 和 second_list 中常见的键值，如下所示，first_list 和 second_list 有 X 和 Z 作为公共(public)键，我想得到 X 和 Z 对应的值并创建一个最终字典，如 EXPECTED OUTPUT BELOW 所示，我有当前和预期的输出，有人可以就这里的错误提供指导吗？

first_list = [{'W':['xyz','abc','def']},{'X':['1','2','3']},{'Y':['4','5','6']},{'Z':['1','5','7']}]

second_list = [{'X':True},{'Z':True}]


first_list_keys,second_list_keys = [],[]
for item in first_list:
    for key,value in item.items():
        print (key,value)
        first_list_keys.append(key)

for item in second_list:
    for key,value in item.items():
        print (key,value)
        second_list_keys.append(key)
print(first_list_keys,second_list_keys)
c = set(first_list_keys).intersection(second_list_keys)

print (c)
final_dict = {}
for item in c:
    item_dict = {}
    for data in first_list:
        for key,value in data.items():
            print (key,value)
            if key == item:
                print ('==================')
                print (value)
                item_dict['deps'] = value


                for item in c:
                    for data in second_list:
                        for key,value in data.items():
                            print (key,value)
                            if key == item:
                                print ('==================')
                                print (value)
                                item_dict['values'] = True
                final_dict[key] = item_dict
print (final_dict)

当前输出:-

{'Z': {'deps': ['1', '5', '7'], 'values': True}}

预期输出:-

{'X':{"deps":['1','2','3'],"values": true },'Z':{"deps":['1','5','7'],"values": true }

更新:-

{'W':{"deps":['xyz','abc','def'],"values": False},'X':{"deps":['1','2','3'],"values": True },'Y':{"deps":['4','5','6'],"values":False},'Z':{"deps":['1','5','7'],"values": True }

Answer 1

这是一种解决方案。请注意，为了能够在第二个列表中查找键，我首先将其转换为字典，然后使用它。此代码假定任一列表中的每个字典都只包含一个键/值对:

first_list = [{'W':['xyz','abc','def']},{'X':['1','2','3']},{'Y':['4','5','6']},{'Z':['1','5','7']}]

second_list = [{'X':True},{'Z':True}]

result ={}

# Create a map from "second_list", where for each entry in "second_list", we pull out the one
# key/value pair contained therin and then add it to the new dictionary
second_list_map = { next(iter(kv.keys())):next(iter(kv.values())) for kv in second_list}

# For each entry in the first list...
for fentry in first_list:
    # Get the one key in the entry
    key = next(iter(fentry.keys()))
    # If this key is also in the second list...
    if key in second_list_map:
        # Add the appropriate entry to the result dictionary
        result[key] = {
            "deps": fentry[key],
            "values": second_list_map[key]
        }

print(result)

结果:

{'X': {'deps': ['1', '2', '3'], 'values': True}, 'Z': {'deps': ['1', '5', '7'], 'values': True}}

这是更新要求的主体，保留第一个列表中的所有值，并注意第二个列表中缺少带有 'values': False 的键:

for fentry in first_list:
    key = next(iter(fentry.keys()))
    result[key] = {
        "deps": fentry[key],
        "values": second_list_map[key] if key in second_list_map else False
    }

结果:

{'W': {'deps': ['xyz', 'abc', 'def'], 'values': False}, 'X': {'deps': ['1', '2', '3'], 'values': True}, 'Y': {'deps': ['4', '5', '6'], 'values': False}, 'Z': {'deps': ['1', '5', '7'], 'values': True}}