将一个字典的值与另一个字典的键匹配 (Python)

Question

大家晚上好，

希望你一切都好。

我的Objective 我正在尝试将一个字典的值与另一个字典的键匹配。

dict1 有键但没有值 dict2 有键和值

dict2 的值可以作为 dict1 中的键找到。我正在尝试编写代码来识别 dict2 中的哪些值与 dict1.

中的键匹配

我的尝试 注释代码如下。

dict1 = {('dict1_key1',): [], ('dict1_key2',): []} #dictionary with keys, but no values;


for i in dict2.keys():  #iterate through the keys of dict2
    for x in dict2[i]:  #reaching every element in tuples in dict2
        if x == dict1.keys():  #if match found in the name of keys in dict1
            print(f"{i} holding {x}.") #print which key and value pair in dict 2 match the keys in dict1

如果我按如下方式编写 for loop，代码就可以工作：


for i in dict2.keys():  #iterate through the keys of dict2
    for x in dict2[i]:  #reaching every element in tuples in dict2
        if x == dict1_key1 or x == dict1_key2():  #if match found in the name of keys in dict1
            print(f"{i} holding {x}.") #print which key and value pair in dict 2 match the keys in dict1

然而，dict1 实际上需要能够包含不同数量的键，这就是为什么我希望 if x == dict1.keys(): 可以工作。

如有任何反馈，我们将不胜感激。

@Mark Meyer

请求的示例值：

dict1 = {('Tower_001',): [], ('Tower_002'): []}

dict2 = {1: 'Block_A', 'Tower_001'] #first key in dict2
        #skipping keys 2 through 13
        {14: ['Block_N', 'Tower_002']#last key in dict2

Answer 1

您可以对dict1.keys和dict2.value中的所有值进行组合。然后只需取这些集合的交集即可找到也是值的键：

dict1 = {('Tower_001',): [], ('Tower_005',): [], ('Tower_002',): []}

dict2 = {1: ['Block_A', 'Tower_001'], #first key in dict2
        14: ['Block_N', 'Tower_002']} #last key in dict2
         
         
set(k for l in dict2.values() for k in l) & set(k for l in dict1.keys() for k in l)
# {'Tower_001', 'Tower_002'}

将一个字典的值与另一个字典的键匹配 (Python)

Matching the values of one dictionary to the keys of another dictionary (Python)

python

dictionary

comparison