python-搜索字典子列表; 将字典键转换为值

Question

说我有以下字典(我正在使用的字典更多,更大):

dict1={1:["item", "word", "thing"], 2:["word", "item"], 3:["thing", "item", "item"]}

并将字典中使用的每个单词存储在列表中:

all_words=["item", "word", "thing"]

我想通过字典子列表运行列表中的每个单词,并返回找到它们的所有子列表的键,将它们存储在元组中.所以我想得到:

dict2={"item":(1, 2, 3), "word":(1, 2), "thing":(1, 3)}

继承人我所拥有的:

dict2={}    
for word in all_words:
    for key, sublist in dict2.items():
        for word in sublist:
            if word not in sublist:
                dict2[word]=dict2[word]+key
            else:
                dict2[word]=key

Answer 1

因此,基于评论的固定程序将如下所示

>>> dict2 = {}
>>> for word in all_words:
...     # Iterate over the dict1's items
...     for key, sublist in dict1.items():
...         # If the word is found in the sublist
...         if word in sublist:
...             # If the current word is found in dict2's keys
...             if word in dict2:
...                 # Append the current key as a one element tuple
...                 dict2[word] += (key,)
...             else:
...                 # Create a one element tuple and assign it to the word
...                 dict2[word] = (key,)
... 
>>> dict2
{'item': (1, 2, 3), 'word': (1, 2), 'thing': (1, 3)}

如果你知道字典理解,那么同样可以写成

>>> {word: tuple(k for k, v in dict1.items() if word in v) for word in all_words}
{'item': (1, 2, 3), 'word': (1, 2), 'thing': (1, 3)}

整个元组创建逻辑,基于dict1每个对应的word,已被挤压为单个生成器表达式并转换为元组与tuple(k for k, v in dict1.items() if word in v)