如何从当前字典中返回一个具有列表值的新字典？

Question

现在，我很难弄清楚如何获得以下输出

所需输出：

 {'Bobby':  [set(), set(), set(), {'movie_2'}, set(),{'movie_1'}], 
  'Carren': [set(), set(), {'movie_4'}, {'movie_2'}, set(), {'movie_1'}],
  'Deric':  [set(), {'movie_1'}, set(), {'movie_2'}, set(),{'movie_3'}],
  'Alec':   [set(), {'movie_1'}, {'movie_3'}, set(), set(), {'movie_4'}] }

我正在使用的词典：

d = {'movie_1': {('Bobby', 5), ('Carren', 5), ('Alec', 1), ('Deric', 1)},
     'movie_2': {('Carren', 3), ('Deric', 3), ('Bobby', 3)},
     'movie_3': {('Alec', 2), ('Deric', 5)},
     'movie_4': {('Carren', 2), ('Alec', 5)} }

在每个值中，元组表示（person，position）。你知道吗

因此，对于字典d，当您查看键'movie_1'时，第一个值是('Bobby', 5)。你知道吗

然后将键'movie_1'放入列表的第5个索引中。你知道吗

我目前掌握的代码：

我创建了一个包含6个set()项的列表，但我不完全确定从这里要做什么才能获得所需的输出。你知道吗

from collections import defaultdict

d = {
    'movie_1': {('Bobby', 5), ('Carren', 5), ('Alec', 1), ('Deric', 1)},
    'movie_2': {('Carren', 3), ('Deric', 3), ('Bobby', 3)},
    'movie_3': {('Alec', 2), ('Deric', 5)},
    'movie_4': {('Carren', 2), ('Alec', 5)},
}

def new_dict(d):
    le_dict = defaultdict(set)
    values = [set(), set(), set(), set(), set(), set()]
    for key, value in d.items():
        for person_score in value:
            le_dict[person_score[0]].add(key)   
    return le_dict

Answer 1

d = {'movie_1': {('Bobby', 5), ('Carren', 5), ('Alec', 1), ('Deric', 1)}, 'movie_2': {('Carren', 3), ('Deric', 3), ('Bobby', 3)}, 'movie_3': {('Alec', 2), ('Deric', 5)}, 'movie_4': {('Carren', 2), ('Alec', 5)} }

def new_dict(d):
    le_dict = {}
    for key, values in d.items():
        for person, score in values:
            if person not in le_dict:
                le_dict[person] = [set(), set(), set(), set(), set(), set()]
            le_dict[person][score].add(key)
    return le_dict

new_dict(d)

结果是：

{'Bobby': [set(), set(), set(), {'movie_2'}, set(), {'movie_1'}],
 'Alec': [set(), {'movie_1'}, {'movie_3'}, set(), set(), {'movie_4'}],
 'Carren': [set(), set(), {'movie_4'}, {'movie_2'}, set(), {'movie_1'}],
 'Deric': [set(), {'movie_1'}, set(), {'movie_2'}, set(), {'movie_3'}]}