Python 到字典的元组列表具有作为元组列表中所有相似项的列表的值

Question

我的数据是这样的：

movies = [
    "movie 1",
    "movie 2",
    "movie 3",
    "movie 4",
    "movie 5",
    "movie 6",
    "movie 7",
    "movie 8",
    "movie 9",
    "movie 10",
    "movie 11",
    "movie 12",
    "movie 13",
    "movie 14",
    "movie 15",
]

list_of_tuples = [
    ("movie 1", "movie 3"),
    ("movie 3", "movie 6"),
    ("movie 6", "movie 9"),
    ("movie 9", "movie 12"),
    ("movie 12", "movie 15"),
    ("movie 2", "movie 4"),
    ("movie 4", "movie 7"),
    ("movie 8", "movie 10"),
    ("movie 10", "movie 5"),
    ("movie 14", "movie 13"),
    ("movie 11", "movie 13"),
]

Output 应该是这样的：

result_dict = {'movie 1' : ['movie 1' , 'movie 3', 'movie 6', 'movie 9', 'movie 12', 'movie 15'],
               'movie 2' : ['movie 2', 'movie 4', 'movie 7'],
               'movie 3' : ['movie 1' , 'movie 3', 'movie 6', 'movie 9', 'movie 12', 'movie 15'],
                ....}

这里元组中的元素是相同的，所以“电影 1”类似于“电影 3”，“电影 3”类似于“电影 6”，“电影 6”类似于“电影 9”，“电影 9”类似于“电影 12” '和'电影12'到'电影15'。

我想得到一个字典，其中包含所有类似的项目作为值。

我试过这样，但我没有得到结果：

result_dict = {movie : list() for movie in movies}

for tup in list_of_tuples:
  mov1, mov2 = tup

  result_dict[mov1].append(mov2)
  result_dict[mov2].append(mov1)

  for x in result_dict[mov2]:
    if x not in result_dict[mov1]:
    result_dict[mov1].append(x)
  
  for x in result_dict[mov1]:
    if x not in result_dict[mov2]:
      result_dict[mov2].append(x )

请帮助我以最小的时间复杂度进行转换。

提前致谢。

感谢@James Lin 帮助获得了这个结果，我在下面发布了代码的外观。

relationships = []
relationship = set()
for tuple_data in list_of_tuples:
    tuple_data = set(tuple_data)
    if tuple_data.intersection(relationship):
       relationship |= tuple_data
    else:
       # broken link
       relationship = set()
       relationship |= tuple_data
       relationships.append(relationship)

for idx in range(len(relationships)):
  relationships[idx] = list(relationships[idx])



result_dict = {movie : list() for movie in movies}

for key in result_dict.keys():
  for item in relationships:
    if key in item:
      result_dict[key] = item

Output 是：

{'movie 1': ['movie 1', 'movie 15', 'movie 12', 'movie 9', 'movie 6', 'movie 3'], 'movie 2': ['movie 7', 'movie 4', 'movie 2'], 'movie 3': ['movie 1', 'movie 15', 'movie 12', 'movie 9', 'movie 6', 'movie 3'], 'movie 4': ['movie 7', 'movie 4', 'movie 2'], 'movie 5': ['movie 10', 'movie 5', 'movie 8'], 'movie 6': ['movie 1', 'movie 15', 'movie 12', 'movie 9', 'movie 6', 'movie 3'], 'movie 7': ['movie 7', 'movie 4', 'movie 2'], 'movie 8': ['movie 10', 'movie 5', 'movie 8'], 'movie 9': ['movie 1', 'movie 15', 'movie 12', 'movie 9', 'movie 6', 'movie 3'], 'movie 10': ['movie 10', 'movie 5', 'movie 8'], 'movie 11': ['movie 14', 'movie 11', 'movie 13'], 'movie 12': ['movie 1', 'movie 15', 'movie 12', 'movie 9', 'movie 6', 'movie 3'], 'movie 13': ['movie 14', 'movie 11', 'movie 13'], 'movie 14': ['movie 14', 'movie 11', 'movie 13'], 'movie 15': ['movie 1', 'movie 15', 'movie 12', 'movie 9', 'movie 6', 'movie 3']}

请帮助我理解整个过程的复杂性。 优化它也很棒。

谢谢

Answer 1

假设您的关系是自上而下排列的，您的描述并不完全清楚，我将尝试给您一些提示：

您需要遍历list_of_tuples以建立每个元素之间的关系

relationships = []
relationship = set()
for tuple_data in list_of_tuples:
    tuple_data = set(tuple_data)
    if tuple_data.intersection(relationship):
       relationship |= tuple_data
    else:
       # broken link
       relationship = tuple_data
       relationships.append(relationship)

print(relationships)

这将打印出：

[{'movie 15', 'movie 12', 'movie 6', 'movie 9', 'movie 3', 'movie 1'}, {'movie 2', 'movie 7', 'movie 4'}, {'movie 8', 'movie 5', 'movie 10'}, {'movie 11', 'movie 14', 'movie 13'}]

从此列表中，您将能够生成所需的字典。

更新：使用 set() 解决与电影 13 相关的电影 11

更新：您可以先尝试分析您的代码，例如。 _ldap.get_option(_ldap.OPT_API_INFO) 升级到 MacOS Mojave 后变慢

Answer 2

您可以使用defaultdict来完成此操作。

from collections import defaultdict

list_of_tuples = [
    ("movie 1", "movie 3"),
    ("movie 3", "movie 6"),
    ("movie 6", "movie 9"),
    ("movie 9", "movie 12"),
    ("movie 12", "movie 15"),
    ("movie 2", "movie 4"),
    ("movie 4", "movie 7"),
    ("movie 8", "movie 10"),
    ("movie 10", "movie 5"),
    ("movie 14", "movie 13"),
    ("movie 11", "movie 13"),
]

result_dict = defaultdict(list)

for k ,v in list_of_tuples:

    #for value in the tuple, find out if this is already part of
    #the existing dictionary. If yes, get the key so you can
    #append to the key else start a new key item

    a = ''.join([x for x, y in result_dict.items() for z in y if z == k])

    #if found, above list comprehension will result in 1 element

    if a = '' : #if not found, then create a new list for key
        result_dict[k].append(v)

    else: # value is part of a key list, so append value to key list 
        result_dict[a].append(v)

result_dict = dict(result_dict)
print (result_dict)

上述代码的output为：

{'movie 1': ['movie 3', 'movie 6', 'movie 9', 'movie 12', 'movie 15'], 'movie 2': ['movie 4', 'movie 7'], 'movie 8': ['movie 10', 'movie 5'], 'movie 14': ['movie 13'], 'movie 11': ['movie 13']}

这是你想要的

Answer 3

您可以随时调用dict(list_of_tuples)来获取这些元组的相应字典。

我不知道这是否是最有效的时间，但我得到了你想要得到的东西 ~ O(n) 使用以下代码：

from collections import defaultdict

movie_dict = dict(list_of_tuples)

index = defaultdict(list)
for key, value in movie_dict.items():
    index[key] += [value]
    index[value] += [key]

output 是：

defaultdict(list,
            {'movie 1': ['movie 3'],
             'movie 3': ['movie 1', 'movie 6'],
             'movie 6': ['movie 3', 'movie 9'],
             'movie 9': ['movie 6', 'movie 12'],
             'movie 12': ['movie 9', 'movie 15'],
             'movie 15': ['movie 12'],
             'movie 2': ['movie 4'],
             'movie 4': ['movie 2', 'movie 7'],
             'movie 7': ['movie 4'],
             'movie 8': ['movie 10'],
             'movie 10': ['movie 8', 'movie 5'],
             'movie 5': ['movie 10'],
             'movie 14': ['movie 13'],
             'movie 13': ['movie 14', 'movie 11'],
             'movie 11': ['movie 13']})

ETA：这会给你一个电影的索引，它与它的相似之处。 如果你想要电影等价类，可以这么说，你需要做一些集合操作。 我会尽快添加更多信息。