Python 到字典的元組列表具有作為元組列表中所有相似項的列表的值

Question

我的數據是這樣的：

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 應該是這樣的：

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'],
                ....}

這里元組中的元素是相同的，所以“電影 1”類似於“電影 3”，“電影 3”類似於“電影 6”，“電影 6”類似於“電影 9”，“電影 9”類似於“電影 12” '和'電影12'到'電影15'。

我想得到一個字典，其中包含所有類似的項目作為值。

我試過這樣，但我沒有得到結果：

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 )

請幫助我以最小的時間復雜度進行轉換。

提前致謝。

感謝@James Lin 幫助獲得了這個結果，我在下面發布了代碼的外觀。

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']}

請幫助我理解整個過程的復雜性。 優化它也很棒。

謝謝

Answer 1

假設您的關系是自上而下排列的，您的描述並不完全清楚，我將嘗試給您一些提示：

您需要遍歷list_of_tuples以建立每個元素之間的關系

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)

這將打印出：

[{'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'}]

從此列表中，您將能夠生成所需的字典。

更新：使用 set() 解決與電影 13 相關的電影 11

更新：您可以先嘗試分析您的代碼，例如。 _ldap.get_option(_ldap.OPT_API_INFO) 升級到 MacOS Mojave 后變慢

Answer 2

您可以使用defaultdict來完成此操作。

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)

上述代碼的output為：

{'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']}

這是你想要的

Answer 3

您可以隨時調用dict(list_of_tuples)來獲取這些元組的相應字典。

我不知道這是否是最有效的時間，但我得到了你想要得到的東西 ~ O(n) 使用以下代碼：

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：這會給你一個電影的索引，它與它的相似之處。 如果你想要電影等價類，可以這么說，你需要做一些集合操作。 我會盡快添加更多信息。