Python中构建演员合作关系字典的实现问题

需求说明 #

实现功能：根据给定的影片演员阵容数据，生成每个演员的合作演员列表，要求同一合作演员即使共同参演多部影片也只保留一次。

示例输入 #

cast_info = [[
  {'name': 'Drew Barrymore'},
  {'name': 'Brian Herzlinger'},
  {'name': 'Corey Feldman'},
  {'name': 'Eric Roberts'}, {'name': 'Griffin Dunne'}, {'name': 'Samuel L. Jackson'}, {'name': 'Matt LeBlanc'}, {'name': "Bill D'Elia"}], [{'name': 'Erich Anderson'}, {'name': 'Judie Aronson'}, {'name': 'Peter Barton'}, {'name': 'Kimberly Beck'}, {'name': 'Tom Everett'}, {'name': 'Flashlight Man'}, {'name': 'Corey Feldman'}]]

预期输出 #

{'Drew Barrymore': ['Brian Herzlinger', 'Corey Feldman', 'Eric Roberts', 'Griffin Dunne', 'Samuel L. Jackson', 'Matt LeBlanc', "Bill D'Elia"],
 'Brian Herzlinger': ['Drew Barrymore', 'Corey Feldman', 'Eric Roberts', 'Griffin Dunne', 'Samuel L. Jackson', 'Matt LeBlanc', "Bill D'Elia"],
 'Corey Feldman': ['Drew Barrymore', 'Brian Herzlinger', 'Eric Roberts', 'Griffin Dunne', 'Samuel L. Jackson', 'Matt LeBlanc', "Bill D'Elia", 'Erich Anderson', 'Judie Aronson', 'Peter Barton', 'Kimberly Beck', 'Tom Everett', 'Flashlight Man'],
 'Eric Roberts': ['Drew Barrymore', 'Brian Herzlinger', 'Corey Feldman', 'Griffin Dunne', 'Samuel L. Jackson', 'Matt LeBlanc', "Bill D'Elia"],
 'Griffin Dunne': ['Drew Barrymore', 'Brian Herzlinger', 'Corey Feldman', 'Eric Roberts', 'Samuel L. Jackson', 'Matt LeBlanc', "Bill D'Elia"],
 'Samuel L. Jackson': ['Drew Barrymore', 'Brian Herzlinger', 'Corey Feldman', 'Eric Roberts', 'Griffin Dunne', 'Matt LeBlanc', "Bill D'Elia"],
 'Matt LeBlanc': ['Drew Barrymore', 'Brian Herzlinger', 'Corey Feldman', 'Eric Roberts', 'Griffin Dunne', 'Samuel L. Jackson', "Bill D'Elia"],
 "Bill D'Elia": ['Drew Barrymore', 'Brian Herzlinger', 'Corey Feldman', 'Eric Roberts', 'Griffin Dunne', 'Samuel L. Jackson', 'Matt LeBlanc'],
 'Erich Anderson': ['Judie Aronson', 'Peter Barton', 'Kimberly Beck', 'Tom Everett', 'Flashlight Man', 'Corey Feldman'],
 'Judie Aronson': ['Erich Anderson', 'Peter Barton', 'Kimberly Beck', 'Tom Everett', 'Flashlight Man', 'Corey Feldman'],
 'Peter Barton': ['Erich Anderson', 'Judie Aronson', 'Kimberly Beck', 'Tom Everett', 'Flashlight Man', 'Corey Feldman'],
 'Kimberly Beck': ['Erich Anderson', 'Judie Aronson', 'Peter Barton', 'Tom Everett', 'Flashlight Man', 'Corey Feldman'],
 'Tom Everett': ['Erich Anderson', 'Judie Aronson', 'Peter Barton', 'Kimberly Beck', 'Flashlight Man', 'Corey Feldman'],
 'Flashlight Man': ['Erich Anderson', 'Judie Aronson', 'Peter Barton', 'Kimberly Beck', 'Tom Everett', 'Corey Feldman']}

注意：演员Corey Feldman的合作演员需包含两部影片中的其他演员。

当前代码的问题 #

你写的代码存在以下几个问题：

1. 逻辑无效：actor_name2 != list_of_actors[0]只是做了一个比较操作，没有对actor_name2做任何修改，所以当actor['name']等于list_of_actors[0]时，依然会把该演员自己添加到列表中；
2. 仅处理单个演员：代码只针对list_of_actors[0]这一个演员处理，没有遍历所有演员生成对应的合作列表；
3. 无去重机制：没有处理重复添加的问题，同一合作演员会被多次添加。

for film_cast in cast_info:
    for actor in film_cast:
        if list_of_actors[0] in actor['name']:
            actor_name2 = actor['name'] 
            actor_name2 != list_of_actors[0]  # 这行代码无实际作用
            costars_lists[0].append(actor_name2) 

解决方案 #

实现思路 #

1. 先提取所有演员的姓名，确保不会遗漏任何需要生成合作列表的对象；
2. 用集合存储每个演员的合作演员，利用集合自动去重的特性，避免手动判断重复；
3. 遍历每一部影片的演员阵容，对阵容中的每个演员，将同阵容的其他演员添加到他的合作集合中；
4. 最后将集合转换为列表，整理成字典格式输出。

代码实现 #

cast_info = [[
  {'name': 'Drew Barrymore'},
  {'name': 'Brian Herzlinger'},
  {'name': 'Corey Feldman'},
  {'name': 'Eric Roberts'}, {'name': 'Griffin Dunne'}, {'name': 'Samuel L. Jackson'}, {'name': 'Matt LeBlanc'}, {'name': "Bill D'Elia"}], [{'name': 'Erich Anderson'}, {'name': 'Judie Aronson'}, {'name': 'Peter Barton'}, {'name': 'Kimberly Beck'}, {'name': 'Tom Everett'}, {'name': 'Flashlight Man'}, {'name': 'Corey Feldman'}]]

# 提取所有演员姓名，用集合避免重复
all_actors = {actor['name'] for film in cast_info for actor in film}
# 初始化合作演员字典，每个演员对应一个空集合
costar_dict = {actor: set() for actor in all_actors}

# 遍历每部影片的演员阵容
for film_cast in cast_info:
    # 提取当前影片的所有演员姓名
    current_film_actors = [actor['name'] for actor in film_cast]
    # 为当前影片的每个演员添加合作演员
    for actor in current_film_actors:
        for costar in current_film_actors:
            if actor != costar:
                costar_dict[actor].add(costar)

# 将集合转换为列表（可选排序，让输出更整齐）
result = {actor: sorted(list(costars)) for actor, costars in costar_dict.items()}

# 打印结果
import pprint
pprint.pprint(result)

代码说明 #

• 用集合存储合作演员，自动解决重复添加的问题，无需手动检查是否已存在；
• 先提取所有演员，确保覆盖所有需要生成合作列表的对象；
• 遍历影片时先提取当前影片的演员姓名列表，避免重复从字典中读取数据，提升效率；
• 最后将集合转为列表并排序，输出格式与预期一致。

内容的提问来源于stack exchange，提问作者PoorProgrammer