我正在尝试根据他们合作过多少部电影来制作演员/女演员的图表。我提取了一个字典people_dict，看起来像：
people_dict = {..., 3: ['American Pimp (1999)',
'Beats, Rhymes & Life: The Travels of a Tribe Called Quest (2011)',
'Gangsta Rap: The Glockumentary (2007)',
'Ghetto Physics (2010)',
'Mac Dre: Legend of the Bay (2014)',
'Menace II Society (1993)',
'Pimpalation: Return of the Trill (2006)',
'Porndogs: The Adventures of Sadie (2009)',
'Rhyme & Reason (1997)',
'Stop Pepper Palmer (2014)',
'Townbiz (2010)',
'Uprising: Hip Hop and the LA Riots (2012)'],...}
其中key是演员/演员ID，value是他/她所演过的所有电影的列表。
现在我想找到演员/女演员之间的合作，因此我需要计算不同键值之间的交集，如下所示：
for people_id, movie_list in people_dict.items():
for other_people_id, other_movie_list in people_dict.items():
if people_id < other_people_id:
intersection = set(movie_list) & set(other_movie_list)
if intersection:
edge_list.append((people_id, other_people_id))
但是，我的实现是嵌套循环，这本词典中有超过100,000名演员/女演员。结果可能需要几天时间。
我可以做些什么来加速我的代码。
编辑：
我意识到我需要提供更多关于我的问题的内容。
假设我有people_dict只有三位演员/演员，因为284和602都在电影'Electric Daisy Carnival Experience (2011)'中演出，但同时也没有共同点660和其他两个人的电影，我只希望将284和602写在284 602这样的文本文件中。
people_dict = {284: ['DGK: Parental Advisory (2012)',
'Electric Daisy Carnival Experience (2011)',
'Malibu Horror Story (2014)'],
602: ['5 Sides of a Coin (2003)',
'As I AM: The Life and Times of DJ AM (2015)',
'Electric Daisy Carnival Experience (2011)',
'Hang the DJ (1998)'],
661: ['Every Boy Needs a Father (2013)',
'The Sketches of Matty Mad Man: Part Dos (2014)',
'Trippy: Spitting Rainbows (2014)',
'Vicky Foxx: Forever in Blue Jeans (2010)']}
答案 0 :(得分：0)
每次都将两个列表设置为集合 - 为什么不从一开始就将movie_list设置为一组？
无论如何，主要问题是复杂度O（n ^ 2）。 你真的需要将每个项目相互比较两次吗？
也许这会引发一些想法：
from collections import defaultdict
import functools
import time
import random
movies = range(100)
actors = range(1000)
def get_random_movies(max=10):
return random.sample(movies, random.randint(1, max))
people_dict = {actor_id: get_random_movies() for actor_id in actors}
def timeit(func):
@functools.wraps(func)
def newfunc(*args, **kwargs):
start_time = time.time()
func(*args, **kwargs)
elapsed_time = time.time() - start_time
print('function [{}] finished in {} ms'.format(
func.__name__, int(elapsed_time * 1000)))
return newfunc
@timeit
def usingDefaultDict():
md = defaultdict(set)
for k, v in people_dict.items():
for m in v:
md[m].add(k)
collaborations = {
actor_id: set([aid for actors in md.values() for aid in actors if actor_id in actors])
for actor_id in people_dict}
print(next(iter(collaborations.items())))
@timeit
def usingNestedLoops():
edge_list = []
for people_id, movie_list in people_dict.items():
for other_people_id, other_movie_list in people_dict.items():
if people_id < other_people_id:
intersection = set(movie_list) & set(other_movie_list)
if intersection:
edge_list.append((people_id, other_people_id))
collaborations = {
actor_id: set([aid for actors in edge_list for aid in actors if actor_id in actors])
for actor_id in people_dict}
print(next(iter(collaborations.items())))
usingNestedLoops()
usingDefaultDict()
导致集合的字典：{actor_id：actors_collaborated_with}
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函数[usingNestedLoops]在66736 ms完成
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函数[usingDefaultDict]在816 ms完成