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python - 处理单个字典的嵌套循环

问题描述

我正在尝试根据他们合作的电影数量来构建演员/女演员图表。我提取了一个people_dict看起来像这样的字典：

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/女演员ID，value是他/她演过的所有电影的列表。

现在我试图找到演员/女演员之间的协作，因此我需要计算不同键的值之间的交集，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，这本字典中有超过 100、000 个演员/女演员。可能需要几天时间才能给我结果。

那么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信息。

假设我有一个people_dict只有三个演员/女演员，因为284和602都在电影中表演，'Electric Daisy Carnival Experience (2011)'同时，另外两个人没有共同的电影660，那么我只是期望它284会602被写在一个文本文件中，如284 602.

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)']}

标签： pythondictionary

解决方案

  • 每次将两个列表转换为集合 - 为什么不从一开始就将 movie_list 设置为集合？

  • 无论如何，主要问题是复杂度 O(n^2)。您真的需要将每个项目相互比较两次吗？

也许这会激发一些想法：

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()

这导致集合的字典：{actor_id：actors_collaborated_with}

(0, {0, 1, 512, 516, 519, 8, 520, 522, 12, 524, 14, 15, 529, 18, 532, 533, 534, 535, 26, 28, 29, 30, 31, 540, 545, 35, 36, 38, 39, 40, 43, 47, 48, 49, 50, 563, 564, 54, 568, 57, 58, 572, 63, 64, 65, 576, 67, 69, 70, 581, 582, 73, 74, 75, 583, 77, 587, 81, 594, 597, 86, 89, 605, 606, 97, 610, 100, 101, 617, 107, 621, 110, 622, 623, 629, 118, 121, 635, 124, 639, 129, 130, 131, 132, 133, 135, 650, 139, 140, 141, 653, 143, 655, 146, 147, 658, 660, 151, 154, 666, 669, 158, 160, 161, 673, 675, 164, 165, 166, 680, 169, 684, 175, 687, 177, 181, 693, 183, 184, 185, 698, 189, 190, 704, 705, 195, 197, 709, 710, 201, 713, 714, 715, 205, 718, 721, 210, 723, 213, 214, 215, 730, 222, 734, 735, 225, 737, 228, 230, 743, 232, 746, 748, 238, 240, 753, 243, 755, 246, 761, 762, 251, 763, 253, 766, 255, 767, 769, 258, 259, 261, 775, 776, 777, 779, 270, 273, 274, 275, 786, 277, 278, 787, 792, 281, 282, 795, 286, 287, 292, 294, 299, 813, 815, 304, 817, 308, 309, 821, 311, 824, 826, 828, 829, 830, 319, 321, 837, 328, 840, 331, 334, 846, 847, 849, 339, 340, 851, 342, 854, 344, 855, 352, 867, 868, 359, 872, 362, 363, 366, 367, 880, 882, 883, 372, 373, 886, 890, 892, 384, 387, 390, 902, 903, 393, 394, 908, 399, 913, 916, 918, 921, 411, 412, 421, 423, 936, 937, 426, 938, 430, 431, 432, 433, 942, 949, 950, 951, 952, 444, 445, 447, 960, 450, 964, 454, 457, 460, 461, 462, 463, 464, 972, 981, 471, 983, 986, 991, 995, 484, 997, 998, 487, 489, 495, 499, 502})

函数 [usingNestedLoops] 在 66736 毫秒内完成

(0, {0, 1, 512, 516, 519, 8, 520, 522, 12, 524, 14, 15, 529, 18, 532, 533, 534, 535, 26, 540, 29, 30, 31, 28, 545, 35, 36, 38, 39, 40, 43, 47, 48, 49, 50, 563, 564, 54, 568, 57, 58, 572, 63, 576, 65, 64, 67, 581, 582, 583, 70, 73, 74, 587, 69, 77, 75, 81, 594, 597, 86, 89, 605, 606, 97, 610, 100, 101, 617, 107, 621, 622, 623, 110, 629, 118, 121, 635, 124, 639, 129, 130, 131, 132, 133, 135, 650, 139, 140, 653, 141, 143, 655, 658, 146, 147, 660, 151, 154, 666, 669, 158, 160, 161, 673, 675, 164, 165, 166, 680, 169, 684, 175, 687, 177, 181, 693, 183, 184, 185, 698, 189, 190, 704, 705, 195, 197, 709, 710, 713, 714, 201, 715, 205, 718, 721, 210, 723, 213, 214, 215, 730, 734, 735, 222, 737, 225, 228, 230, 743, 232, 746, 748, 238, 240, 753, 755, 243, 246, 761, 762, 251, 763, 253, 766, 255, 767, 769, 258, 259, 261, 775, 776, 777, 779, 270, 273, 786, 274, 275, 277, 278, 787, 792, 281, 282, 795, 286, 287, 292, 294, 299, 813, 815, 304, 817, 308, 821, 309, 311, 824, 826, 828, 829, 830, 319, 321, 837, 840, 328, 331, 846, 847, 334, 849, 339, 851, 340, 342, 854, 344, 855, 352, 867, 868, 359, 872, 362, 363, 366, 367, 880, 882, 883, 372, 373, 886, 890, 892, 384, 387, 902, 903, 390, 393, 394, 908, 399, 913, 916, 918, 921, 411, 412, 421, 423, 936, 937, 426, 938, 430, 942, 432, 433, 431, 949, 950, 951, 952, 444, 445, 447, 960, 450, 964, 454, 457, 972, 460, 462, 461, 464, 463, 981, 471, 983, 986, 991, 995, 484, 997, 998, 487, 489, 495, 499, 502})

函数 [usingDefaultDict] 在 816 毫秒内完成

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