计算列表中最常见的项目与字典值列表交叉检查

Question

我有一本字典Songs，以歌曲为键，以歌曲的情绪为值。我还有一本Users字典，里面有听过的歌曲。我现在想检查字典中的哪些情绪Songs对每个用户来说最流行，并将这些情绪放入一个新的字典中，将用户与该情绪耦合起来。这就是我现在所拥有的：from collections import CounterSongs = {    'Song1' : ['techno', 'upbeat'],    'Song2' : ['rock', 'upbeat'],    'Song3' : ['pop', 'sad'],    'Song5' : ['pop', 'calm'],    'Song6' : ['rap', 'upbeat'],    'Song7' : ['rock', 'sad']}Users = {    'User1' : ['Song1', 'Song2', 'Song6'],    'User2' : ['Song3', 'Song7'],    'User3' : ['Song2', 'Song7']}Users_moods = dict.fromkeys(Users)for user, song in Users.items():    for song, mood in Songs.items():        mood = set(mood)        mood_counter = Songs[song]Users_moods = {user: counter.most_common(1)[0][0] for user, counter in Users.items()}print(Users_moods)print(Songs)print(Users)但它给了我错误。Users_mood对于此示例，字典应如下所示：Users_moods = {    'User1' : 'upbeat',    'User2' : 'sad',    'User3' : 'rock'}

Answer 1

def getMoods(user, users, songs):

    userSongs = users[user]  # songs this user listens to

    moods = {}  # count of each mood this user listens to

    for song in userSongs:

        for mood in songs[song]:  # for each mood of that song

            moods.setdefault(mood, 0)

            moods[mood] += 1

    return max(moods, key=moods.get)  # the most prevalent mood

for user in users:

    print(user, 'likes', getMoods(user, users, songs))  # mood with the highest count

Answer 2

您实际上并没有使用Counter并尝试调用most_common列表。让我们尝试将其分解：

1. 创建一个新字典，并为每个用户找到其相关的情绪（总体）。

2. 从该词典中找到每个用户最常见的情绪。

为了更好地处理 1，我们可以使用defaultdict：

from collections import Counter, defaultdict

Users_moods = defaultdict(list)

for user, songs in Users.items():

    for song in songs:

        Users_moods[user].extend(Songs[song])

common_moods = {user: Counter(moods).most_common(1)[0][0] for user, moods in Users_moods.items()}

print(Users_moods)

print(common_moods)

这给出：

defaultdict(<class 'list'>, {'User1': ['techno', 'upbeat', 'rock', 'upbeat', 'rap', 'upbeat'], 'User2': ['pop', 'sad', 'rock', 'sad'], 'User3': ['rock', 'upbeat', 'rock', 'sad']})

{'User1': 'upbeat', 'User2': 'sad', 'User3': 'rock'}

common或者，您可以使用相同的循环动态构建字典，为Counter每个用户创建一个单独的字典：

common_moods = {}

for user, songs in Users.items():

    User_moods = Counter()

    for song in songs:

        User_moods += Counter(Songs[song])

    common_moods[user] = User_moods.most_common(1)[0][0]

print(common_moods)

Answer 3

如果您打算使用 collections.Counter，让我们看看https://docs.python.org/3/library/collections.html#collections.Counter您将需要

mood_counter = Counter(iterable)

据推测，这里的可迭代来自于查看一个用户，然后是他们的所有歌曲，然后是这些歌曲的所有情绪。我们不要尝试制作一个衬垫，而是正常地迭代这些衬垫。

def get_user_mood(user, Users, Songs):

    accumulated_moods = []

    for song_name in Users[user]:

        moods_for_this_song = Songs[song_name]

        accumulated_moods.extend(moods_for_this_song)

    mood_counter = Counter(accumulated_moods)

    return mood_counter.most_common(1)[0][0] # validate?

这使您可以在理解中相当轻松地构建字典

Users_moods = {user: get_user_mood(user, Users, Songs) for user in Users}