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'}
3 回答
GCT1015
TA贡献1827条经验 获得超4个赞
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
喵喵时光机
TA贡献1846条经验 获得超7个赞
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：
创建一个新字典，并为每个用户找到其相关的情绪（总体）。
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)
慕斯王
TA贡献1864条经验 获得超2个赞
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}
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