Question

我有播放列表的数据结构：

users={
  'playlist1': {'Karma Police': 2.0,'Bittersweet Symphony': 3.0,'The Queen Is Dead':4.0,'Song 1': 1.0},
  'playlist2': {'Bittersweet Symphony': 1.0,'Karma Police': 1.0,'The Queen Is Dead': 7.0,'Song 2': 1.0 },
  'playlist3': {'Karma Police': 4.0,'Bittersweet Symphony': 4.0,'The Queen Is Dead':3.0,'Song 3': 1.0}
       }

正在传递给这个函数：

def sim_distance(users,playlist1,playlist2):
  '''
  Returns a distance-based similarity score for 
  playlist1 and playlist2
  '''  
  # Get the list of shared_items
  si={}
  for item in users[playlist1]:
    if item in users[playlist2]:
        si[item]=1
  # if they have no ratings in common, return 0
  if len(si)==0: return 0
  # Add up the squares of all the differences
  sum_of_squares=sum([pow(users[playlist1][item]-users[playlist2][item],2)
                          for item in users[playlist1] if item in users[playlist2]])
  return 1/(1+sum_of_squares)

#print sim_distance(users, 'playlist1', 'playlist2')

最后，上述function是另一个的一部分：

def topMatches(users,playlist,n=2,similarity=sim_distance):

  '''
   Returns the best matches for user from 
   the prefs dictionary.
   Number of results and similarity function 
   are optional params.
  '''

  scores=[(similarity(users,playlist,other),other)
                       for other in users if other!=playlist]
# Sort the list so the highest scores appear at the top 
  scores.sort( )
  scores.reverse( )
  return scores[0:n]

topMatches(users, 'playlist1')

打印：[(0.14285714285714285, 'playlist3'), (0.06666666666666667, 'playlist2')]

但如果我有更多嵌套结构，如下所示：

playlists_user1={'user1':[
               {'playlist1A':{
                    'tracks': [        
                    {'name': 'Karma Police','artist': 'Radiohead', 'count': 1.0},
                    {'name': 'Bitter Sweet Symphony','artist': 'The Verve','count': 2.0},  
                    {'name': 'Song 1a','artist': 'Band 1a','count': 2.0}                  
                     ]
                    }
               },
               {'playlist1B':{
                    'tracks': [        
                    {'name': 'We Will Rock You','artist': 'Queen', 'count': 3.0},
                    {'name': 'Roxanne','artist': 'Police','count': 5.0},
                    {'name': 'Song 1b','artist': 'Band 1b','count': 2.0}                     
                     ]
                  }
                }
              ]
            }   
playlists_user2={'user2':[
               {'playlist2A':{
                    'tracks': [        
                    {'name': 'Karma Police','artist': 'Radiohead', 'count': 1.0},
                    {'name': 'Bitter Sweet Symphony','artist': 'The Verve','count': 4.0},
                    {'name': 'Song 2a','artist': 'Band 2a','count': 2.0}                      
                     ]
                    }
               },
               {'playlist2B':{
                    'tracks': [        
                    {'name': 'We Will Rock You','artist': 'Queen', 'count': 4.0},
                    {'name': 'Roxanne','artist': 'Police','count': 1.0},
                    {'name': 'Song 2b','artist': 'Band 2b','count': 2.0}                     
                     ]
                  }
                }
              ]
            }   
playlists_user3={'user3':[
               {'playlist3A':{
                    'tracks': [        
                    {'name': 'Karma Police','artist': 'Radiohead', 'count': 4.0},
                    {'name': 'Bitter Sweet Symphony','artist': 'The Verve','count': 6.0},
                    {'name': 'Song 3a','artist': 'Band 3a','count': 1.0}                      
                     ]
                    }
               },
               {'playlist3B':{
                    'tracks': [        
                    {'name': 'We Will Rock You','artist': 'Queen', 'count': 8.0},
                    {'name': 'Roxanne','artist': 'Police','count': 3.0},
                    {'name': 'Song 3b','artist': 'Band 3b','count': 4.0}                     
                     ]
                  }
                }
              ]
            }

如何更正：

1） sim_distance(users,playlist1,playlist2)

和

2） topMatches(users,playlist,n=2,similarity=sim_distance)

为了使我的代码适应这个新的嵌套结构？

我在matches和'playlist1A'以及'playlist1B'寻找user2的最高user3和'playlist2A'的最高匹配以及'playlist2B'和user1中的user3等等。

Answer 1

我根据您发布的最新数据做了一些假设。代码未经测试，我不得不从列表理解中分解为丑陋的for循环。这些词典以烦人的方式嵌套，所以我无法使它更清晰

import operator


def sim_distance(playlist1, playlist2):
    """
    Returns a distance-based similarity score for 
    playlist1 and playlist2
    """    
    # Flatten playlists
    playlist1, playlist2 = list(playlist1.values())[0], list(playlist2.values())[0]
    sum_of_squares = 0.
    for i in range(len(playlist1['tracks'])):
        for j in range(len(playlist2['tracks'])):
            if playlist1['tracks'][i]['name'] == playlist2['tracks'][j]['name']:
                sum_of_squares += (playlist1['tracks'][i]['count'] - playlist2['tracks'][j]['count'])**2

    # if they have no ratings in common, return 0
    if (sum_of_squares < 10e-10): return 0.

    return 1/(1+sum_of_squares)


def topMatches(users, playlist, n=2, similarity=sim_distance):

    '''
    Returns the best matches for a playlist of a user
    The candidates are all other playlists of others users.
    users is a list of dictionaries.

    This code assumes that each playlist has a different name
    Number of results and similarity function 
    are optional params.
    '''
    playlist_name = list(playlist.keys())[0] 
    scores = {}
    for user in users:
        for other_playlist in list(user.values())[0]:
            other_name = list(other_playlist.keys())[0]
            # Making sure not to compare the playlist with itself
            if  playlist_name != other_name:
                scores[other_name] = sim_distance(playlist, other_playlist)

    # Sort the list so the highest scores appear at the top 
    sorted_scores = sorted(scores.items(), key=operator.itemgetter(1))
    sorted_scores.reverse()
    return sorted_scores[0:n]

#Playsts_userX as defined in the question
users = [playlists_user1, playlists_user2, playlists_user3]
test_playlist = {'playlist1A': {'tracks': [{'artist': 'Radiohead',
    'count': 1.0,
    'name': 'Karma Police'},
   {'artist': 'The Verve', 'count': 2.0, 'name': 'Bitter Sweet Symphony'},
   {'artist': 'Band 1a', 'count': 2.0, 'name': 'Song 1a'}]}}
    topMatches(users, test_playlist, n=8)

给出

[('playlist2A', 0.07692307692307693),
 ('playlist3A', 0.013157894736842105),
 ('playlist1B', 0.0),
 ('playlist3B', 0.0),
 ('playlist2B', 0.0)]

python - 嵌套字典之间的顶级匹配

1 个答案: