Python在嵌套字典中获取键

Question

我知道如何索引和检索以下dict中的几乎每一项：

playlists={'user':[
               {'playlist':{
                    'tracks': [        
                    {'name': 'Karma Police','artist': 'Radiohead', 'count': "1.0"},
                    {'name': 'Bitter Sweet Symphony','artist': 'The Verve','count': "2.0"}                    
                     ]
                   }
                }
              ]
            }

除了：如何使用key/value关系打印string 'playlist'？在

Answer 1

这棵树看起来很复杂。最好在类中包装它。但不管怎样：

例如，要获取播放列表中第一首曲目的名称：

print(playlists['user'][0]['playlist']['tracks'][0]['name'])

以及字符串“Playlist”：

print(list(playlists['user'][0].keys())[0])

它变得复杂起来，因为您需要键和值，而这并不完全是事物的本来面目，因为通常您希望在知道键的情况下提取值。我会编辑这篇文章，然后在不久的时间内想出一个稍微优雅一点的方法。在

好的，假设您对python中的类有一定的经验，下面是一个示例。如果没有，你可以在其他地方找到大量的信息：

# Class based example

class Playlist(object):
    def __init__(self, name, tracks=[]):
        self.name = name
        self.tracks = tracks

    def GetTrack(self, searchString):
        for T in self.tracks:
            if searchString in T.name:
                return T
        else:
            return  None

    def AddTrack(self, track):
        if isinstance(track, Track):
            self.tracks.append(Track)
        else:
            pass # Or do some exception handling

class Track(object):
    def __init__(self, name, artist, count):
        self.name = name
        self.artist = artist
        self.count = count

    def Play(self):
        pass # Could in theory add some play functionality


# Now you would create a new playlist by going:
MyPlaylist = Playlist("Heavy Metal")

MyPlaylist.AddTrack("SomeSong", "SomeArtist", 1) # etc

# Or

Countrysongs = [Track("SongName", "ArtistName", 12), Track("Bobby", "The Artist", 2)]

AnotherPlaylist = Playlist("Country", Countrysongs)


# And to access the Playlist name or the Song Name
MyPlaylist.GetTrack("SongName")

# And exception handle it
SongToGet = AnotherPlaylist.GetTrack("Sasdjkl")

if not SongToGet:
    print ("Could not find song")

下面是一个例子，它使构建一个更大的库变得更容易一些。它比一本大字典更容易、更快地获取信息，也更容易维护！在

Answer 2

您还可以使用下面的iteritems方法简单地访问key/value对。而不是像下面这样打印，你可以执行你的操作。在

In [14]: user1 = playlists.get('user')[0]

In [15]: for key, value in user1.iteritems():
   ....:     print key
   ....:     print value
   ....:     
playlist
{'tracks': [{'count': '1.0', 'name': 'Karma Police', 'artist': 'Radiohead'}, {'count': '2.0', 'name': 'Bitter Sweet Symphony', 'artist': 'The Verve'}]}