Python 打印对象地址而不是值

Question

我想打印已添加到 tracks[] 列表中的所有曲目。当我尝试这样做时，我得到了该对象在内存中的地址，而不是它的实际值。我显然不明白从一个 class 到另一个 class 的对象 creation/passing 是如何工作的。

class Song:

    def __init__(self, title, artist, album, track_number):
        self.title = title
        self.artist = artist
        self.album = album
        self.track_number = track_number

        artist.add_song(self)


class Album:

    def __init__(self, title, artist, year):
        self.title = title
        self.artist = artist
        self.year = year

        self.tracks = []

        artist.add_album(self)

    def add_track(self, title, artist=None):
        if artist is None:
            artist = self.artist

        track_number = len(self.tracks)

        song = Song(title, artist, self, track_number)

        self.tracks.append(song)
        print(self.tracks)


class Artist:
    def __init__(self, name):
        self.name = name

        self.albums = []
        self.songs = []

    def add_album(self, album):
        self.albums.append(album)

    def add_song(self, song):
        self.songs.append(song)


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

    def add_song(self, song):
        self.songs.append(song)

band = Artist("Bob's Awesome Band")
album = Album("Bob's First Single", band, 2013)
album.add_track("A Ballad about Cheese")
album.add_track("A Ballad about Cheese (dance remix)")
album.add_track("A Third Song to Use Up the Rest of the Space")
playlist = Playlist("My Favourite Songs")


for song in album.tracks:
    playlist.add_song(song)

Answer 1

您似乎在尝试打印数组，而不是数组中的值。 print(self.tracks) 正在打印 self.tracks object，这是一个数组。尝试 print(self.tracks[x])，x 是您要打印的字符串的索引。

如果要打印该数组中的所有 object，请遍历它并打印每个 object。

使用它遍历数组：

for x in range(len(self.tracks)):
    print self.tracks[x].title

或

for track in self.tracks
    print track.title

要获取每首歌曲标题的值 object，请在循环中使用 track.title 对其进行寻址。要获取艺术家或年份，请将其更改为 track.artist 或 track.year。

您可以使用相同的逻辑构建更大的字符串，例如： print("Title " + track.title + ", 艺术家 " + track.artist)

Answer 2

是的，一个对象的 "value" 包括它的类型和内存位置。您需要提取所需的属性值。请注意，您需要对 Song 属性中包含的其他对象执行此操作。

一种方法是在 class、__repr__ 中实施 "representation" 方法。对于您的应用程序，它可能看起来像这样：

def __repr__(self):
    return "\n".join([self.title, self.artist.name, self.album.title,
                      "Track " + str(self.track_number)])

现在，只要您使用语法上需要其字符串表示的 Song 对象，Python 就会使用此方法进行转换。无需任何其他编码，您的程序现在生成：

[A Ballad about Cheese
Bob's Awesome Band
Bob's First Single
Track 0]
[A Ballad about Cheese
Bob's Awesome Band
Bob's First Single
Track 0, A Ballad about Cheese (dance remix)
Bob's Awesome Band
Bob's First Single
Track 1]
[A Ballad about Cheese
Bob's Awesome Band
Bob's First Single
Track 0, A Ballad about Cheese (dance remix)
Bob's Awesome Band
Bob's First Single
Track 1, A Third Song to Use Up the Rest of the Space
Bob's Awesome Band
Bob's First Single
Track 2]

当然，您需要根据自己的列表需求对其进行自定义。

Python 打印对象地址而不是值

Python prints object address instead of values

python

oop

inheritance

init