如何打印此类变量？

Question

如果我尝试打印作为列表的类变量,我会得到一个Python对象.(这些是我在stackoverflow上找到的例子).

 class Contacts:
    all_contacts = []

    def __init__(self, name, email):
       self.name = name
       self.email = email
       Contacts.all_contacts.append(self)

    def __str__(self):
       return '%s, <%s>' % (self.name, self.email)

c1 = Contacts("Grace", "something@hotmail.com")
print(c1.all_contacts)

[<__main__.Contact object at 0x0287E430>, <__main__.Contact object`

但在这个更简单的例子中,它实际上打印:

class Example():
    samplelist= [1,2,3]

test= Example()
print (test.samplelist)
[1, 2, 3]

我认为这一行是罪魁祸首:Contact.all_contacts.append(self)在第一个示例代码中.但我不确定这里发生了什么.

编辑:

一些用户告诉我只是附加self.name而不是仅仅self.

所以当我这样做时:

class Contacts:
   all_contacts = []

   def __init__(self, name, email):
      self.name = name
      self.email = email
      Contacts.all_contacts.append(self.name)
      Contacts.all_contacts.append(self.email)

   def __str__(self):
      return '%s, <%s>' % (self.name, self.email)

   def __repr__(self):
      return str(self)

c1 = Contacts("Paul", "something@hotmail.com")
c2 = Contacts("Darren", "another_thing@hotmail.com")
c3 = Contacts("Jennie", "different@hotmail.com")

print(Contacts.all_contacts)

我明白了:

['Paul', 'something@hotmail.com', 'Darren', 'another_thing@hotmail.com', 'Jennie', 'different@hotmail.com']

代替:

[Paul, <something@hotmail.com>, Darren, <another_thing@hotmail.com>, Jennie, <different@hotmail.com>]

因此,__str__方法中的格式不起作用.

Answer 1

当您打印列表时,它会调用__str__列表,但列表会在内部调用__repr__()其元素.您也应该__repr__()为您的班级实施.示例 -

class Contacts:
    all_contacts = []

    def __init__(self, name, email):
       self.name = name
       self.email = email
       Contacts.all_contacts.append(self)

    def __str__(self):
       return '%s, <%s>' % (self.name, self.email)

    def __repr__(self):
        return str(self)

演示 -

class Contacts:
    all_contacts = []

    def __init__(self, name, email):
       self.name = name
       self.email = email
       Contacts.all_contacts.append(self)

    def __str__(self):
       return '%s, <%s>' % (self.name, self.email)

    def __repr__(self):
        return str(self)

contact1 = Contacts("Grace1", "something1@hotmail.com")
contact2 = Contacts("Grace2", "something2@hotmail.com")
contact3 = Contacts("Grace3", "something3@hotmail.com")
print(Contacts.all_contacts)

结果 -

[Grace1, <something1@hotmail.com>, Grace2, <something2@hotmail.com>, Grace3, <something3@hotmail.com>]

此外,从输出看起来列表实际上有6元素,所以你应该考虑改变__repr__返回.