python - 单个列表元素的属性 setter

Question

所有，我有一个类有几个列表对象，定义如下:

class Device:

  def __init__(self):
    self._channels = [None]*6
    self._outputs = [None]*4

  @property
  def channels(self):
    return self._channels

  @channels.setter
  def channels(self,value):
    print("inside:",self.channels, value)
    self._channels = value

这里的奇怪之处在于调用 device.channels[1] = 'try' 有效，但似乎没有“通过”@setter.channels 函数。下面的输出揭示了奇怪的地方:

device = Device()
print("before:",device.channels)
device.channels[1] = "try"
print("after:",frmdeviced.channels)
device.channels = "try2"
print("check:",frm4d.channels)

输出是:

 before: [None, None, None, None, None, None] <br/>
 after: [None, 'try', None, None, None, None] # accessing single element is achieved<br/>                                              # , but not through @channels.setter!<br/> 
 inside: [None, 'try', None, None, None, None] try # only here we're<br/>
 check: try2                                   # at least the setter works..<br/>

由于我需要在设置 channels 的单个元素时运行逻辑，因此这种行为是有问题的。我想知道导致这种行为的底层 python 机制是什么，它是如何被覆盖的？是否有更pythonic的方式来实现设置/获取特定列表元素的目标？

Answer 1

device.channels[1] = "try" 会先访问"@property" getter方法，返回一个列表，然后进行索引操作在列表上不在设备上。下面的例子演示了它-

>>> class Device:

  def __init__(self):
    self._channels = [None]*6
    self._outputs = [None]*4

  @property
  def channels(self):
    print("inside @property")
    return self._channels

  @channels.setter
  def channels(self,value):
    print("inside:",self.channels, value)
    self._channels = value


>>> device = Device()
>>> device.channels[1] = "try"
inside @property