的属性设置为一个单独的列表元素
问题描述:
所有,我有一个几个列表对象类,定义如下:的属性设置为一个单独的列表元素
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
这里奇怪的是,在调用device.channels[1] = 'try'
作品,但似乎并没有为'通过@ setter.channels函数。从下面的输出显示古怪:
device = Device()
print("before:",device.channels)
device.channels[1] = "try"
print("after:",frmdeviced.channels)
device.channels = "try2"
print("check:",frm4d.channels)
并且输出是:
before: [None, None, None, None, None, None]
after: [None, 'try', None, None, None, None] # accessing single element is achieved
# , but not through @channels.setter!
inside: [None, 'try', None, None, None, None] try # only here we're
check: try2 # at least the setter works..
由于我需要逻辑时的channels
单个元件被设置为运行,这种行为是有问题的。 我想知道导致这种行为的基础python机制是什么,它是如何被覆盖?是否有更多pythonic方式来实现设置/获取特定列表元素的目标?
答
device.channels[1] = "try"
首先要访问"@property"
getter方法,它返回一个列表,然后索引操作将在不在设备上的列表上执行。下面的例子证明它 -
>>> 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
+0
@ user2357112,我以前看不到python3.x标记,现在我已经修改了我的答案。感谢您指出它 – AlokThakur
答
要仪器列表,创建一个仪表列表类。 Jupyter会话示例:
In [23]: class MonitoredList(list):
def __setitem__(self, index, value):
# run special logic here
print("setting index {} to {}".format(index, value))
super(MonitoredList, self).__setitem__(index, value)
In [24]: zz = MonitoredList([None]*6)
In [25]: zz
Out[25]: [None, None, None, None, None, None]
In [26]: zz[3] = 42
setting index 3 to 42
In [27]: zz
Out[27]: [None, None, None, 42, None, None]
您需要在'channels'引用的对象上实现'__setitem__',在'Device'上执行它是没有意义的。 – jonrsharpe
设置列表的一个元素是列表的一个操作,而不是交给列表的'Device'的操作。 – user2357112