在python上子类化列表 [英] Subclassing a list on python
问题描述
我正在在线上学习教程,代码是这样的:
I am following a tutorial online and the code is this:
class Hands(list):
def __init__(self, size=0, die_class=None, *args, **kwargs):
if not die_class:
raise ValueError("You must provide a die class")
super().__init__()
for _ in range(size):
self.append(die_class())
基本上,它模拟一个具有多个骰子( size
)和他们持有的骰子( die_class
)的玩家.
It basically models a player with a number of dice (size
) and what dice they are holding (die_class
).
我的困惑是为什么我们需要调用 super().__ init __
?我尝试在没有它的情况下运行代码,并且效果很好!为什么需要打这个电话?
My confusion is why do we need to call super().__init__
? I tried running the code without it and it worked fine! Why is the call necessary?
推荐答案
如果基类要确保运行任何初始化代码,则需要调用 __ init __()
.没有呼叫就可以工作(似乎)是一个巧合,或者您可能还没有解决由此产生的问题.即使它在您当前使用的Python版本和实现中均能正常工作,也不能保证在不调用基类的 __ init __
方法的情况下其他版本和实现也能正常工作.
You need to call the __init__()
if the base class to be sure any initialisation code there is run. That it (seems) to work without that call can be a coincidence, or you may simply haven't hit the resulting problem yet. Even if it works consistently in the Python version and implementation you are using currently, it isn't guaranteed for other versions and implementations to work without calling the base class' __init__
method.
您实际上还可以使用该调用来用骰子对象填充列表:
Also you can actually use that call to populate the list with your dice objects:
class Hands(list):
def __init__(self, size=0, die_factory=None):
if not die_factory:
raise ValueError('You must provide a die factory')
super().__init__(die_factory() for _ in range(size))
我已将 die_class
重命名为 die_factory
,因为可以在此处使用产生新die对象的任何可调用对象.
I've renamed die_class
to die_factory
as any callable that produces a new die object can be used here.
注意:这里可能违反 Hands
和 list
之间的 is-a 关系,除非 Hands
对象实际上是一个 list
,即 all 的所有方法和列表的行为对于 Hands
对象也是有意义的.
Note: You may violate the is-a relationship between Hands
and list
here unless a Hands
object actually is a list
, i.e. all methods and behaviour of lists also make sense for Hands
objects.
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