[英]Python: How do you chain methods on lists of lists of objects?
我希望能夠在具有子節點和孫子節點的節點實例上鏈接方法。 當我在該頂級節點上調用方法時，我希望能夠返回該對象的子孫。 我還希望能夠從最頂層的節點獲得這些子孫的屬性。 我有以下數據模型：
class GrandParent(object):
def __init__(self,name='',age=0,is_retired=True,children=[]):
self.name = name
self.age = age
self.is_retired = is_retired
if children:
self.children = children
def print_mychildren(self):
for child in self.children:
print "Child: ",child
def mychildren(self):
return self.children
def __str__(self):
return "name: %s age: %s is_retired:%s" %(self.name,self.age,self.is_retired)
class Parent(object):
def __init__(self,name='',age=0,has_mortgage=True,children=[]):
self.name = name
self.age = age
self.has_mortgage = has_mortgage
if children:
self.children = children
def print_mychildren(self):
for child in self.children:
print "Child: ",child
def __str__(self):
return "name: %s age: %s has_mortgage: %s" %(self.name,self.age,self.has_mortgage)
class Child(object):
def __init__(self,name='',age=0,has_toys=True):
self.name = name
self.age = age
self.has_toys = has_toys
def __str__(self):
return "name: %s age: %s has_toys:%s" %(self.name,self.age,self.has_toys)
if __name__ == '__main__':
beaver = Child('Beaver',12,True)
ward = Child('Ward',16,False)
june = Parent('June',38,True,[beaver,ward])
grandpa = GrandParent('Grandpa',61,True,[june])
print grandpa
grandpa.print_mychildren() # print June
# Doesn't work
grandpa.mychildren().print_mychildren() # I want it to print Ward and Beaver
# Doesn't work
print grandpa.mychildren().mychild('Beaver').age # I want it to return an age
請注意，我希望將GrandParent，Parent和Child類分開，因為我想為每個類提供不同的屬性，例如has_mortgage或is_retired。
從上面的數據模型中，我希望能夠鏈接方法以遍歷頂級節點的子節點。 它看起來像這樣：
grandpa.children # returns a list of children
print grandpa.mychild('June').has_mortgage # returns the value
grandpa.mychildren().mychildren() # prints a list of grandchildren
print grandpa.mychildren().mychild('Ward').has_toys # return the value
換句話說，我可以發表聲明：
print grandpa.mychildren().mychildren()
表現得像：
for child in grandpa.children:
for grandchild in child.children:
print grandchild
我很感激你的回答。 謝謝。
保羅
伊利諾伊州芝加哥
你可以做你想做的事，但這需要一些工作。 您需要將Parent和Grandparent類的children值設置為將方法和成員變量的訪問映射到其內容的自定義容器。
這是一個可能的實現。 它可能有一些奇怪的角落案例，我還沒有解決，但它適用於我嘗試過的基本內容：
from operator import attrgetter
class mappingContainer(object):
def __init__(self, contents):
self.contents = contents
# sequence protocol methods
def __getitem__(self, index):
return self.contents[index]
def __setitem__(self, index, value):
self.contents[index] = value
def __iter__(self):
return iter(self.contents)
def __contains__(self, o):
return o in self.contents
def __len__(self):
return len(self.contents)
# map attribute access and method calls
def __getattr__(self, name):
return mappingContainer(map(attrgetter(name), self.contents))
def __call__(self, *args, **kwargs):
return mappingContainer([o(*args, **kwargs) for o in self.contents])
# string conversions
def __repr__(self):
return "mappingContainer(%s)" % repr(self.contents)
這是一個簡單的“人”類，我曾用它來測試它。 每個人在mappingContainer都有一個名稱，零個或多mappingContainer以及一個任意命名的方法，它打印一些東西並返回一個值。
class Person(object):
def __init__(self, name, children = None):
self.name = name
if not children:
children = []
self.children = mappingContainer(children)
def __repr__(self):
return self.name
def someMethod(self):
print "%s's someMethod()" % str(self)
return "%s's return value" % str(self)
這是我測試它的方式（長線包裹，顯示在這里）：
>>> c1 = Person("Adam")
>>> c2 = Person("Albert")
>>> c3 = Person("Alice")
>>> c4 = Person("Agnes")
>>> p1 = Person("Bob", [c1, c2])
>>> p2 = Person("Betty", [c3,c4])
>>> gp = Person("Charles", [p1, p2])
>>> gp
Charles
>>> gp.children
mappingContainer([Bob, Betty])
>>> gp.children.children
mappingContainer([mappingContainer([Adam, Albert]),
mappingContainer([Alice, Agnes])])
>>> gp.children.children.someMethod()
Adam's someMethod()
Albert's someMethod()
Alice's someMethod()
Agnes's someMethod()
mappingContainer([mappingContainer(["Adam's return value",
"Albert's return value"]),
mappingContainer(["Alice's return value",
"Agnes's return value"])])
現在這可能不是你想要的，因為你不能通過gp.children.children直接迭代孫子gp.children.children （雖然函數調用會一直向下冒泡，但它們的結果仍然會嵌套在兩層容器中） 。 我認為mappingContainer可能會以某種方式進行調整以解決這個問題，但我不確定它是否值得。
如果出現任何問題，這類鏈式調用將成為調試的噩夢（除非你是超人，否則你的代碼中的某些內容在開發的某些時候總會出錯）。 如果您明確地迭代或遞歸您擁有的關系層次結構，那么您的代碼將更容易編寫和理解。
一種方法是為列表集合創建一個包裝器（與jQuery處理方法鏈接的方式略有相似）
碼：
class ChainList(list):
def __getattribute__(self, name):
try:
return object.__getattribute__(self, name)
except:
pass
return ChainList([getattr(child, name)
for child in self if name in dir(child)])
def __call__(self, *args, **kwargs):
return ChainList([child(*args, **kwargs)
for child in self if callable(child)])
class GrandParent(object):
def __init__(self,name='',age=0,is_retired=True,children=[]):
self.name = name
self.age = age
self.is_retired = is_retired
if children:
self.children = ChainList(children)
def print_mychildren(self):
for child in self.children:
print "Child: ",child
def mychildren(self):
return self.children
def __str__(self):
return "name: %s age: %s is_retired:%s" %(self.name,self.age,self.is_retired)
class Parent(object):
def __init__(self,name='',age=0,has_mortgage=True,children=[]):
self.name = name
self.age = age
self.has_mortgage = has_mortgage
if children:
self.children = ChainList(children)
def print_mychildren(self):
for child in self.children:
print "Child: ",child
def __str__(self):
return "name: %s age: %s has_mortgage: %s" %(self.name,self.age,self.has_mortgage)
def mychild(self, name):
for child in self.children:
if child.name == name:
return child
class Child(object):
def __init__(self,name='',age=0,has_toys=True):
self.name = name
self.age = age
self.has_toys = has_toys
def __str__(self):
return "name: %s age: %s has_toys:%s" %(self.name,self.age,self.has_toys)
if __name__ == '__main__':
beaver = Child('Beaver',12,True)
ward = Child('Ward',16,False)
june = Parent('June',38,True,[beaver,ward])
grandpa = GrandParent('Grandpa',61,True,[june])
print grandpa
grandpa.print_mychildren() # print June
# Doesn't work
grandpa.mychildren().print_mychildren() # I want it to print Ward and Beaver
# Doesn't work
print grandpa.mychildren().mychild('Beaver').age # I want it to return an age
主要思想是ChainList類並重寫__getattribute__和__call__並包裝需要與此類鏈接的所有列表。
我還將mychild（）添加到Parent以使示例正常工作。
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