實現一個檢查python 2.7的add方法中的類型的正確方法是什麼？

Question

我想限制集合的元素是某個類的實例。我應該繼承set並重寫add方法嗎？我將如何做到這一點？

Answer 1

首先，在構建一組自定義對象時，您可能需要查看this question及其答案。簡單地說，你需要定義方法，如__hash__()和__eq__()，這樣就可以將它們添加到組：

class Foo: 
    def __init__(self, value=0): 
     self.value = value 
    def __hash__(self): 
     return self.value 
    def __eq__(self, other): 
     return isinstance(other, Foo) and self.value == other.value 

現在你可以比較的對象，與set能以及：

In [19]: a = Foo() 

In [20]: b = Foo() 

In [21]: c = Foo(1) 

In [22]: a == b 
Out[22]: True 

In [23]: b == c 
Out[23]: False 

In [24]: s = set([a, b, c]) 

In [25]: s 
Out[25]: 
set([<__main__.Foo instance at 0x267f758>, 
    <__main__.Foo instance at 0x267f950>]) 

In [26]: s.add(Foo()) 

In [27]: s 
Out[27]: 
set([<__main__.Foo instance at 0x267f758>, 
    <__main__.Foo instance at 0x267f950>]) 

問題是，你仍然可以添加一些東西到一套完全不同的：

In [28]: s.add(1) 

In [29]: s 
Out[29]: 
set([<__main__.Foo instance at 0x267f758>, 
    <__main__.Foo instance at 0x267f950>, 
    1]) 

一種方式是覆蓋按照您的建議使用set的方法：

In [30]: class FooSet(set): 
    ....:  def add(self, elem): 
    ....:   if isinstance(elem, Foo): 
    ....:    set.add(self, elem) 
    ....:   else: 
    ....:    raise TypeError('%s is not a Foo' % elem) 
    ....:    # or just put "pass" here for silent behavior 

In [31]: s = FooSet([a, b, c]) 

In [32]: s 
Out[32]: 
set([<__main__.Foo instance at 0x267f758>, 
    <__main__.Foo instance at 0x267f950>]) 

In [33]: s.add(Foo()) 

In [34]: s 
Out[34]: 
set([<__main__.Foo instance at 0x267f758>, 
    <__main__.Foo instance at 0x267f950>]) 

In [35]: s.add(Foo(2)) 

In [36]: s 
Out[36]: 
set([<__main__.Foo instance at 0x267f758>, 
    <__main__.Foo instance at 0x267f950>, 
    <__main__.Foo instance at 0x26808c0>]) 

In [37]: s.add(2) 
--------------------------------------------------------------------------- 
TypeError         Traceback (most recent call last) 

... 

TypeError: 2 is not a Foo