python - 在 python 中嵌套列表

Question

不幸的是，我正在尝试解决一个超出我能力范围的问题。我有一系列嵌套列表，在遍历它们时，如果下一个元素是列表，我想将其附加为当前元素的属性。像往常一样，一个例子比我糟糕的英语更好(这里有一些代码可以复制和粘贴):

class T(object):
    def __init__(self, id, children):
         self.id = id 
         self.children = children or []

    def __repr__(self):
         return u"T(id={0}, children={1})".format(self.id, self.children) 


# first a short example
l0 = [T(id=1, children=[]), T(id=2, children=[]), [T(id=3, children=[]), 
      T(id=4, children=[]), [T(id=5, children=[])]]]

如您所见，l0 有 3 个元素，最后一个是三个元素的列表:我需要的是将最后一个列表附加到不是列表的前一个元素(并且递归地)预期输出:

l1 = magic(l0)    
[T(id=1, children=[]), T(id=2, children=[T(id=3, children=[]), T(id=4, children=[T(id=5, children=[])])])]

希望有人可以分享一些建议来解决这个问题，我已经投入了很多时间，但我什至没有接近解决这个问题。

编辑

为了完整起见，这里有一个更复杂的例子

l0 = [T(children=[], id=1),
      T(children=[], id=2),
      T(children=[], id=3),
      [T(children=[], id=40),
       T(children=[], id=41),
       T(children=[], id=42),
       T(children=[], id=43),
       T(children=[], id=44),
       T(children=[], id=45),
       [T(children=[], id=50),
        T(children=[], id=51),
        T(children=[], id=52),
        T(children=[], id=54),
        [T(children=[], id=60),
         T(children=[], id=61),
         T(children=[], id=62),
         T(children=[], id=63),
         [T(children=[], id=70)],
         T(children=[], id=64)]]],
      T(children=[], id=8),
      T(children=[], id=9)]

我使用@rik-poggi 函数构建了一个 doctest 作为示例，到目前为止它似乎没问题:

>>> from magic_bag import magic
>>> class T(object):                                                        
...     def __init__(self, id, children):                                   
...         self.id = id                                                    
...         self.children = children or []                                  
...                                                                         
...     def __repr__(self):                                                 
...         return u"T(children={0}, id={1})".format(self.children, self.id)
...                                                                         
>>> l0 = [T(id=1, children=[]), T(id=2, children=[]), T(id=3, children=[]), 
... [T(id=40, children=[]), T(id=41, children=[]), T(id=42, children=[]),   
... T(id=43, children=[]), T(id=44, children=[]), T(id=45, children=[]),    
... [T(id=50, children=[]), T(id=51, children=[]), T(id=52, children=[]),   
... T(id=54, children=[]), [T(id=60, children=[]), T(id=61, children=[]),   
... T(id=62, children=[]), T(id=63, children=[])]]], T(id=8, children=[]),  
... T(id=9, children=[])]                                                   
>>> l1 = magic(l0)                                              
>>> l1[0]                                                                   
T(children=[], id=1)                                                        
>>> l1[1]                                                                   
T(children=[], id=2)                                                        
>>> l1[3]                                                                   
T(children=[], id=8)                                                        
>>> l1[4]                                                                   
T(children=[], id=9)                                                        
>>> l1[5]                                                                   
Traceback (most recent call last):                                          
    ...                                                                     
IndexError: list index out of range                                         
>>> l1[2].children[5].children[3]                                           
T(children=[T(children=[], id=60), T(children=[], id=61), T(children=[], id=62), T(children=[], id=63)], id=54)
>>> l0 = [T(id=1, children=[]), T(id=2, children=[]), T(id=3, children=[]), 
... [T(id=40, children=[]), T(id=41, children=[]), T(id=42, children=[]),   
... T(id=43, children=[]), T(id=44, children=[]), T(id=45, children=[]),    
... [T(id=50, children=[]), T(id=51, children=[]), T(id=52, children=[]),   
... T(id=54, children=[]), [T(id=60, children=[]), T(id=61, children=[]),   
... T(id=62, children=[]), T(id=63, children=[]), [T(id=70, children=[])],  
... T(id=64, children=[])]]], T(id=8, children=[]), T(id=9, children=[])]   
>>> l1 = magic(l0)                                              
>>> l1[2].children[5].children[0].id                                        
50                                                                          
>>> len(l1[2].children[5].children[3].children)                             
5                                                                           
>>> l1[2].children[5].children[3].children[3].children                      
[T(children=[], id=70)]                                                     
>>> l1[2].children[5].children[3].children[4].id==64                        
True                                                                  

使用@rob-wouters 替代方案，它也通过了相同的测试，因此对于我尝试过的测试用例，两者都可以正常工作。我会保留 Rik 的，因为我认为当我需要这种行为时，独立函数会更方便。

Answer 1

我会这样做:

class T(object):
    def __init__(self, id, children):
         self.id = id 
         self.children = children or []

    def add_children(self, children):
        for child in children:
            if isinstance(child, list):
                self.children[-1].add_children(child)
            else:
                self.children.append(child)

    def __repr__(self):
         return u"T(id={0}, children={1})".format(self.id, self.children) 


l0 = [T(id=1, children=[]),
      T(id=2, children=[]), [T(id=3, children=[]), T(id=4, children=[]),
                            [T(id=5, children=[])]]]

root = T(id=0, children=[])
root.add_children(l0)
print(root.children)

如果您真的想要一个独立的方法，而不是让两个函数处理相同的情况，您可以使用以下方法:

def add_children(node, children):
    for child in children:
        if hasattr(child, '__iter__'):
            add_children(node.children[-1], child)
        else:
            node.children.append(child)

def create_tree(lst):
    root = T(id=0, children=[])
    add_children(root, lst)
    return root.children

print(create_tree(l0))

这更优雅一些，因为与具有几乎相同的两个函数相比，它避免了很多重复代码。我确实更改了我的 isinstance 检查，以支持检查 __iter__，这样可以更灵活地存储子项列表。