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在python中嵌套列表
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我试图解决一个问题,不幸的是,这超出了我的能力范围.
我有一系列嵌套列表,并在迭代它们时,如果下一个元素是一个列表,我想将它作为当前元素的属性追加.
像往常一样,一个例子比我糟糕的英语(这里有一些代码可以复制和粘贴)更好：
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=[])]]]
正如你所看到的,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追加到前一个不是列表的元素(递归)
预期产量：
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,我甚至都没有解决它.
编辑
为了完整起见,这是一个更复杂的例子
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,它也会通过相同的测试,因此对于我尝试过的测试用例都可以正常工作.我会保留Rik的,因为我认为当我需要这种行为时,一个独立的函数可以更方便.
解决方法:
我想出了这个：
def append_children(parent, iterable):
last = None
for i in iterable:
if hasattr(i, '__iter__'):
append_children(last, i)
else:
parent.children.append(i)
last = i
def magic(lst):
result = []
for i in lst:
if hasattr(i, '__iter__'):
append_children(result[-1], i)
else:
result.append(i)
return result
例：
>>> l_in = [T(id=1, children=[]), T(id=2, children=[]), [T(id=3, children=[]),
... T(id=4, children=[]), [T(id=5, children=[])]]]
>>> l_expected = [T(id=1, children=[]),
... T(id=2, children=[T(id=3, children=[]),
... T(id=4, children=[T(id=5, children=[])])])]
>>> l_ouput = magic(l_in)
>>> repr(l_output) == repr(l_expected)
True
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