我有这个元组列表
list_of_tuples = [('0', '1'), ('1', '1.1'), ('1', '1.2'), ('1', '1.3'), ('1', '1.4'), ('0', '3'), ('3', '3.1'), ('3', '3.2'), ('3', '3.3'), ('3', '3.4'), ('3', '3.5'), ('0', '4'), ('4', '4.1'), ('4', '4.2'), ('4', '4.3'), ('4', '4.4'), ('4', '4.5'), ('4', '4.6'), ('4', '4.7'), ('4', '4.8'), ('4', '4.9'), ('0', '5'), ('5', '5.1'), ('5', '5.2'), ('5', '5.3'), ('5', '5.4'), ('0', '6'), ('6', '6.1'), ('6', '6.2'), ('6', '6.3'), ('6', '6.4'), ('6', '6.5'), ('0', '7'), ('7', '7.1'), ('7', '7.2'), ('7', '7.3'), ('7.3', '7.3.1'), ('7.3', '7.3.2'), ('7.3', '7.3.3'), ('0', '8'), ('8', '8.1.1'), ('8', '8.1.2'), ('8', '8.2'), ('0', '9'), ('9', '9.1'), ('9', '9.2'), ('0', '10'), ('10', '10.1'), ('10', '10.2'), ('10', '10.3'), ('10.3', '10.3.2'), ('10.3', '10.3.3'), ('10.3', '10.3.4'), ('10.3', '10.3.5'), ('10.3', '10.3.6')
元组中的第一个值是父元素，第二个值是值。
(parent,child)
我希望输出像这样。
{ '0':
{'1':
{'1.1':None,
'1.2':None......
}
{'3':
{'3.1':None/[]..
}
}
}
我可以将它添加到字典中，但我希望它像一个有多个孩子的树一样嵌套。
d = defaultdict(list)
for k, v in h:
d[k].append(v)
感谢任何帮助。
答案 0 :(得分：5)
我要打开一个快速的树类，可以接受这个输入并用它构建一个树，然后我将为该类编写一个方法，将它们转换回字典
class Tree:
def __init__(self, name, parent=None): #parent is None to detect root
self.name = name
self.parent = parent
self.children = []
def add(self, target , child):
'''
Does DFS until it finds Tree with name target. Creates a Tree(child)
as child of Tree name
'''
if self.name == target:
self.children.append(Tree(child, self))
return True
else:
for subtree in self.children:
if subtree.add(target, child):
return True
if self.parent:
return False
raise ValueError('Bad Parent no such node {}'.format(target))
def dictify(self):
d = {}
for child in self.children:
d.update(child.dictify())
return {self.name: d}
list_of_tuples = [('0', '1'), ('1', '1.1'), ('1', '1.2'), ('1', '1.3'), ('1', '1.4'), ('0', '3'), ('3', '3.1'), ('3', '3.2'), ('3', '3.3'), ('3', '3.4'), ('3', '3.5'), ('0', '4'), ('4', '4.1'), ('4', '4.2'), ('4', '4.3'), ('4', '4.4'), ('4', '4.5'), ('4', '4.6'), ('4', '4.7'), ('4', '4.8'), ('4', '4.9'), ('0', '5'), ('5', '5.1'), ('5', '5.2'), ('5', '5.3'), ('5', '5.4'), ('0', '6'), ('6', '6.1'), ('6', '6.2'), ('6', '6.3'), ('6', '6.4'), ('6', '6.5'), ('0', '7'), ('7', '7.1'), ('7', '7.2'), ('7', '7.3'), ('7.3', '7.3.1'), ('7.3', '7.3.2'), ('7.3', '7.3.3'), ('0', '8'), ('8', '8.1.1'), ('8', '8.1.2'), ('8', '8.2'), ('0', '9'), ('9', '9.1'), ('9', '9.2'), ('0', '10'), ('10', '10.1'), ('10', '10.2'), ('10', '10.3'), ('10.3', '10.3.2'), ('10.3', '10.3.3'), ('10.3', '10.3.4'), ('10.3', '10.3.5'), ('10.3', '10.3.6')]
root = Tree('0')
for parent, child in list_of_tuples:
root.add(parent, child)
print(root.dictify())
这是pprint（漂亮印刷）
{'0': {'1': {'1.1': {}, '1.2': {}, '1.3': {}, '1.4': {}},
'10': {'10.1': {},
'10.2': {},
'10.3': {'10.3.2': {},
'10.3.3': {},
'10.3.4': {},
'10.3.5': {},
'10.3.6': {}}},
'3': {'3.1': {}, '3.2': {}, '3.3': {}, '3.4': {}, '3.5': {}},
'4': {'4.1': {},
'4.2': {},
'4.3': {},
'4.4': {},
'4.5': {},
'4.6': {},
'4.7': {},
'4.8': {},
'4.9': {}},
'5': {'5.1': {}, '5.2': {}, '5.3': {}, '5.4': {}},
'6': {'6.1': {}, '6.2': {}, '6.3': {}, '6.4': {}, '6.5': {}},
'7': {'7.1': {},
'7.2': {},
'7.3': {'7.3.1': {}, '7.3.2': {}, '7.3.3': {}}},
'8': {'8.1.1': {}, '8.1.2': {}, '8.2': {}},
'9': {'9.1': {}, '9.2': {}}}}
如果您希望这些空字母为None，只需将dictify更改为return {self.name: d if d else None}
编辑：schwobaseggl对插入复杂性提出了一个很好的观点。这是一个利用有序输入的Tree类的版本
class Tree:
def __init__(self, name, parent=None): #parent is None to detect root
self.name = name
self.parent = parent
self.children = []
def add(self, target , child):
'''
Accepts additions in DFS order. Relies on the fact that every node will
be the direct descendant of the previous node or one of its ancestors.
'''
if self.name == target:
kiddo = Tree(child, self)
self.children.append(kiddo)
return kiddo
elif self.parent:
return self.parent.add(target, child)
else:
raise ValueError('Bad Parent no such node {}'.format(target))
def dictify(self):
d = {}
for child in self.children:
d.update(child.dictify())
return {self.name: d}
list_of_tuples = [('0', '1'), ('1', '1.1'), ('1', '1.2'), ('1', '1.3'), ('1', '1.4'), ('0', '3'), ('3', '3.1'), ('3', '3.2'), ('3', '3.3'), ('3', '3.4'), ('3', '3.5'), ('0', '4'), ('4', '4.1'), ('4', '4.2'), ('4', '4.3'), ('4', '4.4'), ('4', '4.5'), ('4', '4.6'), ('4', '4.7'), ('4', '4.8'), ('4', '4.9'), ('0', '5'), ('5', '5.1'), ('5', '5.2'), ('5', '5.3'), ('5', '5.4'), ('0', '6'), ('6', '6.1'), ('6', '6.2'), ('6', '6.3'), ('6', '6.4'), ('6', '6.5'), ('0', '7'), ('7', '7.1'), ('7', '7.2'), ('7', '7.3'), ('7.3', '7.3.1'), ('7.3', '7.3.2'), ('7.3', '7.3.3'), ('0', '8'), ('8', '8.1.1'), ('8', '8.1.2'), ('8', '8.2'), ('0', '9'), ('9', '9.1'), ('9', '9.2'), ('0', '10'), ('10', '10.1'), ('10', '10.2'), ('10', '10.3'), ('10.3', '10.3.2'), ('10.3', '10.3.3'), ('10.3', '10.3.4'), ('10.3', '10.3.5'), ('10.3', '10.3.6')]
root = Tree('0')
curr = root
for parent, child in list_of_tuples:
curr = curr.add(parent, child)
print(root.dictify())
答案 1 :(得分：4)
Patrick的解决方案是通用的，面向对象的。但是，它是O(N^2)（N是边数），因为它遍历每个边的整个树。既然您知道自己获得了深度优先的边缘，那么通过记忆树中的当前位置，插入正确的位置，然后返回，您可以节省很多时间（对于大树：很多！）如果需要的话。
以下内容更简洁，O(N)没有您自己的类和额外转换的额外开销：
from pprint import pprint
d = {}
crnt = d # memo the crnt subtree
stck = [] # stack of (sub)trees along current path
for k, v in list_of_tuples:
while stck and k not in crnt:
crnt = stck.pop()
if k not in crnt:
crnt[k] = {}
stck.append(crnt)
crnt = crnt[k]
crnt[v] = {}
pprint(d)
{'0': {'1': {'1.1': {}, '1.2': {}, '1.3': {}, '1.4': {}},
'10': {'10.1': {},
'10.2': {},
'10.3': {'10.3.2': {},
'10.3.3': {},
'10.3.4': {},
'10.3.5': {},
'10.3.6': {}}},
'3': {'3.1': {}, '3.2': {}, '3.3': {}, '3.4': {}, '3.5': {}},
'4': {'4.1': {},
'4.2': {},
'4.3': {},
'4.4': {},
'4.5': {},
'4.6': {},
'4.7': {},
'4.8': {},
'4.9': {}},
'5': {'5.1': {}, '5.2': {}, '5.3': {}, '5.4': {}},
'6': {'6.1': {}, '6.2': {}, '6.3': {}, '6.4': {}, '6.5': {}},
'7': {'7.1': {},
'7.2': {},
'7.3': {'7.3.1': {}, '7.3.2': {}, '7.3.3': {}}},
'8': {'8.1.1': {}, '8.1.2': {}, '8.2': {}},
'9': {'9.1': {}, '9.2': {}}}}