如何在Python中实现通用树?这有内置的数据结构吗?
当前回答
你可以试试:
from collections import defaultdict
def tree(): return defaultdict(tree)
users = tree()
users['harold']['username'] = 'hrldcpr'
users['handler']['username'] = 'matthandlersux'
建议在这里:https://gist.github.com/2012250
其他回答
另一个基于Bruno回答的树的实现:
class Node:
def __init__(self):
self.name: str = ''
self.children: List[Node] = []
self.parent: Node = self
def __getitem__(self, i: int) -> 'Node':
return self.children[i]
def add_child(self):
child = Node()
self.children.append(child)
child.parent = self
return child
def __str__(self) -> str:
def _get_character(x, left, right) -> str:
if x < left:
return '/'
elif x >= right:
return '\\'
else:
return '|'
if len(self.children):
children_lines: Sequence[List[str]] = list(map(lambda child: str(child).split('\n'), self.children))
widths: Sequence[int] = list(map(lambda child_lines: len(child_lines[0]), children_lines))
max_height: int = max(map(len, children_lines))
total_width: int = sum(widths) + len(widths) - 1
left: int = (total_width - len(self.name) + 1) // 2
right: int = left + len(self.name)
return '\n'.join((
self.name.center(total_width),
' '.join(map(lambda width, position: _get_character(position - width // 2, left, right).center(width),
widths, accumulate(widths, add))),
*map(
lambda row: ' '.join(map(
lambda child_lines: child_lines[row] if row < len(child_lines) else ' ' * len(child_lines[0]),
children_lines)),
range(max_height))))
else:
return self.name
还有一个如何使用它的例子:
tree = Node()
tree.name = 'Root node'
tree.add_child()
tree[0].name = 'Child node 0'
tree.add_child()
tree[1].name = 'Child node 1'
tree.add_child()
tree[2].name = 'Child node 2'
tree[1].add_child()
tree[1][0].name = 'Grandchild 1.0'
tree[2].add_child()
tree[2][0].name = 'Grandchild 2.0'
tree[2].add_child()
tree[2][1].name = 'Grandchild 2.1'
print(tree)
它应该输出:
Root node / / \ Child node 0 Child node 1 Child node 2 | / \ Grandchild 1.0 Grandchild 2.0 Grandchild 2.1
我推荐任何树(我是作者)。
例子:
from anytree import Node, RenderTree
udo = Node("Udo")
marc = Node("Marc", parent=udo)
lian = Node("Lian", parent=marc)
dan = Node("Dan", parent=udo)
jet = Node("Jet", parent=dan)
jan = Node("Jan", parent=dan)
joe = Node("Joe", parent=dan)
print(udo)
Node('/Udo')
print(joe)
Node('/Udo/Dan/Joe')
for pre, fill, node in RenderTree(udo):
print("%s%s" % (pre, node.name))
Udo
├── Marc
│ └── Lian
└── Dan
├── Jet
├── Jan
└── Joe
print(dan.children)
(Node('/Udo/Dan/Jet'), Node('/Udo/Dan/Jan'), Node('/Udo/Dan/Joe'))
anytree也有一个强大的API:
简单的树创建 简单树修改 预序树迭代 后序树迭代 解析相对节点路径和绝对节点路径 从一个节点移动到另一个节点。 树渲染(参见上面的例子) 节点连接/分离连接
class Tree(dict):
"""A tree implementation using python's autovivification feature."""
def __missing__(self, key):
value = self[key] = type(self)()
return value
#cast a (nested) dict to a (nested) Tree class
def __init__(self, data={}):
for k, data in data.items():
if isinstance(data, dict):
self[k] = type(self)(data)
else:
self[k] = data
作为一个字典,但提供尽可能多的嵌套字典。 试试下面的方法:
your_tree = Tree()
your_tree['a']['1']['x'] = '@'
your_tree['a']['1']['y'] = '#'
your_tree['a']['2']['x'] = '$'
your_tree['a']['3'] = '%'
your_tree['b'] = '*'
将传递一个嵌套的字典…就像树一样。
{'a': {'1': {'x': '@', 'y': '#'}, '2': {'x': '$'}, '3': '%'}, 'b': '*'}
... 如果你已经有字典了,它会把每一层都投射到一棵树上:
d = {'foo': {'amy': {'what': 'runs'} } }
tree = Tree(d)
print(d['foo']['amy']['what']) # returns 'runs'
d['foo']['amy']['when'] = 'now' # add new branch
这样,你就可以随心所欲地编辑/添加/删除每个词典级别。 遍历等所有dict方法仍然适用。
Treelib也很方便完成这项任务。文档可以在treelib找到。
from treelib import Node, Tree
tree = Tree() # creating an object
tree.create_node("Harry", "harry") # root node
tree.create_node("Jane", "jane", parent="harry") #adding nodes
tree.create_node("Bill", "bill", parent="harry")
tree.create_node("Diane", "diane", parent="jane")
tree.create_node("Mary", "mary", parent="diane")
tree.create_node("Mark", "mark", parent="jane")
tree.show()
Harry
├── Bill
└── Jane
├── Diane
│ └── Mary
└── Mark
如果有人需要一个更简单的方法,树只是一个递归嵌套的列表(因为set是不可哈希的):
[root, [child_1, [[child_11, []], [child_12, []]], [child_2, []]]]
每个分支都是一对:[object, [children]] 每个叶子是一对:[object, []]
但是如果你需要一个带有方法的类,你可以使用任何树。
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