【发布时间】：2011-01-22 10:00:37
【问题描述】：
我正在尝试构建一棵通用树。
Python中有没有内置的数据结构来实现它？
【问题讨论】：
标签： python data-structures tree
【发布时间】：2011-01-22 10:00:37
【问题描述】：
我推荐anytree（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 还有一个强大的 API：
- 简单的树创建
- 简单的树修改
- 预排序树迭代
- 后序树迭代
- 解析相对和绝对节点路径
- 从一个节点走到另一个节点。
- 树渲染（见上例）
- 节点附加/分离连接
【讨论】：
-
只是最好的答案，其他人正在重新发明轮子。
-
在回答中披露您是您推荐的软件包的作者是一种很好的形式。
-
@c0fec0de 我爱你！！！！这个库很棒，甚至还有可视化功能
-
@Ondrej 其他答案是依赖较少的，最初的问题确实询问了内置数据结构。虽然
anytree可能是一个很棒的库，但这是一个 python 问题，而不是 Node.js 问题。 -
我通过谷歌找到了这个答案。这个图书馆真的很不错。我特别喜欢使用 mixin 类制作任何对象的树的能力！
Python 没有 Java 那样广泛的“内置”数据结构。但是，因为 Python 是动态的，所以很容易创建通用树。例如，二叉树可能是：
class Tree:
def __init__(self):
self.left = None
self.right = None
self.data = None
你可以这样使用它：
root = Tree()
root.data = "root"
root.left = Tree()
root.left.data = "left"
root.right = Tree()
root.right.data = "right"
如果每个节点需要任意数量的子节点，则使用子节点列表：
class Tree:
def __init__(self, data):
self.children = []
self.data = data
left = Tree("left")
middle = Tree("middle")
right = Tree("right")
root = Tree("root")
root.children = [left, middle, right]
【讨论】：
-
这并没有真正解释如何制作一个有用的树实现。
-
问题是用Python3标记的，那么不需要从对象中派生
class Tree -
@cfi 从
object派生有时只是一个指导原则：如果一个类没有继承自其他基类，则显式继承自对象。这也适用于嵌套类。 请参阅 Google Python Style Guide -
@platzhirsch：请完整阅读并引用该指南：Google 明确指出这是 Python 2 代码按预期工作所必需的，并建议这样做以提高与 Py3 的兼容性。这里我们谈论的是 Py3 代码。无需进行额外的传统打字。
-
这是一棵二叉树，而不是要求的一般树。
通用树是具有零个或多个子节点的节点，每个子节点都是正确的（树）节点。它与二叉树不同，它们是不同的数据结构，尽管两者共享一些术语。
Python 中没有任何用于泛型树的内置数据结构，但可以通过类轻松实现。
class Tree(object):
"Generic tree node."
def __init__(self, name='root', children=None):
self.name = name
self.children = []
if children is not None:
for child in children:
self.add_child(child)
def __repr__(self):
return self.name
def add_child(self, node):
assert isinstance(node, Tree)
self.children.append(node)
# *
# /|\
# 1 2 +
# / \
# 3 4
t = Tree('*', [Tree('1'),
Tree('2'),
Tree('+', [Tree('3'),
Tree('4')])])
【讨论】：
-
太棒了，这也可以很容易地用作图表！我看到的唯一问题是：如何区分左节点和右节点？
-
按儿童索引。在这种情况下，Left 将永远是 children[0]。
你可以试试：
from collections import defaultdict
def tree(): return defaultdict(tree)
users = tree()
users['harold']['username'] = 'hrldcpr'
users['handler']['username'] = 'matthandlersux'
【讨论】：
-
如果您想扩展到任意数量的级别，请检查：stackoverflow.com/a/43237270/511809
class Node:
"""
Class Node
"""
def __init__(self, value):
self.left = None
self.data = value
self.right = None
class Tree:
"""
Class tree will provide a tree as well as utility functions.
"""
def createNode(self, data):
"""
Utility function to create a node.
"""
return Node(data)
def insert(self, node , data):
"""
Insert function will insert a node into tree.
Duplicate keys are not allowed.
"""
#if tree is empty , return a root node
if node is None:
return self.createNode(data)
# if data is smaller than parent , insert it into left side
if data < node.data:
node.left = self.insert(node.left, data)
elif data > node.data:
node.right = self.insert(node.right, data)
return node
def search(self, node, data):
"""
Search function will search a node into tree.
"""
# if root is None or root is the search data.
if node is None or node.data == data:
return node
if node.data < data:
return self.search(node.right, data)
else:
return self.search(node.left, data)
def deleteNode(self,node,data):
"""
Delete function will delete a node into tree.
Not complete , may need some more scenarion that we can handle
Now it is handling only leaf.
"""
# Check if tree is empty.
if node is None:
return None
# searching key into BST.
if data < node.data:
node.left = self.deleteNode(node.left, data)
elif data > node.data:
node.right = self.deleteNode(node.right, data)
else: # reach to the node that need to delete from BST.
if node.left is None and node.right is None:
del node
if node.left == None:
temp = node.right
del node
return temp
elif node.right == None:
temp = node.left
del node
return temp
return node
def traverseInorder(self, root):
"""
traverse function will print all the node in the tree.
"""
if root is not None:
self.traverseInorder(root.left)
print(root.data)
self.traverseInorder(root.right)
def traversePreorder(self, root):
"""
traverse function will print all the node in the tree.
"""
if root is not None:
print(root.data)
self.traversePreorder(root.left)
self.traversePreorder(root.right)
def traversePostorder(self, root):
"""
traverse function will print all the node in the tree.
"""
if root is not None:
self.traversePostorder(root.left)
self.traversePostorder(root.right)
print(root.data)
def main():
root = None
tree = Tree()
root = tree.insert(root, 10)
print(root)
tree.insert(root, 20)
tree.insert(root, 30)
tree.insert(root, 40)
tree.insert(root, 70)
tree.insert(root, 60)
tree.insert(root, 80)
print("Traverse Inorder")
tree.traverseInorder(root)
print("Traverse Preorder")
tree.traversePreorder(root)
print("Traverse Postorder")
tree.traversePostorder(root)
if __name__ == "__main__":
main()
【讨论】：
-
你能添加一些注释来介绍你的代码和你的实现吗？
-
感谢使用实用函数的完整二叉树实现。由于它是 Python 2，我为需要 Python 3 版本的人创建了 Binary Tree implementation (Py3) 的要点。
【讨论】：
-
仅供参考：互联网上充斥着对 Boost 的仇恨。显然，处理它应该是一个巨大的痛苦，特别是因为对它的支持已经停止。所以我建议远离那个
-
谢谢。我个人没有遇到任何麻烦，但我不想误导，所以我删除了该引用。
我将根树实现为字典{child:parent}。例如，对于根节点0，一棵树可能看起来像这样：
tree={1:0, 2:0, 3:1, 4:2, 5:3}
这种结构很容易沿着从任何节点到根的路径向上移动，这与我正在处理的问题相关。
【讨论】：
-
这是我考虑的方式，直到我看到答案。虽然由于一棵树是有两个孩子的父母，如果你想下去，你可以做
{parent:[leftchild,rightchild]}。 -
另一种方法是使用列表的列表，其中列表中的第一个（或更多）元素是节点值，以下嵌套的两个列表表示其左右子树（或更多用于 n- ary 树）。
Greg Hewgill 的回答很棒，但如果您需要每个级别的更多节点，您可以使用列表|字典来创建它们：然后使用方法按名称或顺序（如 id）访问它们
class node(object):
def __init__(self):
self.name=None
self.node=[]
self.otherInfo = None
self.prev=None
def nex(self,child):
"Gets a node by number"
return self.node[child]
def prev(self):
return self.prev
def goto(self,data):
"Gets the node by name"
for child in range(0,len(self.node)):
if(self.node[child].name==data):
return self.node[child]
def add(self):
node1=node()
self.node.append(node1)
node1.prev=self
return node1
现在只需创建一个根并构建它： 例如：
tree=node() #create a node
tree.name="root" #name it root
tree.otherInfo="blue" #or what ever
tree=tree.add() #add a node to the root
tree.name="node1" #name it
root
/
child1
tree=tree.add()
tree.name="grandchild1"
root
/
child1
/
grandchild1
tree=tree.prev()
tree=tree.add()
tree.name="gchild2"
root
/
child1
/ \
grandchild1 gchild2
tree=tree.prev()
tree=tree.prev()
tree=tree.add()
tree=tree.name="child2"
root
/ \
child1 child2
/ \
grandchild1 gchild2
tree=tree.prev()
tree=tree.goto("child1") or tree=tree.nex(0)
tree.name="changed"
root
/ \
changed child2
/ \
grandchild1 gchild2
这应该足以让您开始弄清楚如何进行这项工作
【讨论】：
-
此答案中缺少某些内容，过去 2 天我一直在尝试此解决方案，我认为您在对象添加方法中有一些逻辑流程。我将提交我对这个问题的答案，请检查一下，如果我能提供帮助，请告诉我。
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
作为字典工作，但提供尽可能多的嵌套字典。 请尝试以下操作：
your_tree = Tree()
your_tree['a']['1']['x'] = '@'
your_tree['a']['1']['y'] = '#'
your_tree['a']['2']['x'] = '$'
your_tree['a']['3'] = '%'
your_tree['b'] = '*'
将提供一个嵌套的 dict ... 确实像一棵树一样工作。
{'a': {'1': {'x': '@', 'y': '#'}, '2': {'x': '$'}, '3': '%'}, 'b': '*'}
...如果您已经有一个字典，它会将每个级别转换为一棵树：
d = {'foo': {'amy': {'what': 'runs'} } }
tree = Tree(d)
print(d['foo']['amy']['what']) # returns 'runs'
d['foo']['amy']['when'] = 'now' # add new branch
通过这种方式，您可以根据需要保持编辑/添加/删除每个字典级别。 用于遍历等的所有 dict 方法仍然适用。
【讨论】：
-
你选择扩展
dict而不是defaultdict有什么原因吗？根据我的测试，扩展defaultdict而不是 dict 然后将self.default_factory = type(self)添加到 init 的顶部应该以相同的方式运行。 -
我可能在这里遗漏了一些东西，你如何浏览这个结构？例如，从孩子到父母或兄弟姐妹似乎很难
如果有人需要更简单的方法来做到这一点，一棵树只是一个递归嵌套的列表（因为集合是不可散列的）：
[root, [child_1, [[child_11, []], [child_12, []]], [child_2, []]]]
每个分支都是一对：[ object, [children] ]
每片叶子都是一对：[ object, [] ]
但是如果你需要一个有方法的类，你可以使用anytree。
【讨论】：
我已经使用嵌套字典实现了树。这很容易做到，并且对我来说非常大的数据集很有效。我在下面发布了一个示例，您可以在Google code查看更多内容
def addBallotToTree(self, tree, ballotIndex, ballot=""):
"""Add one ballot to the tree.
The root of the tree is a dictionary that has as keys the indicies of all
continuing and winning candidates. For each candidate, the value is also
a dictionary, and the keys of that dictionary include "n" and "bi".
tree[c]["n"] is the number of ballots that rank candidate c first.
tree[c]["bi"] is a list of ballot indices where the ballots rank c first.
If candidate c is a winning candidate, then that portion of the tree is
expanded to indicate the breakdown of the subsequently ranked candidates.
In this situation, additional keys are added to the tree[c] dictionary
corresponding to subsequently ranked candidates.
tree[c]["n"] is the number of ballots that rank candidate c first.
tree[c]["bi"] is a list of ballot indices where the ballots rank c first.
tree[c][d]["n"] is the number of ballots that rank c first and d second.
tree[c][d]["bi"] is a list of the corresponding ballot indices.
Where the second ranked candidates is also a winner, then the tree is
expanded to the next level.
Losing candidates are ignored and treated as if they do not appear on the
ballots. For example, tree[c][d]["n"] is the total number of ballots
where candidate c is the first non-losing candidate, c is a winner, and
d is the next non-losing candidate. This will include the following
ballots, where x represents a losing candidate:
[c d]
[x c d]
[c x d]
[x c x x d]
During the count, the tree is dynamically updated as candidates change
their status. The parameter "tree" to this method may be the root of the
tree or may be a sub-tree.
"""
if ballot == "":
# Add the complete ballot to the tree
weight, ballot = self.b.getWeightedBallot(ballotIndex)
else:
# When ballot is not "", we are adding a truncated ballot to the tree,
# because a higher-ranked candidate is a winner.
weight = self.b.getWeight(ballotIndex)
# Get the top choice among candidates still in the running
# Note that we can't use Ballots.getTopChoiceFromWeightedBallot since
# we are looking for the top choice over a truncated ballot.
for c in ballot:
if c in self.continuing | self.winners:
break # c is the top choice so stop
else:
c = None # no candidates left on this ballot
if c is None:
# This will happen if the ballot contains only winning and losing
# candidates. The ballot index will not need to be transferred
# again so it can be thrown away.
return
# Create space if necessary.
if not tree.has_key(c):
tree[c] = {}
tree[c]["n"] = 0
tree[c]["bi"] = []
tree[c]["n"] += weight
if c in self.winners:
# Because candidate is a winner, a portion of the ballot goes to
# the next candidate. Pass on a truncated ballot so that the same
# candidate doesn't get counted twice.
i = ballot.index(c)
ballot2 = ballot[i+1:]
self.addBallotToTree(tree[c], ballotIndex, ballot2)
else:
# Candidate is in continuing so we stop here.
tree[c]["bi"].append(ballotIndex)
【讨论】：
我在我的网站上发布了一个 Python 3 树实现：https://web.archive.org/web/20120723175438/www.quesucede.com/page/show/id/python_3_tree_implementation
代码如下：
import uuid
def sanitize_id(id):
return id.strip().replace(" ", "")
(_ADD, _DELETE, _INSERT) = range(3)
(_ROOT, _DEPTH, _WIDTH) = range(3)
class Node:
def __init__(self, name, identifier=None, expanded=True):
self.__identifier = (str(uuid.uuid1()) if identifier is None else
sanitize_id(str(identifier)))
self.name = name
self.expanded = expanded
self.__bpointer = None
self.__fpointer = []
@property
def identifier(self):
return self.__identifier
@property
def bpointer(self):
return self.__bpointer
@bpointer.setter
def bpointer(self, value):
if value is not None:
self.__bpointer = sanitize_id(value)
@property
def fpointer(self):
return self.__fpointer
def update_fpointer(self, identifier, mode=_ADD):
if mode is _ADD:
self.__fpointer.append(sanitize_id(identifier))
elif mode is _DELETE:
self.__fpointer.remove(sanitize_id(identifier))
elif mode is _INSERT:
self.__fpointer = [sanitize_id(identifier)]
class Tree:
def __init__(self):
self.nodes = []
def get_index(self, position):
for index, node in enumerate(self.nodes):
if node.identifier == position:
break
return index
def create_node(self, name, identifier=None, parent=None):
node = Node(name, identifier)
self.nodes.append(node)
self.__update_fpointer(parent, node.identifier, _ADD)
node.bpointer = parent
return node
def show(self, position, level=_ROOT):
queue = self[position].fpointer
if level == _ROOT:
print("{0} [{1}]".format(self[position].name,
self[position].identifier))
else:
print("\t"*level, "{0} [{1}]".format(self[position].name,
self[position].identifier))
if self[position].expanded:
level += 1
for element in queue:
self.show(element, level) # recursive call
def expand_tree(self, position, mode=_DEPTH):
# Python generator. Loosly based on an algorithm from 'Essential LISP' by
# John R. Anderson, Albert T. Corbett, and Brian J. Reiser, page 239-241
yield position
queue = self[position].fpointer
while queue:
yield queue[0]
expansion = self[queue[0]].fpointer
if mode is _DEPTH:
queue = expansion + queue[1:] # depth-first
elif mode is _WIDTH:
queue = queue[1:] + expansion # width-first
def is_branch(self, position):
return self[position].fpointer
def __update_fpointer(self, position, identifier, mode):
if position is None:
return
else:
self[position].update_fpointer(identifier, mode)
def __update_bpointer(self, position, identifier):
self[position].bpointer = identifier
def __getitem__(self, key):
return self.nodes[self.get_index(key)]
def __setitem__(self, key, item):
self.nodes[self.get_index(key)] = item
def __len__(self):
return len(self.nodes)
def __contains__(self, identifier):
return [node.identifier for node in self.nodes
if node.identifier is identifier]
if __name__ == "__main__":
tree = Tree()
tree.create_node("Harry", "harry") # root node
tree.create_node("Jane", "jane", parent = "harry")
tree.create_node("Bill", "bill", parent = "harry")
tree.create_node("Joe", "joe", parent = "jane")
tree.create_node("Diane", "diane", parent = "jane")
tree.create_node("George", "george", parent = "diane")
tree.create_node("Mary", "mary", parent = "diane")
tree.create_node("Jill", "jill", parent = "george")
tree.create_node("Carol", "carol", parent = "jill")
tree.create_node("Grace", "grace", parent = "bill")
tree.create_node("Mark", "mark", parent = "jane")
print("="*80)
tree.show("harry")
print("="*80)
for node in tree.expand_tree("harry", mode=_WIDTH):
print(node)
print("="*80)
【讨论】：
如果您已经在使用networkx 库，那么您可以使用它来实现一棵树。否则，您可以尝试其他答案之一。
NetworkX 是一个用于创建、操作和研究的 Python 包 复杂网络的结构、动力学和功能。
因为“树”是（通常有根的）连通无环图的另一个术语，这些在 NetworkX 中称为“树状结构”。
您可能想要实现一个平面树（也称为有序树），其中每个兄弟节点都有一个唯一的等级，这通常通过标记节点来完成。
但是，graph 语言看起来与 tree 语言不同，通常使用有向图来“生根”树状结构，因此，虽然有一些非常酷的功能和相应的可视化可用，如果您还没有使用 networkx，它可能不是一个理想的选择。
构建树的示例：
import networkx as nx
G = nx.Graph()
G.add_edge('A', 'B')
G.add_edge('B', 'C')
G.add_edge('B', 'D')
G.add_edge('A', 'E')
G.add_edge('E', 'F')
库使每个节点为any hashable object，并且每个节点的子节点数量没有限制。
该库还提供与树相关的graph algorithms 和visualization 功能。
【讨论】：
您好，可以试试itertree（我是作者）。
该包朝着 anytree 包的方向发展，但重点有所不同。在大树（>100000 个项目）上的性能要好得多，并且它处理迭代器以具有有效的过滤机制。
>>>from itertree import *
>>>root=iTree('root')
>>># add some children:
>>>root.append(iTree('Africa',data={'surface':30200000,'inhabitants':1257000000}))
>>>root.append(iTree('Asia', data={'surface': 44600000, 'inhabitants': 4000000000}))
>>>root.append(iTree('America', data={'surface': 42549000, 'inhabitants': 1009000000}))
>>>root.append(iTree('Australia&Oceania', data={'surface': 8600000, 'inhabitants': 36000000}))
>>>root.append(iTree('Europe', data={'surface': 10523000 , 'inhabitants': 746000000}))
>>># you might use __iadd__ operator for adding too:
>>>root+=iTree('Antarktika', data={'surface': 14000000, 'inhabitants': 1100})
>>># for building next level we select per index:
>>>root[0]+=iTree('Ghana',data={'surface':238537,'inhabitants':30950000})
>>>root[0]+=iTree('Niger', data={'surface': 1267000, 'inhabitants': 23300000})
>>>root[1]+=iTree('China', data={'surface': 9596961, 'inhabitants': 1411780000})
>>>root[1]+=iTree('India', data={'surface': 3287263, 'inhabitants': 1380004000})
>>>root[2]+=iTree('Canada', data={'type': 'country', 'surface': 9984670, 'inhabitants': 38008005})
>>>root[2]+=iTree('Mexico', data={'surface': 1972550, 'inhabitants': 127600000 })
>>># extend multiple items:
>>>root[3].extend([iTree('Australia', data={'surface': 7688287, 'inhabitants': 25700000 }), iTree('New Zealand', data={'surface': 269652, 'inhabitants': 4900000 })])
>>>root[4]+=iTree('France', data={'surface': 632733, 'inhabitants': 67400000 }))
>>># select parent per TagIdx - remember in itertree you might put items with same tag multiple times:
>>>root[TagIdx('Europe'0)]+=iTree('Finland', data={'surface': 338465, 'inhabitants': 5536146 })
可以渲染创建的树：
>>>root.render()
iTree('root')
└──iTree('Africa', data=iTData({'surface': 30200000, 'inhabitants': 1257000000}))
└──iTree('Ghana', data=iTData({'surface': 238537, 'inhabitants': 30950000}))
└──iTree('Niger', data=iTData({'surface': 1267000, 'inhabitants': 23300000}))
└──iTree('Asia', data=iTData({'surface': 44600000, 'inhabitants': 4000000000}))
└──iTree('China', data=iTData({'surface': 9596961, 'inhabitants': 1411780000}))
└──iTree('India', data=iTData({'surface': 3287263, 'inhabitants': 1380004000}))
└──iTree('America', data=iTData({'surface': 42549000, 'inhabitants': 1009000000}))
└──iTree('Canada', data=iTData({'surface': 9984670, 'inhabitants': 38008005}))
└──iTree('Mexico', data=iTData({'surface': 1972550, 'inhabitants': 127600000}))
└──iTree('Australia&Oceania', data=iTData({'surface': 8600000, 'inhabitants': 36000000}))
└──iTree('Australia', data=iTData({'surface': 7688287, 'inhabitants': 25700000}))
└──iTree('New Zealand', data=iTData({'surface': 269652, 'inhabitants': 4900000}))
└──iTree('Europe', data=iTData({'surface': 10523000, 'inhabitants': 746000000}))
└──iTree('France', data=iTData({'surface': 632733, 'inhabitants': 67400000}))
└──iTree('Finland', data=iTData({'surface': 338465, 'inhabitants': 5536146}))
└──iTree('Antarktika', data=iTData({'surface': 14000000, 'inhabitants': 1100}))
例如过滤可以这样完成：
>>>item_filter = Filter.iTFilterData(data_key='inhabitants', data_value=iTInterval(0, 20000000))
>>>iterator=root.iter_all(item_filter=item_filter)
>>>for i in iterator:
>>> print(i)
iTree("'New Zealand'", data=iTData({'surface': 269652, 'inhabitants': 4900000}), subtree=[])
iTree("'Finland'", data=iTData({'surface': 338465, 'inhabitants': 5536146}), subtree=[])
iTree("'Antarktika'", data=iTData({'surface': 14000000, 'inhabitants': 1100}), subtree=[])
【讨论】：
另一个松散地基于Bruno's answer的树实现：
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
以及如何使用它的示例：
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)
应该输出：
根节点 // \ 子节点 0 子节点 1 子节点 2 | / \ 孙子 1.0 孙子 2.0 孙子 2.1【讨论】：
-
干净简单。您在
__str__函数中缺少一些定义，即accumulation和add。但是树实现本身效果很好。 -
应该是
itertools.accumulate和operator.add，抱歉省略了。
如果要创建树数据结构，则首先必须创建 treeElement 对象。如果您创建 treeElement 对象，那么您可以决定您的树的行为方式。
下面是 TreeElement 类：
class TreeElement (object):
def __init__(self):
self.elementName = None
self.element = []
self.previous = None
self.elementScore = None
self.elementParent = None
self.elementPath = []
self.treeLevel = 0
def goto(self, data):
for child in range(0, len(self.element)):
if (self.element[child].elementName == data):
return self.element[child]
def add(self):
single_element = TreeElement()
single_element.elementName = self.elementName
single_element.previous = self.elementParent
single_element.elementScore = self.elementScore
single_element.elementPath = self.elementPath
single_element.treeLevel = self.treeLevel
self.element.append(single_element)
return single_element
现在，我们必须使用这个元素来创建树，在这个例子中我使用的是 A* 树。
class AStarAgent(Agent):
# Initialization Function: Called one time when the game starts
def registerInitialState(self, state):
return;
# GetAction Function: Called with every frame
def getAction(self, state):
# Sorting function for the queue
def sortByHeuristic(each_element):
if each_element.elementScore:
individual_score = each_element.elementScore[0][0] + each_element.treeLevel
else:
individual_score = admissibleHeuristic(each_element)
return individual_score
# check the game is over or not
if state.isWin():
print('Job is done')
return Directions.STOP
elif state.isLose():
print('you lost')
return Directions.STOP
# Create empty list for the next states
astar_queue = []
astar_leaf_queue = []
astar_tree_level = 0
parent_tree_level = 0
# Create Tree from the give node element
astar_tree = TreeElement()
astar_tree.elementName = state
astar_tree.treeLevel = astar_tree_level
astar_tree = astar_tree.add()
# Add first element into the queue
astar_queue.append(astar_tree)
# Traverse all the elements of the queue
while astar_queue:
# Sort the element from the queue
if len(astar_queue) > 1:
astar_queue.sort(key=lambda x: sortByHeuristic(x))
# Get the first node from the queue
astar_child_object = astar_queue.pop(0)
astar_child_state = astar_child_object.elementName
# get all legal actions for the current node
current_actions = astar_child_state.getLegalPacmanActions()
if current_actions:
# get all the successor state for these actions
for action in current_actions:
# Get the successor of the current node
next_state = astar_child_state.generatePacmanSuccessor(action)
if next_state:
# evaluate the successor states using scoreEvaluation heuristic
element_scored = [(admissibleHeuristic(next_state), action)]
# Increase the level for the child
parent_tree_level = astar_tree.goto(astar_child_state)
if parent_tree_level:
astar_tree_level = parent_tree_level.treeLevel + 1
else:
astar_tree_level += 1
# create tree for the finding the data
astar_tree.elementName = next_state
astar_tree.elementParent = astar_child_state
astar_tree.elementScore = element_scored
astar_tree.elementPath.append(astar_child_state)
astar_tree.treeLevel = astar_tree_level
astar_object = astar_tree.add()
# If the state exists then add that to the queue
astar_queue.append(astar_object)
else:
# Update the value leaf into the queue
astar_leaf_state = astar_tree.goto(astar_child_state)
astar_leaf_queue.append(astar_leaf_state)
您可以从对象中添加/删除任何元素，但要确保结构完整。
【讨论】：