如何在Python中实现通用树?这有内置的数据结构吗?
当前回答
bigtree是一个Python树实现,集成了Python列表、字典和pandas DataFrame。它是python式的,易于学习,并可扩展到许多类型的工作流。
bigtree有很多组成部分,即
从列表、字典和熊猫数据框架构建树 遍历树 修改树(移位/复制节点) 搜索树 辅助方法(克隆树,修剪树,获取两个树之间的差异) 导出树(打印到控制台,导出树到字典,熊猫数据框架,图像等) 其他树结构:二叉树! 其他图结构:有向无环图(dag)!
我还能说什么呢……是的,这也是有据可查的。
一些例子:
from bigtree import list_to_tree, tree_to_dict, tree_to_dot
# Create tree from list, print tree
root = list_to_tree(["a/b/d", "a/c"])
print_tree(root)
# a
# ├── b
# │ └── d
# └── c
# Query tree
root.children
# (Node(/a/b, ), Node(/a/c, ))
# Export tree to dictionary / image
tree_to_dict(root)
# {
# '/a': {'name': 'a'},
# '/a/b': {'name': 'b'},
# '/a/b/d': {'name': 'd'},
# '/a/c': {'name': 'c'}
# }
graph = tree_to_dot(root, node_colour="gold")
graph.write_png("tree.png")
来源/免责声明:我是bigtree的创造者;)
其他回答
如果您想要创建树数据结构,那么首先必须创建treeElement对象。如果您创建了treeElement对象,那么您可以决定树的行为。
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)
你可以从对象中添加/删除任何元素,但要使结构为完整的。
我已经在我的网站https://web.archive.org/web/20120723175438/www.quesucede.com/page/show/id/python_3_tree_implementation上发布了一个Python 3树的实现
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)
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。
Greg Hewgill的回答很好,但如果你每层需要更多的节点,你可以使用列表|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建立它: 例:
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
这应该足够让你开始思考如何让它工作了
您可以使用Python中的dataclasses模块创建Tree数据结构。
iter方法可用于使树可迭代,允许您通过改变yield语句的顺序来遍历树。
contains方法可用于检查树中是否存在特定值。
from dataclasses import dataclass
# A
# / \
# B C
# / \ \
# D E F
# / \
# G H
@dataclass
class Node:
data: str
left: Node = None
right: Node = None
def __iter__(self):
if self.left:
yield from self.left
yield self
if self.right:
yield from self.right
def __contains__(self, other):
for node in self:
if node.data == other:
return True
return False
t = Node(
'A',
Node(
'B',
Node(
'D',
Node('G'),
Node('H'),
),
Node('E'),
),
Node(
'C',
right=Node('F'),
),
)
assert ('A' in t) is True
assert ('I' in t) is not True
for node in t:
print(node.data, ' -> ', end='')
# G -> D -> H -> B -> E -> A -> C -> F ->
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