如何在Python中实现通用树?这有内置的数据结构吗?


当前回答

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。

其他回答

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

泛型树是一个具有零个或多个子节点的节点,每个子节点都是一个合适的(树)节点。它与二叉树不同,它们是不同的数据结构,尽管它们都有一些相同的术语。

Python中没有任何用于泛型树的内置数据结构,但很容易通过类实现。

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')])])

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。

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  ->