如何在Python中实现通用树?这有内置的数据结构吗?
当前回答
Python不像Java那样具有相当广泛的“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]
其他回答
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。
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
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。
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。
Python不像Java那样具有相当广泛的“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]
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