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
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。
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()
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。
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]
并没有内置树,但是可以通过从List继承Node类型并编写遍历方法来轻松地构造一个树。如果你这样做,我发现平分法很有用。
您还可以浏览PyPi上的许多实现。
如果我没记错的话,Python标准库不包含树数据结构,原因和。net基类库不包含树数据结构是一样的:内存的局部性降低了,导致缓存丢失更多。在现代处理器上,将大量内存放入缓存通常会更快,而“指针丰富”的数据结构会抵消这种好处。
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