如何在Python中实现通用树?这有内置的数据结构吗?
当前回答
另一个基于Bruno回答的树的实现:
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
还有一个如何使用它的例子:
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)
它应该输出:
Root node / / \ Child node 0 Child node 1 Child node 2 | / \ Grandchild 1.0 Grandchild 2.0 Grandchild 2.1
其他回答
嗨,你可以试试itertree(我是作者)。
该包与任何树包的方向相同,但关注点略有不同。在巨大的树(>100000个项目)上的性能要好得多,它处理迭代器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 })
创建的树可以被渲染:
>>>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}))
过滤可以这样做:
>>>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=[])
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。
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。
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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