如何在Python中实现树?

我试图构造一个一般树。

Python中有任何内置的数据结构来实现它吗?

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小开

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，我发现平分很有用。

你也可以浏览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。

小开

Python没有相当广泛的“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]
小开

我使用嵌套字典实现了树。这很容易做到，而且对我来说，它在相当大的数据集上很有效。我在下面发布了一个示例，你可以在谷歌代码看到更多

  def addBallotToTree(self, tree, ballotIndex, ballot=""):
"""Add one ballot to the tree.


The root of the tree is a dictionary that has as keys the indicies of all
continuing and winning candidates.  For each candidate, the value is also
a dictionary, and the keys of that dictionary include "n" and "bi".
tree[c]["n"] is the number of ballots that rank candidate c first.
tree[c]["bi"] is a list of ballot indices where the ballots rank c first.


If candidate c is a winning candidate, then that portion of the tree is
expanded to indicate the breakdown of the subsequently ranked candidates.
In this situation, additional keys are added to the tree[c] dictionary
corresponding to subsequently ranked candidates.
tree[c]["n"] is the number of ballots that rank candidate c first.
tree[c]["bi"] is a list of ballot indices where the ballots rank c first.
tree[c][d]["n"] is the number of ballots that rank c first and d second.
tree[c][d]["bi"] is a list of the corresponding ballot indices.


Where the second ranked candidates is also a winner, then the tree is
expanded to the next level.


Losing candidates are ignored and treated as if they do not appear on the
ballots.  For example, tree[c][d]["n"] is the total number of ballots
where candidate c is the first non-losing candidate, c is a winner, and
d is the next non-losing candidate.  This will include the following
ballots, where x represents a losing candidate:
[c d]
[x c d]
[c x d]
[x c x x d]


During the count, the tree is dynamically updated as candidates change
their status.  The parameter "tree" to this method may be the root of the
tree or may be a sub-tree.
"""


if ballot == "":
# Add the complete ballot to the tree
weight, ballot = self.b.getWeightedBallot(ballotIndex)
else:
# When ballot is not "", we are adding a truncated ballot to the tree,
# because a higher-ranked candidate is a winner.
weight = self.b.getWeight(ballotIndex)


# Get the top choice among candidates still in the running
# Note that we can't use Ballots.getTopChoiceFromWeightedBallot since
# we are looking for the top choice over a truncated ballot.
for c in ballot:
if c in self.continuing | self.winners:
break # c is the top choice so stop
else:
c = None # no candidates left on this ballot


if c is None:
# This will happen if the ballot contains only winning and losing
# candidates.  The ballot index will not need to be transferred
# again so it can be thrown away.
return


# Create space if necessary.
if not tree.has_key(c):
tree[c] = {}
tree[c]["n"] = 0
tree[c]["bi"] = []


tree[c]["n"] += weight


if c in self.winners:
# Because candidate is a winner, a portion of the ballot goes to
# the next candidate.  Pass on a truncated ballot so that the same
# candidate doesn't get counted twice.
i = ballot.index(c)
ballot2 = ballot[i+1:]
self.addBallotToTree(tree[c], ballotIndex, ballot2)
else:
# Candidate is in continuing so we stop here.
tree[c]["bi"].append(ballotIndex)
小开

我在我的网站上发布了一个Python 3树的实现:https://web.archive.org/web/20120723175438/www.quesucede.com/page/show/id/python_3_tree_implementation

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)
小开

你可以试试:

from collections import defaultdict
def tree(): return defaultdict(tree)
users = tree()
users['harold']['username'] = 'hrldcpr'
users['handler']['username'] = 'matthandlersux'

如上所示:https://gist.github.com/2012250

小开

我实现了一个根树作为字典{child:parent}。例如，对于根节点0，树可能是这样的:

tree={1:0, 2:0, 3:1, 4:2, 5:3}

这种结构使得沿着一条路径从任意节点向上到根结点非常容易，这与我正在处理的问题有关。

小开

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
现在只需要创建一个根并建立它: 例:< / p > 
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

小开

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')])])
小开 
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()
小开

我推荐anytree(我是作者)。

例子:

from anytree import Node, RenderTree


udo = Node("Udo")
marc = Node("Marc", parent=udo)
lian = Node("Lian", parent=marc)
dan = Node("Dan", parent=udo)
jet = Node("Jet", parent=dan)
jan = Node("Jan", parent=dan)
joe = Node("Joe", parent=dan)


print(udo)
Node('/Udo')
print(joe)
Node('/Udo/Dan/Joe')


for pre, fill, node in RenderTree(udo):
print("%s%s" % (pre, node.name))
Udo
├── Marc
│   └── Lian
└── Dan
├── Jet
├── Jan
└── Joe


print(dan.children)
(Node('/Udo/Dan/Jet'), Node('/Udo/Dan/Jan'), Node('/Udo/Dan/Joe'))

anytree也有一个强大的API:

  • 简单的树创建
  • 简单树修改
  • 预序树迭代
  • 后序树迭代
  • 解析相对节点路径和绝对节点路径
  • 从一个节点移动到另一个节点。
  • 树渲染(参见上面的例子)
  • 节点连接/分离连接
小开 
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，但是提供了你想要的任意数量的嵌套字典。 

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，你可以随心所欲地编辑/添加/删除每个字典级别。 遍历等的所有dict方法仍然适用
小开

另一个松散地基于布鲁诺的回答的树实现:

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
小开

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。

小开

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] ]
并且每个叶子是一对:[ object, [] ]

但是如果你需要一个带有方法的类，你可以使用任何树。

小开

如果你已经在使用networkx库，那么你可以使用它来实现一个树。

NetworkX是一个用于创建、操作和研究的Python包 复杂网络的结构、动力学和功能。

因为“树”是(通常根)连接无环图的另一个术语，这些在NetworkX中被称为“树状图”。

你可能想要实现一个平面树(又名有序树)，其中每个兄弟姐妹都有一个唯一的秩，这通常通过标记节点来完成。

然而，图语言看起来不同于树语言，并且“根”树的方法通常是使用有向图来完成的，因此，虽然有一些非常酷的功能和相应的数据可视化可用，但如果你还没有使用networkx，这可能不是一个理想的选择。

一个构建树的例子:

import networkx as nx
G = nx.Graph()
G.add_edge('A', 'B')
G.add_edge('B', 'C')
G.add_edge('B', 'D')
G.add_edge('A', 'E')
G.add_edge('E', 'F')

标准库允许每个节点为任何可哈希对象，并且每个节点的子节点数没有限制。

小开

嗨，你可以给itertree一个尝试(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项)上的性能要好得多，并且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=[])
小开

Treelib也很方便完成这项任务。文档可以在treelib中找到。

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
小开

bigtree是一个Python树实现，它集成了Python列表、字典和pandas数据框架。它是python的，使其简单易学和可扩展到许多类型的工作流。

bigtree有不同的组件，即

  • 从列表，字典和熊猫数据框架构造树
  • 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!
  • 其他结构图:有向无环图(dag)!

我还能说什么呢……啊，是的，它也是证据确凿的。

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")

enter image description here

来源/免责声明:我是bigtree的创建者;)


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