如何在Python中实现通用树?这有内置的数据结构吗?


当前回答

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

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

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

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

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)

bigtree是一个Python树实现,集成了Python列表、字典和pandas DataFrame。它是python式的,易于学习,并可扩展到许多类型的工作流。

bigtree有很多组成部分,即

从列表、字典和熊猫数据框架构建树 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! 其他图结构:有向无环图(dag)!

我还能说什么呢……是的,这也是有据可查的。

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")

来源/免责声明:我是bigtree的创造者;)

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