如何在Python中实现通用树?这有内置的数据结构吗?


当前回答

我推荐任何树(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

其他回答

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;)

我将根树实现为字典{child:parent}。比如根节点为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。

如果您已经在使用networkx库,那么您可以使用它实现一个树。

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,“扎根”树的方法通常是使用有向图,因此,虽然有一些非常酷的功能和相应的可视化可用,但如果你还没有使用networkx,它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')

该库允许每个节点是任何可哈希对象,并且不限制每个节点拥有的子节点的数量。

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。