从具有重复子树的嵌套列表构造树视图（使用 anytree/treelib）

Question

我有一个如下所示的嵌套列表：

lst = [['a', 'b', 'e'],      # this e is the branch of b
       ['a', 'f', 'e'],      # this e is the branch of f,
       ['a', 'h', 'i i i']]  # string with spaces

我想构建一棵树：

a
├── b
│   └── e
├── f
|   └── e
└── h
    └── i i i

我想使用两个包中的任何一个：treelib and anytree。我已经阅读了很多帖子并尝试了很多不同的方法，但没有成功。

更新：

我想到了下面的方法，但是我现在遇到的问题是

不保证分支的垂直顺序（例如，"b"、"f"、"h"）（当我在一个列表中有很多列表时）。
"e" 作为 "f" 的分支不会出现

from treelib import Node, Tree

# make list flat
lst = sum([i for i in lst], [])

tree = Tree()
tree_dict = {}

# create root node
tree_dict[lst[0]] = tree.create_node(lst[0])

for index, item in enumerate(lst[1:], start=1):
    if item not in tree_dict.keys():
        partent_node = tree_dict[lst[index-1]]
        tree_dict[item] = tree.create_node(item, parent=partent_node)

tree.show()

Answer 1

我调查了 anytree 并得出以下结论：

from anytree import Node, RenderTree

lst = [["a", "b", "c", "e"], ["a", "b", "f"], ["a", "b", "c", "g", "h"], ["a", "i"]]


def list_to_anytree(lst):
    root_name = lst[0][0]
    root_node = Node(root_name)
    nodes = {root_name: root_node}  # keeping a dict of the nodes
    for branch in lst:
        assert branch[0] == root_name
        for parent_name, node_name in zip(branch, branch[1:]):
            node = nodes.setdefault(node_name, Node(node_name))
            parent_node = nodes[parent_name]
            if node.parent is not None:
                assert node.parent.name == parent_name
            else:
                node.parent = parent_node
    return root_node


anytree = list_to_anytree(lst)
for pre, fill, node in RenderTree(anytree):
    print(f"{pre}{node.name}")

这里没有发生太多事情。我只是将您的列表转换为任何树节点（并且 assert 列表表示在执行此操作时有效）。我保留了 nodes.

中已有节点的字典

输出确实是

a
├── b
│   ├── c
│   │   ├── e
│   │   └── g
│   │       └── h
│   └── f
└── i

如果您有多个同名节点，则不能使用上面的dict；您需要从根节点遍历子节点：

def list_to_anytree(lst):
    root_name = lst[0][0]
    root_node = Node(root_name)
    for branch in lst:
        parent_node = root_node
        assert branch[0] == parent_node.name
        for cur_node_name in branch[1:]:
            cur_node = next(
                (node for node in parent_node.children if node.name == cur_node_name),
                None,
            )
            if cur_node is None:
                cur_node = Node(cur_node_name, parent=parent_node)
            parent_node = cur_node
    return root_node

你的例子

lst = [
    ["a", "b", "e"],  # this e is the branch of b
    ["a", "f", "e"],  # this e is the branch of f,
    ["a", "h", "i i i"],
]

anytree = list_to_anytree(lst)
for pre, fill, node in RenderTree(anytree):
    print(f"{pre}{node.name}")

然后给出：

a
├── b
│   └── e
├── f
│   └── e
└── h
    └── i i i

从具有重复子树的嵌套列表构造树视图（使用 anytree/treelib）

constructing a tree view from a nested list with duplicate subtrees (using anytree/treelib)

python

treeview

tree

python-3.x

anytree