从Python中的邻接表构建菜单树

Question

Consider a basic adjacency list; a list of nodes represent by a Node class, with properties id , parent_id , and name . The parent_id of top-level nodes = None.

What would be a Pythonic way of transforming the list into an un-ordered html menu tree, eg:

• node name
• node name
  • sub-node name
  • sub-node name

Answer 1

Assuming you've got something like this:

data = [

    { 'id': 1, 'parent_id': 2, 'name': "Node1" },
    { 'id': 2, 'parent_id': 5, 'name': "Node2" },
    { 'id': 3, 'parent_id': 0, 'name': "Node3" },
    { 'id': 4, 'parent_id': 5, 'name': "Node4" },
    { 'id': 5, 'parent_id': 0, 'name': "Node5" },
    { 'id': 6, 'parent_id': 3, 'name': "Node6" },
    { 'id': 7, 'parent_id': 3, 'name': "Node7" },
    { 'id': 8, 'parent_id': 0, 'name': "Node8" },
    { 'id': 9, 'parent_id': 1, 'name': "Node9" }
]

This function iterates through the list and creates the tree, collecting children of each node is the sub list:

def list_to_tree(data):
    out = { 
        0: { 'id': 0, 'parent_id': 0, 'name': "Root node", 'sub': [] }
    }

    for p in data:
        out.setdefault(p['parent_id'], { 'sub': [] })
        out.setdefault(p['id'], { 'sub': [] })
        out[p['id']].update(p)
        out[p['parent_id']]['sub'].append(out[p['id']])

    return out[0]

Example:

tree = list_to_tree(data)
import pprint
pprint.pprint(tree)

If parent ids are None's (not 0's), modify the function like this:

def list_to_tree(data):
    out = {
        'root': { 'id': 0, 'parent_id': 0, 'name': "Root node", 'sub': [] }
    }

    for p in data:
        pid = p['parent_id'] or 'root'
        out.setdefault(pid, { 'sub': [] })
        out.setdefault(p['id'], { 'sub': [] })
        out[p['id']].update(p)
        out[pid]['sub'].append(out[p['id']])

    return out['root']
    # or return out['root']['sub'] to return the list of root nodes

Answer 2

This is how I ended up implementing it- @thg435's way is elegant, but builds a list of dictionaries to print. This one will print an actual HTML UL menu tree:

nodes = [ 
{ 'id':1, 'parent_id':None, 'name':'a' },
{ 'id':2, 'parent_id':None, 'name':'b' },
{ 'id':3, 'parent_id':2, 'name':'c' },
{ 'id':4, 'parent_id':2, 'name':'d' },
{ 'id':5, 'parent_id':4, 'name':'e' },
{ 'id':6, 'parent_id':None, 'name':'f' }
]

output = ''

def build_node(node):
    global output
    output += '<li><a>'+node['name']+'</a>'
    build_nodes(node['id']
    output += '</li>'

def build_nodes(node_parent_id):
    global output
    subnodes = [node for node in nodes if node['parent_id'] == node_parent_id]
    if len(subnodes) > 0 : 
        output += '<ul>'
        [build_node(subnode) for subnode in subnodes]
        output += '</ul>'

build_nodes(None) # Pass in None as a parent id to start with top level nodes

print output

You can see it here: http://ideone.com/34RT4

Mine uses recursion (cool) and a global output string (not cool)

Someone could surely improve on this, but it's working for me right now..