Python将字典的平面列表转换为层次树

Question

我在尝试转换以下列表时遇到问题：

lst = [
    {"id": 0, "job": "CEO", "ManagerID": 0, "name": "John Smith"},
    {"id": 1, "job": "Medical Manager", "ManagerID": 0, "name": "Medic 1"},
    {"id": 2, "job": "Medical Assist", "ManagerID": 1, "name": "Medic 2"},
    {"id": 3, "job": "ICT Manager", "ManagerID": 0, "name": "ICT 1"},
    {"id": 4, "job": "ICT Assist", "ManagerID": 3, "name": "ICT 2"},
    {"id": 5, "job": "ICT Junior", "ManagerID": 4, "name": "ICT 3"}
]

进入格式如

output = [
    {"id": 0, "job": "CEO", "ManagerID": 0, "name": "John Smith", "children" : [
        { "id":1, "job": "Medical Manager", "name": "Medic 1", "children" : [
            {"id": 2, "job": "Medical Assist", "name": "Medic 2"}
            ]
        },
        {"id": 3, "job": "ICT Manager", "name": "ICT 1", "children":[
            {"id": 4, "job": "ICT Assist", "name": "ICT 2", "children" : [
                {"id": 5, "job": "ICT Junior", "name": "ICT 3"}
            ]}
        ]}
    ],
}]

如果有一个根节点（ManagerID = 0），其他所有节点都会分支。

我试图从另一个问题改编代码，但我无法生成这种所需的格式

我一直在使用的代码如下，但这仍然有父节点的重复

classes = [] #everyones id
for item in lst:
    name = item['id']
    if name not in classes:
        classes.append(name)

treenodes = {}
root_node = None

for item in lst: # Create  tree nodes
    item['children'] = []
    name = item['id']
    treenodes[name] = item
    parent = item['ManagerID']
    if parent not in classes: # parent is root node, create
        if parent not in treenodes:
            node = {}
            node['ManagerID'] = 0 #set manager to root
            node['children'] = []
            node['id'] = parent
            root_node = node
            treenodes[parent] = node

# Connect parents and children
for item in lst: # Create  tree nodes
    parent = item['ManagerID']
    parent_node = treenodes[parent]
    parent_node['children'].append(item)

output = treenodes

非常感谢任何帮助。

Answer 1

这是用于构建层次结构的递归版本。

递归版本

from pprint import pprint


def to_lookup(employees):
    employee_lookup = dict()
    for employee in employees:
        if employee["id"] != employee["ManagerID"]:
            manager_id = employee["ManagerID"]
            children = employee_lookup.get(manager_id)
            if not children:
                children = employee_lookup[manager_id] = list()
            children.append(employee.copy())
        else:
            manager = employee.copy()
    return manager, employee_lookup


def build_hierarchy(manager, employee_lookup):
    employees = employee_lookup.get(manager["id"], list())
    for employee in employees:
        build_hierarchy(employee, employee_lookup)
    if employees:
        manager['children'] = employees
    return manager


employees = [
    {"id": 0, "job": "CEO", "ManagerID": 0, "name": "John Smith"},
    {"id": 1, "job": "Medical Manager", "ManagerID": 0, "name": "Medic 1"},
    {"id": 2, "job": "Medical Assist", "ManagerID": 1, "name": "Medic 2"},
    {"id": 3, "job": "ICT Manager", "ManagerID": 0, "name": "ICT 1"},
    {"id": 4, "job": "ICT Assist", "ManagerID": 3, "name": "ICT 2"},
    {"id": 5, "job": "ICT Junior", "ManagerID": 4, "name": "ICT 3"}
]

manager, employee_lookup = to_lookup(employees)
hierarchy = build_hierarchy(manager, employee_lookup)
pprint(hierarchy)

输出

{'ManagerID': 0,
 'children': [{'ManagerID': 0,
               'children': [{'ManagerID': 1,
                             'id': 2,
                             'job': 'Medical Assist',
                             'name': 'Medic 2'}],
               'id': 1,
               'job': 'Medical Manager',
               'name': 'Medic 1'},
              {'ManagerID': 0,
               'children': [{'ManagerID': 3,
                             'children': [{'ManagerID': 4,
                                           'id': 5,
                                           'job': 'ICT Junior',
                                           'name': 'ICT 3'}],
                             'id': 4,
                             'job': 'ICT Assist',
                             'name': 'ICT 2'}],
               'id': 3,
               'job': 'ICT Manager',
               'name': 'ICT 1'}],
 'id': 0,
 'job': 'CEO',
 'name': 'John Smith'}

性能测试

hierarchy_size = 2000000

employees = [
    {"id": 0, "ManagerID": 0},
]
for idx in range(1, hierarchy_size):
    manager_id = random.randint(0, idx - 1)
    employees.append({"id": idx, "ManagerID": manager_id})

start = datetime.datetime.now()

manager, employee_lookup = to_lookup(employees)
hierarchy = build_hierarchy(manager, employee_lookup)

print(datetime.datetime.now() - start)