Python將字典的平面列表轉換為層次樹

Question

我在嘗試轉換以下列表時遇到了麻煩：

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"}
]

變成喜歡的格式

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"}
            ]}
        ]}
    ],
}]

是否有一個根節點（ManagerID = 0），其他所有分支都分支了。

我嘗試改寫其他問題的代碼，但無法生成此所需格式

我一直在使用的代碼如下，但這仍然重復了父節點

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

任何幫助，不勝感激。

Answer 1

這是用於構建層次結構的遞歸版本。

遞歸版本

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)

產量

{'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'}

性能測試

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)

Answer 2

您的代碼實際上是有效的，但是您需要使用treenodes[0]條目（CEO）。 treenodes中其余的鍵值對僅用於簿記，以便輕松找到給定員工條目的給定經理。

如果您不能指望0作為根節點的ID，那么您可以利用CEO被標記為自我管理這一事實； 根節點是管理員ID指向自己的ID的節點。 更為常見的情況是根節點根本沒有父ID。

您還將CEO添加到了自己的children列表中（CEO的經理ID是他們自己的ID），因此您在樹中有一個遞歸引用。

您找到的代碼不是最清晰或最有效的。 我會從id到復制的對象建立一個字典（這樣您的原始lst字典就不會改變），然后遍歷該結構並將條目添加到其經理id條目中。 我正在使用“根節點自引用”規則（因此管理器ID等於他們自己的ID）：

employees = {}
managers = set()
root_id = None
for emp in lst:
    id, mid = emp['id'], emp['ManagerID']
    # create a copy of emp, and add a "children" list
    employees[id] = {**emp, 'children': []}
    managers.add(mid)
    if id == mid:
        # the root of the tree references itself as the manager
        root_id = id

# add empty manager entries for missing manager IDs, reporting to root ID.
for id in managers - employees.keys():
    employees[id] = {
        'id': id, 'ManagerID': root_id, 'children': [],
        'job': None, 'name': None
    }

for id, emp in employees.items():
    manager = employees[emp.pop('ManagerID')]
    if id != root_id:  # don't add the root to anything
        manager['children'].append(emp)

output = employees[root_id]

上面的代碼使用一組來跟蹤看到過哪些經理ID，因此您可以輕松添加丟失的經理條目（在這種情況下，報告給CEO）。

對於您的輸入，將產生：

{'id': 0, 'job': 'CEO', 'name': 'John Smith', 'children':
    [{'id': 1, 'job': 'Medical Manager', 'name': 'Medic 1', 'children':
        [{'id': 2, 'job': 'Medical Assist', 'name': 'Medic 2', 'children': []}],
     },
     {'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', 'children': []}]
         }]
     }]
}