我需要在 python 中转换分层嵌套字典中的数据列表。结构(父亲 - child )。
这是我的数据。
list_data = [
{
"id": 2,
"father_id": 0,
"desc": "Oficial de Negocios Senior",
"name": "PEDRO MARTIN SOTO ROSALES"
},
{
"id": 4,
"father_id": 2,
"desc": "Ejecutivo comercial",
"name": "Adriana Paredez"
},
{
"id": 5,
"father_id": 2,
"desc": "Ejecutivo comercial",
"name": "Hugo Miranda"
},
{
"id": 3,
"father_id": 2,
"desc": "Ejecutivo comercial",
"name": "Mario Azcona"
},
{
"id": 6,
"father_id": 3,
"desc": "vendedor",
"name": "Diana Diaz"
}
]
我已经尝试过这个递归函数,并且我得到了正确的结构,但是它聚合了前三个 child 的 2 个副本,我真的不需要它。根 father 是 father_id = 0 的元素
def build(loc_key):
children = {row['id']: {'name': row['name'], 'desc': row['desc'],
'child':[]} for row in list_data if row['father_id'] == loc_key}
data = {}
for key, value in children.items():
data[key] = value
for item in list_data:
if item['father_id'] == key:
data[key]['child'].append(build(key))
return data
print(build(0))
这基本上就是我需要得到的
data = {
2: {'desc': 'Oficial de Negocios Senior',
'name': 'PEDRO MARTIN SOTO ROSALES',
'child':
[
{3: {'desc': 'Ejecutivo comercial',
'name': 'Mario Azcona',
'child': [
{6: {'desc': 'vendedor',
'name': 'Diana Diaz',
'child': []}}]},
4: {'desc': 'Ejecutivo comercial',
'name': 'Adriana Paredez',
'child': []},
5: {'desc': 'Ejecutivo comercial',
'name': 'Hugo Miranda',
'child': []}
PD:我需要以动态方式支持它,因为用户可以在数据库中添加 child 。
最佳答案
我认为问题在于您的 build
函数将节点列表作为输入,并且不会成对地对单个节点或其他一些“较小列表”类型的节点进行操作方式。因此,递归在这里并没有真正的意义。 OTOH,一旦您构建了树(顺便说一下,您正在尝试构建树),那么递归将对解析结果结构非常有帮助。但是,它对构建树的帮助不是很大。
这是构建树的一种方法。它是 O(n)
计算时间和内存,但在操作中确实存储了列表的多个副本,因此可能会有一些优化。
import pprint
list_data = [
{
"id": 2,
"father_id": 0,
"desc": "Oficial de Negocios Senior",
"name": "PEDRO MARTIN SOTO ROSALES"
},
{
"id": 4,
"father_id": 2,
"desc": "Ejecutivo comercial",
"name": "Adriana Paredez"
},
{
"id": 5,
"father_id": 2,
"desc": "Ejecutivo comercial",
"name": "Hugo Miranda"
},
{
"id": 3,
"father_id": 2,
"desc": "Ejecutivo comercial",
"name": "Mario Azcona"
},
{
"id": 6,
"father_id": 3,
"desc": "vendedor",
"name": "Diana Diaz"
}
]
def tree_structure(list_data):
#build the requisite data structure in a "flat" way... you can initialize this "as you go" in the loop below if more optimization is needed.
data = {row["id"]: {"desc": row["desc"], "name": row["name"], "child": {}} for row in list_data}
root_id = None
for row in list_data:
if row["father_id"] != 0:
data[row["father_id"]]["child"][row["id"]] = data[row["id"]] #note that this stores only a reference to the child dictionary, so it is O(1) memory
else:
root_id = row["id"] #we need this later
return {root_id: data[root_id]}
pprint.pprint(tree_structure(list_data))
关于python - 使用 python 创建分层数据。基于结果列表,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/54395289/