在 Python 字典中有一个嵌套列表答案

【问题标题】：Have a nested List in a Python Dictionary在 Python 字典中有一个嵌套列表
【发布时间】：2013-03-26 07:06:01
【问题描述】：

我有一个如下的数据库表。数据采用树的形式，带有

            CREATE  TABLE IF NOT EXISTS DOMAIN_HIERARCHY (
                COMPONENT_ID        INT             NOT NULL ,
                LEVEL               INT             NOT NULL ,
                COMPONENT_NAME      VARCHAR(127)    NOT NULL ,
                PARENT              INT             NOT NULL ,
                PRIMARY KEY ( COMPONENT_ID ) 
                );

以下数据在表格中

                (1,1,'A',0)
                (2,2,'AA',1)
                (3,2,'AB',1)
                (4,3,'AAA',2)
                (5,3,'AAB',2)
                (6,3,'ABA',3)
                (7,3,'ABB',3)

我必须检索数据并存储在 python 字典中

在下面的代码中

                conx = sqlite3.connect( 'nameofdatabase.db' )
                curs = conx.cursor()
                curs.execute( 'SELECT COMPONENT_ID, LEVEL, COMPONENT_NAME, PARENT FROM DOMAIN_HIERARCHY' )
                rows = curs.fetchall()                    
                hrcy = {}                 
                for row in rows:
                    entry = ( row[2], {} )
                    cmap[row[0]] = entry
                    if row[1] == 1:
                        hrcy = {entry[0]: entry[1]}
                        hrcy['status'] = 0
                for row in rows:
                    item = cmap[row[0]]
                    parent = cmap.get( row[3], None )
                    if parent:
                        parent[1][row[2]] = item[1]
                        parent[1]['status'] = 0
                print json.dumps( hrcy, indent = 4 )

输出是这样的

                {
                    "status": 0, 
                    "A": {
                        "status": 0, 
                        "AA": {
                            "status": 0, 
                            "AAA": {}, 
                            "AAB": {}
                        }, 
                        "AB": {
                            "status": 0, 
                            "ABA": {}, 
                            "ABB": {}
                        }
                    }
                }

我想要这样的输出

                {
                    "component": "A",
                    "status": 0,
                    "children": [
                        {
                            "component": "AA",
                            "status": 0,
                            "children": [
                                {
                                    "component": "AAA",
                                    "status": 0,
                                    "children": []
                                },
                                {
                                    "component": "AAB",
                                    "status": 0,
                                    "children": []
                                }
                            ]
                        },
                        {
                            "component": "AB",
                            "status": 0,
                            "children": [
                                {
                                    "component": "ABA",
                                    "status": 0,
                                    "children": []
                                },
                                {
                                    "component": "ABB",
                                    "status": 0,
                                    "children": []
                                }
                            ]
                        }
                    ]
                }

谁能告诉我应该做些什么改变？

【问题讨论】：

标签： python sql json

【解决方案1】：

只需构造你想要的字典，而不是其他东西：

for row in rows:
    entry = {"component": row[2], 
             "children": [], 
             "status": 0}
    cmap[row[0]] = entry
    if row[1] == 1:
        hrcy = entry
for row in rows:
    item = cmap[row[0]]
    parent = cmap.get( row[3], None )
    if parent:
        parent["children"].append(item)

【讨论】：

不敢相信这个简单的小代码解决了我的“无法解决”的问题，非常感谢 Janne Karela

【解决方案2】：

我会选择这样的：

import pprint

class Node(dict):
    def __init__(self, *args):
        dict.__init__(self)
        self['id'] = args[0]
        self['group'] = args[1]
        self['component'] = args[2]
        self['parent'] = args[3]
        self['status'] = 0
        self['children'] = []

    def add_node(self, node):
        for child in self['children']:
            if child.is_parent(node):
                child.add_node(node)
                break
        else:
            self['children'].append(node)

    def is_parent(self, node):
        return node['component'].startswith(self['component'])

class RootNode(Node):
    def __init__(self):
        Node.__init__(self, *[0, 0, "Root", -1])

    def is_parent(self, node):
        return True

def make_tree(nodes):
    root = RootNode()
    for data in nodes:
        node = Node(*data)
        root.add_node(node)
    return root

nodes =  [
        (1, 1, 'A', 0),
        (2, 2, 'AA', 1),
        (3, 2, 'AB', 1),
        (4, 3, 'AAA', 2),
        (5, 3, 'AAB', 2),
        (6, 3, 'ABA', 3),
        (7, 3, 'ABB', 3),
      ]

if __name__ == "__main__":
    tree = make_tree(nodes)
    pprint.pprint(tree['children'][0])

外出

{'children': [{'children': [{'children': [],
                             'component': 'AAA',
                             'group': 3,
                             'id': 4,
                             'parent': 2,
                             'status': 0},
                            {'children': [],
                             'component': 'AAB',
                             'group': 3,
                             'id': 5,
                             'parent': 2,
                             'status': 0}],
               'component': 'AA',
               'group': 2,
               'id': 2,
               'parent': 1,
               'status': 0},
              {'children': [{'children': [],
                             'component': 'ABA',
                             'group': 3,
                             'id': 6,
                             'parent': 3,
                             'status': 0},
                            {'children': [],
                             'component': 'ABB',
                             'group': 3,
                             'id': 7,
                             'parent': 3,
                             'status': 0}],
               'component': 'AB',
               'group': 2,
               'id': 3,
               'parent': 1,
               'status': 0}],
 'component': 'A',
 'group': 1,
 'id': 1,
 'parent': 0,
 'status': 0}

【讨论】：