前戏：

1. 列表添加：
   

   v1 = [1,2,3,4]v1.append(123)print(v1)data = [    [11,22,33],    [44,55,66]]data[0].append(data[1])print(data)# data = [#     [11,22,33, [44,55,66]],#     [44,55,66]# ]data[1].append(77)print(data)# data = [#     [11,22,33, [44,55,66,77]],#     [44,55,66,77]# ]print(data[0][3])#[44, 55, 66, 77]

   列表添加元素练习
2. 字典添加：
   

   v1 = {‘k1‘:‘v1‘}v1[‘k2‘] = ‘v2‘print(v1)data = [    {‘k1‘:‘v1‘},    {‘k2‘:‘v2‘}]for item in data:    item[‘kk‘] = ‘vv‘print(data)

   字典添加元素练习

案例：

　　如果parent_id非0，并且parent_id等于id 就添加到id值相同得那行：

# msg_list = [
#     {‘id‘:1,‘content‘:‘xxx‘,‘parent_id‘:None},
#     {‘id‘:2,‘content‘:‘xxx‘,‘parent_id‘:None},
#     {‘id‘:3,‘content‘:‘xxx‘,‘parent_id‘:None},
#     {‘id‘:4,‘content‘:‘xxx‘,‘parent_id‘:1},
#     {‘id‘:5,‘content‘:‘xxx‘,‘parent_id‘:4},
#     {‘id‘:6,‘content‘:‘xxx‘,‘parent_id‘:2},
#     {‘id‘:7,‘content‘:‘xxx‘,‘parent_id‘:5},
#     {‘id‘:8,‘content‘:‘xxx‘,‘parent_id‘:3},
# ]
"""
msg_list = [
      {
         ‘id‘:1,‘content‘:‘xxx‘,parent_id:None,child:[
            {‘id‘:4,‘content‘:‘xxx‘,parent_id:1},
            {‘id‘:5,‘content‘:‘xxx‘,parent_id:1,child:[
               {‘id‘:7,‘content‘:‘xxx‘,parent_id:5},
            ]}]
         },
         {‘id‘:2,‘content‘:‘xxx‘,parent_id:None,child:[
            {‘id‘:6,‘content‘:‘xxx‘,parent_id:2},
         ]},
         {‘id‘:3,‘content‘:‘xxx‘,parent_id:None,child:[
            {‘id‘:8,‘content‘:‘xxx‘,parent_id:3},
         ]},
   ]
"""

案例代码：

msg_list = [    {‘id‘:1,‘content‘:‘xxx‘,‘parent_id‘:None},    {‘id‘:2,‘content‘:‘xxx‘,‘parent_id‘:None},    {‘id‘:3,‘content‘:‘xxx‘,‘parent_id‘:None},    {‘id‘:4,‘content‘:‘xxx‘,‘parent_id‘:1},    {‘id‘:5,‘content‘:‘xxx‘,‘parent_id‘:4},    {‘id‘:6,‘content‘:‘xxx‘,‘parent_id‘:2},    {‘id‘:7,‘content‘:‘xxx‘,‘parent_id‘:5},    {‘id‘:8,‘content‘:‘xxx‘,‘parent_id‘:3},]# v = [ row.setdefault(‘child‘,[]) for row in msg_list]# 列表生成式msg_list_dict = {}for item in msg_list:    item[‘child‘]=[]    msg_list_dict[item[‘id‘]] = itemresult = []for item in msg_list:    pid = item[‘parent_id‘]    if pid:        msg_list_dict[pid][‘child‘].append(item)    else:        result.append(item)for i in result:    print(i)

解题思路：

1. 通过冒泡算法进行解答。【推荐】
2. 通过for嵌套for循环亦可以实现，但是效率较低。

知识扩展：

　　算法练习

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