python数据类型小测试

# 1. 有两个列表,其中一个列表a,每一项映射到另一个列表b每一项,先对a排序,要求b的中映射关系位置保持不变,给b也按照排序, (b的8对应a的[1,2], 7对应[3,4] ... )

a = [[1, 2], [3, 4], [5, 6], [7, 8], [9, 0]]
b = [8, 7, 9, 7, 9]

def relation(a,b):
    relation = {}
    for index, item in enumerate(a):
        relation[tuple(item)] = b[index]
    new_a = [list(i) for i in sorted(relation, key=lambda x: relation[x])]
    new_b = sorted(b)
    return  new_a, new_b

a,b = relation(a,b)
print(a)
print(b)

"""
2. 将一维列表根据父节点,折叠起来
menu_list = [
        {'id':1,'title':'菜单1','pid':None},
        {'id':2,'title':'菜单2','pid':None},
        {'id':3,'title':'菜单3','pid':None},
        {'id':4,'title':'菜单1.1','pid':1},
        {'id':5,'title':'菜单1.2','pid':1},
        {'id':6,'title':'菜单2.1','pid':2},
        {'id':7,'title':'菜单3.1','pid':3},
        {'id':8,'title':'菜单1.1.1','pid':4},
        {'id':9,'title':'菜单1.2.1','pid':5},
        {'id':10,'title':'菜单4','pid':None},
    ]
结果：
    result = [
        {'id':1,'title':'菜单1','pid':None,children:[
            {'id':4,'title':'菜单1.1','pid':1,children:[
                {'id':8,'title':'菜单1.1.1','pid':4,children:[]},
            ]},
            {'id':5,'title':'菜单1.2','pid':1,children:[
                {'id':9,'title':'菜单1.2.1','pid':5},
            ]},
        ]},
        {'id':2,'title':'菜单2','pid':None..},
        {'id':3,'title':'菜单3','pid':None..},
        {'id':10,'title':'菜单4','pid':None..},
    ]
"""

# 方法一
new_dic = {}
for i in menu_list:
    new_dic[i['id']] = i

for v in new_dic.values():
    pid = v['pid']
    if pid:
        new_dic[pid].setdefault('childern', [])
        new_dic[pid]['childern'].append(v)

result = [i for i in menu_list if not i['pid']]

# 方法二
temp_list = menu_list.copy()

for i in menu_list:
    i['children'] = []
    pid = i['pid']
    if i['pid']:
        temp_list[pid-1]['children'].append(i)

result = [i for i in temp_list if not i['pid']]