003 几个python编程例子

作业

1. 有如下值集合[11,22,33,44,55,66,77,88,99,90...]，将所有大于 66 的值保存至字典的第一个key中，将小于 66 的值保存至第二个key的值中，结果为{'k1': 大于66的所有值, 'k2': 小于66的所有值}。

   kema = {11 ,22 ,33 ,44 ,55 ,66 ,77 ,88 ,99 ,99 ,90}
   
   kee = {'k1':[] ,'k2':[]}
   for i in kema:
       if i> 66:
           kee['k1'].append(i)
   
       else:
           kee['k2'].append(i)
   print(kee)
   有难度
   
   第二种答案：自己搜集的答案
   li =[11,22,33,44,55,66,77,88,99,90]
   dic={}
   list_1 =[]  #大于66
   list_2 =[]  #小于66
   for i in li:
       if i ==66: continue
       if i > 66:
            list_1.append(i)
       else:
            list_2.append(i)
   dic.setdefault('k1',list_1)
   dic.setdefault('k2',list_2)
   print(dic)
   

第三种答案：
lis = [11,22,33,44,55,66,77,88,99,90]
dic = dict()
dic['k1'] = list()
dic['k2'] = list()

print(dic)

for i in lis:
    if i > 66:
        dic['k1'].append(i)
    elif i < 66:
        dic['k2'].append(i)
        
print(dic)

   
   
1. 统计`s='hello jason tank tank jason sean say hello dsb dsb sb'`中每个单词的个数，结果为`{'hello': 2, 'alex': 2, 'say': 1, 'sb': 2}`。

   ```python
   s='hello jason tank tank jason sean say hello dsb dsb sb'
   s = s.split( )
   kee = {}
   for key in s:
       if key in kee:
           kee[key] += 1
       else:
           kee[key] = 1
   print(kee)
   
   

2. 有如下两个集合，pythons是报名python课程的学员名字集合，linuxs是报名linux课程的学员名字集合

pythons={'alex','egon','yuanhao','wupeiqi','gangdan','biubiu'}

linuxs={'wupeiqi','oldboy','gangdan'}

1. 求出即报名python又报名linux课程的学员名字集合
2. 求出所有报名的学生名字集合
3. 求出只报名python课程的学员名字
4. 求出没有同时这两门课程的学员名字集合

print(pythons&linuxs) # print(f"{pythons.intersection(linuxs)}")
print(pythons|linuxs) # print(f"{pythons.union(linuxs)}")
print(pythons-linuxs) # print(f"{pythons.difference(linuxs)}")
print(pythons^linuxs) # print(f"{pythons.symmetric_difference(linuxs)}")

4. 有如下列表，列表元素为不可hash类型，去重，得到新列表，且新列表一定要保持列表原来的顺序

l=[
    {'name':'egon','age':18,'sex':'male'},
    {'name':'alex','age':73,'sex':'male'},
    {'name':'egon','age':20,'sex':'female'},
    {'name':'egon','age':18,'sex':'male'},
    {'name':'egon','age':18,'sex':'male'},
]




```第一种答案：
ln = []
for i in l:
    if i not in ln:
        ln.append(i)

print(ln)
第二种答案：自己搜集

new_stu_info_list = []
for stu_info in stu_info_list:
    if stu_info not in new_stu_info_list:
        new_stu_info_list.append(stu_info)

for new_stu_info in new_stu_info_list:
    print(new_stu_info)

1.城市扁平列表转树状

理解值引用和位置引用，实现一次遍历生成结果，时间复杂度为O(n)

1.1

输入：
area_list = [{"code": "1", "parent_code": "0"},
             {"code": "11", "parent_code": "1"},
             {"code": "111", "parent_code": "11"},
             {"code": "1111", "parent_code": "111"},
             {"code": "11111", "parent_code": "1111"},
             {"code": "111111", "parent_code": "111111"},
             {"code": "2", "parent_code": "0"},
             {"code": "22", "parent_code": "2"},
             {"code": "222", "parent_code": "22"}]
输出：
result = {
    "sub":[
        {
            "code":"1",
            "sub":[
                {
                    "code":"11",
                    "sub":[
                        {
                            "code":"111",
                            "sub":[
                                {
                                    "code":"1111",
                                    "sub":[
                                        {
                                            "code":"11111",
                                            "sub":[

                                            ],
                                            "parent_code":"1111"
                                        }
                                    ],
                                    "parent_code":"111"
                                }
                            ],
                            "parent_code":"11"
                        }
                    ],
                    "parent_code":"1"
                }
            ],
            "parent_code":"0"
        },
        {
            "code":"2",
            "sub":[
                {
                    "code":"22",
                    "sub":[
                        {
                            "code":"222",
                            "sub":[

                            ],
                            "parent_code":"22"
                        }
                    ],
                    "parent_code":"2"
                }
            ],
            "parent_code":"0"
        }
    ]
}

解决方案：

def solution(lists):
    result = {}
    for a in lists:
        a.setdefault("parent_code")
        result.setdefault(a['code'], {"sub": []})
        result.setdefault(a['parent_code'], {"sub": []})
        result[a['code']].update(a)
        result[a['parent_code']]['sub'].append(result[a['code']])
    return result["0"]

1.2

输入：

industry_list = [
    {
        "pid": "女装",
        "id": "连衣裙"
    },
    {
        "id": "女装"
    },
    {
        "pid": "女装",
        "id": "半身裙"
    },
    {
        "pid": "女装",
        "id": "A字裙"
    },
    {
        "id": "数码"
    },
    {
        "pid": "数码",
        "id": "电脑配件"
    },
    {
        "pid": "电脑配件",
        "id": "内存"
    },
]

输出：

result = {
    "数码": {
        "电脑配件": {
            "内存": {}
        }
    },
    "女装": {
        "连衣裙": {},
        "半身裙": {},
        "A字裙": {}
    }
}

解决方案

def solution(lists):
    data2 = {}
    for a in lists:
        a.setdefault("pid", "")
        data2.setdefault(a['id'], {})
        data2.setdefault(a['pid'], {})
        data2[a['pid']][a['id']] = data2[a['id']]
    return data2[""]

**作业答案可参考：https://www.cnblogs.com/linhaifeng/articles/7133357.html#_label8`