003 几个python编程例子
作业
-
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)
- 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)}")
- 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`
