内置方法练习：

1.去重下列列表并保留数据值原来的顺序
	eg: [1,2,3,2,1] 去重之后 [1,2,3]
	l1 = [2,3,2,1,2,3,2,3,4,3,4,3,2,3,5,6,5]    [2, 3, 1, 4, 5, 6]
'''
lst = [1, 2, 3, 2, 1]
lst1 = [2, 3, 2, 1, 2, 3, 2, 3, 4, 3, 4, 3, 2, 3, 5, 6, 5]
# 用来接收lst1初次循环出来的数据
# lst2 = []
# print(lst1)
#
# print(list(set(lst)))
#
# 转集合自动去重，升序排列 解决办法有两种  sort()方法和sorted()方法
# print(list(set(lst1)).sort(key=lst1.index))  # sort() 方法没有返回值 返回None
# lst3 = sorted(list(set(lst1)), key=lst1.index)
# print(lst3)
# print(lst3)
# lst3.sort(key=lst1.index)
# print(lst3)
# sorted(lst3, key=lst1.index)
# print(lst3)
# lst3.sort(key=lst1.index)
# print(lst3)
# print(lst1)


# for i in lst1:
#     if i not in lst2:
#         lst2.append(i)
# print(lst2)

# 列表推导式    和for循环原理一样，都不会改变去重前的顺序
# [lst2.append(i) for i in lst1 if i not  in lst2]
# print(lst2)


'''
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={'jason','oscar','kevin','ricky','gangdan','biubiu'}
　　linuxs={'kermit','tony','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
'''
pythons = {'jason', 'oscar', 'kevin', 'ricky', 'gangdan', 'biubiu'}
linuxs = {'kermit', 'tony', 'gangdan'}

# | 求所有的值，重复的只显示一个
# all_student = pythons | linuxs
# print(all_student)

#
# for i in  pythons:
    # print(i)
    # if i in linuxs:
        # only_pythons = pythons.remove(i)
# print(only_pythons)

# 1. 求出即报名python又报名linux课程的学员名字集合
# for i in  pythons:
#     for j in linuxs:
#         if i == j:
#             all_students = i
#             print(i)

# all_student = pythons & linuxs
# print(all_student)

# 2. 求出所有报名的学生名字集合
# all_stdent = pythons | linuxs
# print(all_stdent)

# 3. 求出只报名python课程的学员名字
# python_student = pythons ^ linuxs
# print(python_student)

# 4. 求出没有同时这两门课程的学员名字集合
# all_student = pythons - linuxs
# all_student1 = linuxs - pythons
# all_student2 = all_student | all_student1
# print(all_student2)

# print(python_student)
'''
3.统计列表中每个数据值出现的次数并组织成字典战士
	eg: l1 = ['jason','jason','kevin','oscar']
      结果:{'jason':2,'kevin':1,'oscar':1}
	真实数据
	l1 = ['jason','jason','kevin','oscar','kevin','tony','kevin']
'''
l1 = ['jason', 'jason', 'kevin', 'oscar']

dict1 = {}
# count()方法可以统计数据值出现的次数
# print(l1.count('jason'))
# for i in l1:
#     print(i)
#     dict1[i] = l1.count(i)
#     print(dict1)
# print(dict1)