列表的内置方法

# l1 = [11, 22, 33, 44, 55, 66]
# l1.reverse()
# print(l1)

# l1 = [99, 11, 22, 33, 44, 55, 66]
# l1.sort(reverse=True)  # 默认是升序排列
# print(l1)

# 列表比较
# l1 = [999, 888]
# l2 = [111, 222, 333, 444, 555]
# print(l1 > l2)

# 列表切片
l = [1, 2, 3, 4, 5, 6, 7, 8, 9]
# print(l[1])
# print(l[1:5])
# print(l[1:])  # 冒号右边不写，代表从开始位置一直切到末尾
# print(l[:5])  # 冒号左边不写，代表从头开始一直切断索引指定位置
# print(l[1:8:2]) #
# print(l[-1])  # -1位置取得就是末尾数据
# print(l[-8:-1:2])
# print(l[::-1])  # 冒号左右两边都不写，代表全都要

# s = 'helloworld'
# print(s[1:])
# print(s[:6])

字典的内置方法

# 1. 定义字典
d = {'username': 'ly', 'age': 12}

# 2. 第二种方式dict
# d1 = dict(name='ly', age=18, gender='male')
# print(d1)

# 了解
# info = dict([['name', 'tony'], ('age', 18)])
# print(info)

# dic = {
#     'name': 'xxx',
#     'age': 18,
#     'hobbies': ['play game', 'basketball']
# }

# 1. 取值
# print(dic['name1111'])
# print(dic['hobbies'])

# 2. 第二种方式, 掌握
# print(dic.get('name1111', 666))  # None
# print(dic.get('name', 666))  # None
# print(dic.get('hobbies'))

# 3. 修改值
# dic['name'] = 'ly'     # k值 存在，直接进行修改操作
# dic['pwd'] = '123456'  # k值不存在，会往字典中添加一个k:v
# print(dic)

# l = [1, 2, 3]  # 0-2
# # l[4] = 666
# print(l[4])

# 4. 求长度
# l = [1, 2, 3, 4, 5, 6]
# print(len(l))

# print(len(dic))

# 5. 成员运算
# print('name' in dic)
# print('name' not in dic)

# 6. 删除
# 第一种方式
# del dic['name']
# del dic['hobbies']
# print(dic)

# 第二种方式
# dic.pop('name')
# dic.pop('hobbies')
# print(dic)

dic = {
    'name': 'xxx',
    'age': 18,
    'hobbies': ['play game', 'basketball']
}
# 7. 字典三剑客
# print(dic.keys())  # dict_keys(['name', 'age', 'hobbies']) => 列表
# print(dic.values())  # dict_values(['xxx', 18, ['play game', 'basketball']]) => 列表
# print(dic.items())  # dict_items([('name', 'xxx'), ('age', 18), ('hobbies', ['play game', 'basketball'])])

# 8. 循环字典
# for i in dic:
#     print(i)
#     print(dic[i])
#     print(dic.get(i))

# [('name', 'xxx'), ('age', 18), ('hobbies', ['play game', 'basketball'])]
# k,v = ('name', 'xxx')

# for k, v in dic.items():
#     print(k, v)

for i in dic.keys():
    print(i)

for j in dic.values():
    print(j)
字典需要了解的方法
dic = {
    'name': 'xxx',
    'age': 18,
    'hobbies': ['play game', 'basketball']
}

# print(dic.popitem())
# print(dic)
# dic1 = {
#     'name':'aaa'
# }
# dic1.update({'name': 'ly', 'pwd': 123})
# print(dic1)

# dic1 = dict.fromkeys(['k1', 'k2', 'k3'], [])
# print(dic1)
# dic1['k1'] = []
# dic1['k1'].append(666)
# # dic1['k1'].append(777)
# # dic1['k1'].append(888)
# print(dic1)


# setdefault
# print(dic.setdefault('name1111', 666))
# print(dic)

元组的内置方法

# 1.类型转换
关键字：tuple
tuple(111)   # 不行
tuple(1.11)  # 不行
tuple('helloworld') # 行
...


# 支持for循环的数据类型都可以转为元组

# 第一道笔试题：
t1 = (111)
t2 = (1.22)
t3 = ('helloworld')
t4 = ('a', 'b')
t5 = ('c', )
'''当元组中只有一个元素的时候，也要加逗号'''
print(type(t1))  # <class 'int'>
print(type(t2))  # <class 'float'>
print(type(t3))  # <class 'str'>
print(type(t4))  # <class 'tuple'>
print(type(t5))  # <class 'tuple'>


# 求长度
len(tuple1) 

# 第二道笔试题
t1 = (111, 222, [444, 555, 666])
# print(t1)
# print(t1[2])
t1[2].append(777)
print(t1)

集合的内置方法

而集合类型既没有索引也没有key与值对应，所以无法取得单个的值，而且对于集合来说，主要用于去重与关系元素，根本没有取出单个指定值这种需

d = {} # 默认是空字典 
s = set() # 这才是定义空集合

# 第二道题：去重，并且保留原来的顺序
ll = [11, 22, 22, 22, 33, 33, 44, 44, 55, 66, 77, 77, 88]

# ss = set(ll)
# l1 = list(ss)
# print(l1)

new_list = []
for i in ll:
    if i not in new_list:
        new_list.append(i)
print(new_list)

集合的运算关系

friends1 = {"zero", "kevin", "jason", "egon"}
friends2 = {"Jy", "ricky", "jason", "egon"}
# 求合集，并集
print(friends1 | friends2)

# 求交集
print(friends1 & friends2)

# 求friends1差集
print(friends1 - friends2)

# 求friends2差集
print(friends2 - friends1)

# 求对称差集
print(friends1 ^ friends2)

# 求父集，子集
print(friends1 > friends2)
print(friends1 < friends2)

 

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