一、list类中提供的方法
**********************灰魔法**************************
1. 原来值最后追加 append()
li = [11, 22, 33, 22, 44]li.append(5)li.append("alex")li.append([1234,2323])print(li) # [11, 22, 33, 22, 44, 5, 'alex', [1234, 2323]] 2. 清空列表 clear()
li = [11, 22, 33, 22, 44]li.clear()print(li) # []
3. 计算元素出现的次数 count()
li = [1,2,2,3,42,3,2]v = li.count(2)print(v) # 3
4. 扩展原列表，参数：可迭代对象 append(), extend()
li = [11, 22, 33, 22, 44]li.append([9898,"不得了"])print(li) # [11, 22, 33, 22, 44, [9898, '不得了']]
li = [11, 22, 33, 22, 44]li.extend([9898,"不得了"])print(li) # [11, 22, 33, 22, 44, 9898, '不得了']
5. 根据值获取当前值索引位置（左边优先）index()
li = [11, 22, 33, 22, 44]v= li.index(22)print(v) # 1
6. 在指定索引位置插入元素 insert()
li = [11, 22, 33, 22, 44]li.insert(0,99)print(li) # [99, 11, 22, 33, 22, 44]
7. 删除某个值(1.指定索引；2. 默认最后一个)，并获取删除的值 pop()
li = [11, 22, 33, 22, 44]# v = li.pop()# print(li) # [11, 22, 33, 22]v = li.pop(2)print(li) #[11, 22, 22, 44]
8. 删除列表中的指定值，左边优先 remove()
li = [11, 22, 33, 22, 44]li.remove(22)print(li) # [11, 33, 22, 44]
9. 将当前列表进行翻转 reverse()
li = [11, 22, 33, 22, 44]li.reverse()print(li) # [44, 22, 33, 22, 11]
10. 列表的排序(默认为 False) sort()
li = [11, 44, 22, 33, 22]li.sort()li.sort(reverse=True)print(li) # [44, 33, 22, 22, 11]
*************************深灰魔法**************************
1. 索引取值
print( li [3] ) 2. 切片，切片结果也是列表print( li [3 : -1])3. for循环
for item in li:
print(item)4. 列表元素修改
li = [11, 44, 22, 33, 22]li[1] = 120print(li) # [11, 120, 22, 33, 22]li[1] = [11,22,33,44]print(li) # [11, [11, 22, 33, 44], 22, 33, 22]
5. 删除 del()
li = [11, 44, 22, 33, 22]del li[1]print(li) # [11,22,33,22]
6. 切片
li = [11, 44, 22, 33, 22]li[1:3] = [120,90]print(li) # [11, 120, 90, 33, 22] del li[2:4] print(li) # [11, 120, 22]
7. in 操作
li = [1, 12, 9, "age", ["石振文", ["19", 10], "庞麦郎"], "alex", True]v1 = "石振文" in liprint(v1) # Falsev2 = "age" in liprint(v2) # True
8. 操作
li = [1, 12, 9, "age", ["石振文", ["19", 10], "庞麦郎"], "alex", True]print(li[4][1][0]) # 19print(li[4][1]) # ['19', 10]
9. 字符串转换为列表
a = 123a = str(a)new_a = list(a) # 字符串可以转换为列表，数值不可以print(new_a) # ['1', '2', '3']
10. 列表转换为字符串
li = [11,22,33,"123","alex"]s = ""for i in li: s = s + str(i)print(s) # 112233123alex
或者用 join 的方法
li = ["123","alex"]v = "".join(li)print(v) # 123alex
11. 字符串修改
v = "alex"v = v.replace('l','el')print(v) # aelex
二、元组类中提供的方法
1. 索引
v = tu[0] print(v)2. 切片
v = tu[0:2] print(v)3. 可以被 for 循环，可迭代对象
for item in tu:
print(item)
4. 字符串转换为元组
s = "asdfas"li = ["asdf","asdfas"]tu = ("asdf","as")v = tuple(s)print(v) # ('a', 's', 'd', 'f', 'a', 's')v = tuple(li)print(v) # ('asdf', 'asdfas')v = list(tu)print(v) # ['asdf', 'as']v = "_".join(tu)print(v) # asdf_asli = ["asdf","asdfasdf"]li.extend((11,22,33,))print(li) # ['asdf', 'asdfasdf', 11, 22, 33] 5. 元组的一级元素不可修改/删除/增加
tu = (111,"alex",(11,22),[(33,44)],True,33,44,)# 元组，有序。v = tu[3][0][0]print(v) # 33v=tu[3]print(v) # [(33, 44)]tu[3][0] = 567print(tu) # (111, 'alex', (11, 22), [567], True, 33, 44)tu[3] = '12' # 报错
三、字典类中提供的方法
基本结构
1、字典的 value 可以是任何值——列表、元组、字典、布尔值等
2、布尔值(1,0)、列表、字典不能作为字典的 key，但元组可以
3、索引方式找到字典的元素
4、字典支持 del 删除
info = { "k1": 18, "k3": [ 11, [], (), { 'kk1': 'vv1', 'kk2': 'vv2', 'kk3': (11,22), } ], 2: (11,22,33)}del info['k1']print(info) # {2: (11, 22, 33), 'k3': [11, [], (), {'kk2': 'vv2', 'kk1': 'vv1', 'kk3': (11, 22)}]}del info['k3'][3]['kk1']print(info) # {2: (11, 22, 33), 'k3': [11, [], (), {'kk2': 'vv2', 'kk3': (11, 22)}]} 5、 for 循环
info = { 2: True, "k3": [ 11, [], (), { 'kk1': 'vv1', 'kk2': 'vv2', 'kk3': (11,22), } ], 4: (11,22,33,44)}for item in info: print(item) # 2 # 4 # k3for item in info.keys(): print(item) # 2 # 4 # k3for item in info.values(): print(item) # 11, [], (), {'kk2': 'vv2', 'kk1': 'vv1', 'kk3': (11, 22)}] # True # (11, 22, 33, 44)for k,v in info.items(): print(k,v) # 2 True # 4 (11, 22, 33, 44) # k3 [11, [], (), {'kk1': 'vv1', 'kk3': (11, 22), 'kk2': 'vv2'}] 6. 根据序列，创建字典，并指定统一的值 fromkeys()
v = dict.fromkeys(["k1",123,"999"],123)print(v) # {123: 123, '999': 123, 'k1': 123} v = dict.fromkeys(["k1",123,"999"])print(v) # {123: None, '999': None, 'k1': None} 7. 根据Key获取值，key不存在时，可以指定默认值（None）
dic = { "k1":"v1"}v = dic['k1']print(v) # v1v = dic.get('k11111')print(v) # None 如果用dic则会报错v = dic.get('k11111','asdf')print(v) # asdf 如果不存在，就传递asdf 8. 删除并获取值 pop(), popitem()
dic = { "k1": 'v1', "k2": 'v2'}v = dic.pop('k1',90)print(dic,v) # {'k2': 'v2'} v1 如果没有k1，就传递90k,v = dic.popitem() # 随机删除print(dic,k,v) # {} k2 v2 9. 设置值 setdefault()
dic = { "k1": 'v1', "k2": 'v2'}v = dic.setdefault('k1','123') # 已存在，不设置，获取当前key对应的值print(dic,v) # {'k1': 'v1', 'k2': 'v2'} v1v = dic.setdefault('k3','123') # 不存在，设置，获取当前key对应的值print(dic,v) # {'k1': 'v1', 'k2': 'v2', 'k3': '123'} 123 10. 更新（两种写法都可以） update()
dic = { "k1": 'v1', "k2": 'v2'}dic.update({ 'k1': '111111','k3': 123})print(dic) # {'k2': 'v2', 'k1': '111111', 'k3': 123}dic.update(k1=123,k3=345,k5="asdf")print(dic) # {'k5': 'asdf', 'k2': 'v2', 'k1': 123, 'k3': 345}