lst = ["黄俊"] lst.append("王振") print(lst)  # ["黄俊","王振"] 

lst = ['黄俊', '王振'] lst.insert(1,"刘伟") print(lst)  # 黄军，刘伟 

写作只能添加另一个序列，不能直接在方法中写多个值

lst = ['黄俊', '王振']
tup = ("宋普东","王赢钱","陈宁波")
lst.extend(tup)
print(lst)		# ['黄俊','王振','宋普东','王赢钱','陈宁波']

注意：.pop()会返回被删除值

• 默认删除最后一个

lst = ["魏富强","宋清零","黎建忠","朱家熠"]
res = lst.pop()
print(res)		# "朱家熠"
print(lst)		# ["魏富强","宋清零","黎建忠"]

• 可以指定下标删除

lst = ["魏富强","宋清零","黎建忠","朱家熠"]
res = lst.pop(1)
print(res)		# '宋清零'
print(lst)		# ["魏富强","黎建忠","朱家熠"]

lst = ["魏富强","宋清零","黎建忠","黎建忠","朱家熠"]
lst.remove("黎建忠")
print(lst)		# ["魏富强","宋清零","黎建忠","朱家熠"]

lst = ["魏富强","宋清零","黎建忠","朱家熠"]
lst.clear()
print(lst)		# []

• 当这个值不存在时 , 会报错

lst = ["魏富强","宋清零","黎建忠","朱家熠","魏富强","魏富强"]
res = lst.index("宋清零")		# 1
# res = lst.index("宋清零123") # error
print(res)

lst = ["魏富强","宋清零","黎建忠","朱家熠","魏富强","魏富强"]
res = lst.count("魏富强")
print(res)		# 3

lst = [100,200,90,-3,0,12]
lst.sort()
print(lst)		# [-3, 0, 12, 90, 100, 200]

从大到小排列:

lst = [100,200,90,-3,0,12]
lst.sort(reverse = True)
print(lst)		# [200, 100, 90, 12, 0, -3]

字母是一位一位的进行比较,如果第一位相同,再去比较第二位,依次类推

lst = ["kobe","james","oneal","jordon","iverson"]
lst.sort()
print(lst)		# ['iverson', 'james', 'jordon', 'kobe', 'oneal']

lst = ["王闻","周杰伦","罗志祥","蔡徐坤","王振"]
lst.sort()
print(lst)		# ['周杰伦', '王振', '王闻', '罗志祥', '蔡徐坤']

lst = [100 ,200,300,-3]
lst.reverse()
print(lst)		# [-3, 300, 200, 100]

可以对元组中的列表进行列表函数的操作：

tup = (1,2,3,4,[5,6,7])
tup[-1].append(8)
print(tup)		# (1, 2, 3, 4, [5, 6, 7, 8])

tup = (1,2,3,[4,5,6,(7,8,9)])
tup[-1][-1] = 100	# 将列表中的元组(7,8,9)这个元素改为100
print(tup)		# (1, 2, 3, [4, 5, 6, 100])

我们先来看下下组试验：

a = 19
b = a
print(id(b),id(a))	# 双方id相同
a = 20
print(b)	# 19 # 变更了a的赋值，b不变

lst1 = [1,2,3]
lst2 = lst1
lst1.append(4)		# 变更了lst1的赋值
print(lst2)			# lst2也跟着变了

容器类型中不可变数据的内存地址不会因容器的内存地址改变而改变。
不可变数据: int float bool complex str tuple

语法：.copy()

lst = [1,2,3]
lst2 = lst.copy()
lst.append(5)
print(lst)		# [1, 2, 3, 5]
print(lst2)		# [1, 2, 3]

lst2[0] = 2
print( id(lst[0]))
print( id(lst2[0]))		# 对应元素的值不同，双方id不同，反之则相同

语法：copy.copy()
copy模块.copy函数()

import copy 
lst = [1,2,3]
lst2 = copy.copy(lst)
lst.append(10)
print(lst)
print(lst2)

语法：copy.deepcopy()

import copy
lst = [1,2,3,[4,5,6]]
lst2 = copy.deepcopy(lst)
lst[-1].append(888)
print(lst)		# [1, 2, 3, [4, 5, 6, 888]]
print(lst2)		# [1, 2, 3, [4, 5, 6]]

print(id(lst[-1])  ,  id(lst2[-1]))		# 不相同

• 深拷贝可以应用在字典当中：

import copy
dic = { 
        "a":1,"b":[1,2,3]}
dic2 = copy.deepcopy(dic)
dic["b"].append(4)
print(dic)		# {'a': 1, 'b': [1, 2, 3, 4]}
print(dic2)		# {'a': 1, 'b': [1, 2, 3]}

浅拷贝: 只拷贝第一层级所有的元素, copy.copy()
深拷贝: 拷贝所有层级的元素,都单独开启新的空间 copy.deepcopy()

• (地址:[不可变数据]会暂时指向原数据,[可变数据]会独立开辟新空间)
  可变数据: list set dict
  不可变数据: int float bool complex str tuple

推荐写法：

dic = { 
        }
dic["top"] = "韩国the shy"
dic["middle"] = "肉鸡rookie"
dic["bottom"] = "a_water"
dic["jungle"] = "鞋王"
dic["support"] = "bao_blue"
print(dic)

fromkeys() 使用一组键和默认值创建字典：

lst = ["a","b","c"]
dic = { 
        }.fromkeys(lst,None)		# 将列表所有元素转换成键
print(dic)

出错注意点：

lst = ["a","b","c"]
dic2 = { 
        }.fromkeys(lst,[1,2])	# 将列表所有元素转换成键，并统一赋予对应值[1,2]
dic2["a"].append(3)		# 给键'a'对应的列表增加元素3，别的键对应的列表也都增加了元素3
print(dic2)				# {'a': [1, 2, 3], 'b': [1, 2, 3], 'c': [1, 2, 3]}

改进方式：使用推荐写法

dic3 = { 
        }
dic3["a"] = [1,2]
dic3["b"] = [1,2]
dic3["c"] = [1,2]
dic3["a"].append(3)		# 只给键a对应的列表增加了元素3
print(dic3)				# {'a': [1, 2, 3], 'b': [1, 2], 'c': [1, 2]}

删除键值对：

dic = { 
        'top': '韩国the shy', 'middle': '肉鸡rookie', 'bottom': 'a_water', 'jungle': '鞋王', 'support': 'bao_blue'}
res = dic.pop("bottom")
print(res)		# a_water 返回的是被删除的键对应的值
print(dic)		# {'top': '韩国the shy', 'middle': '肉鸡rookie', 'jungle': '鞋王', 'support': 'bao_blue'}

可以在pop当中设置默认值,预防报错：

dic = { 
        'top': '韩国the shy', 'middle': '肉鸡rookie', 'bottom': 'a_water', 'jungle': '鞋王', 'support': 'bao_blue'}
res = dic.pop("bottom123","没有")		
print(res)		# 没有 没有对应值，返回设置的默认值

dic = { 
        'top': '韩国the shy', 'middle': '肉鸡rookie', 'bottom': 'a_water', 'jungle': '鞋王', 'support': 'bao_blue'}
res = dic.popitem()
print(dic)		# {'top': '韩国the shy', 'middle': '肉鸡rookie', 'bottom': 'a_water', 'jungle': '鞋王'}
print(res)		# ('support', 'bao_blue') 返回的是被删除的键值对

dic = { 
        'top': '韩国the shy', 'middle': '肉鸡rookie', 'bottom': 'a_water', 'jungle': '鞋王', 'support': 'bao_blue'}
dic.clear()
print(dic)		# {}

方法一(推荐)：

dic = { 
        "ww":"大帅哥","spd":"渣男","sql":"负心男"}
newdic = { 
        "ww":"美男子","wz":"一表人才","yh":"未来的土豪"}
dic.update(newdic)
print(dic)		# {'ww': '美男子', 'spd': '渣男', 'sql': '负心男', 'wz': '一表人才', 'yh': '未来的土豪'}

方法二：

dic = { 
        "ww":"大帅哥","spd":"渣男","sql":"负心男"}
dic.update(hj="未来战士",wyq="警察")		# 括号中输入键值对时，用等号连接键值对
print(dic)		# {'ww': '大帅哥', 'spd': '渣男', 'sql': '负心男', 'hj': '未来战士', 'wyq': '警察'}

get方法可以预防报错,默认返回None：

dic = { 
        "ww":"大帅哥","spd":"渣男","sql":"负心男"}
# res = dic["www"] 
res = dic.get("www")	# 返回None

get方法可以设置默认值:

dic = { 
        "ww":"大帅哥","spd":"渣男","sql":"负心男"}
res = dic.get("www","该键不存在")
print(res)		# 该键不存在

可迭代对象: 容器类型数据 , range对象 , 迭代器

dic = { 
        "wy":"智多星","lzs":"花和尚","zs":"浪里白条"}
res = dic.keys()
print(res)		# dict_keys(['wy', 'lzs', 'zs'])

dic = { 
        "wy":"智多星","lzs":"花和尚","zs":"浪里白条"}
res = dic.values()
print(res)		# dict_values(['智多星', '花和尚', '浪里白条'])

dic = { 
        "wy":"智多星","lzs":"花和尚","zs":"浪里白条"}
res = dic.items()
print(res)		# dict_items([('wy', '智多星'), ('lzs', '花和尚'), ('zs', '浪里白条')])

dic = { 
        "wy":"智多星","lzs":"花和尚","zs":"浪里白条"}
res = dic.items()
for k,v in res:
	print(k,v)
''' 输出为： wy 智多星 lzs 花和尚 zs 浪里白条 '''

同时拥有的元素

set1 = { 
        "the shy" , "rookie" , "jacklove" , "神秘男孩"}
set2 = { 
        "蔡国庆","倪萍","赵忠祥","神秘男孩"}
res = set1.intersection(set2)
print(res)			# {'神秘男孩'}

# 简写 &
res = set1 & set2
print(res)

某一方与另一方相比没有的元素，取第一个集合里独有的元素

set1 = { 
        "the shy" , "rookie" , "jacklove" , "神秘男孩"}
set2 = { 
        "蔡国庆","倪萍","赵忠祥","神秘男孩"}
res = set1.difference(set2)
print(res)			# {'rookie', 'jacklove', 'the shy'}

# 简写 -
res = set1 - set2
print(res)

双方所有元素的合集（去重）

set1 = { 
        "the shy" , "rookie" , "jacklove" , "神秘男孩"}
set2 = { 
        "蔡国庆","倪萍","赵忠祥","神秘男孩"}
res = set1.union(set2)		# {'神秘男孩', 'the shy', '蔡国庆', '倪萍', '赵忠祥', 'rookie', 'jacklove'}
print(res)

# 简写 |
res = set1 | set2
print(res)

双方的合集去掉交集后的剩余部分

set1 = { 
        "the shy" , "rookie" , "jacklove" , "神秘男孩"}
set2 = { 
        "蔡国庆","倪萍","赵忠祥","神秘男孩"}
res = set1.symmetric_difference(set2)
print(res)			# {'赵忠祥', '蔡国庆', 'the shy', 'rookie', '倪萍', 'jacklove'}

## 简写 ^
res = set1 ^ set2
print(res)

子集：集合内所有元素对方都有
父集：拥有对方所有元素并且还有更多

set1 = { 
        "张恒","杨