1.列表 list[] 存储，修改

修改一

name = ["GIS","xiaomao","农业","123","....."]name[2]) # 农业name[0:3] #切片：顾头不顾尾['GIS', 'xiaomao', '农业']name[-1] #最后一值.....name[-3:-1] #['农业', '123']name[-3:] #['农业', '123', '.....']

修改二

name = ["GIS","xiaomao","农业","123",".....","123"]name.append("fuck") #追加name.insert(3,"zhou") #插入name[0]="ming" #替换name.remove("ming") #删除del name[1]name.pop() #默认空值，则删除最后一个name.pop(1)name.index("123") #获取索引name[name.index("123")] #根据索引获取要素name.count("123") #统计要素个数name.reverse() #反转列表#name.sore() #要素按ASCll表排序name2 = ["F","U","C","K"]name.extend(name2) #合并

 

实例low

book_list = [("ArcGIS",70),("ENVI", 65),("javascriptGIS", 80),("WebGIS", 90),("pythonGIS", 120),("IOSGIS", 150),("PcGIS", 180),("AndroidGIS", 190),("javaGIS", 88),]shoppingCart = []salary = input("你的零钱:")if salary.isdigit():salary = int(salary)while True:for index,item in enumerate(book_list): #获取索引及要素 print(book_list.index(item),item)print(index,item)userChoice = input('选择的商品：')if userChoice.isdigit():userChoice = int(userChoice)if userChoice < len(book_list) and book_list >= 0:p_item = book_list[userChoice]if p_item[1] <= salary: #零钱够付shoppingCart.append(p_item)salary -= p_item[1]print("你的%s已加入购物车,你的余额为------%s------"%(p_item,salary))else:print("余额不足，只剩---[%s]---"%salary)else:print("%s商品不存在"%userChoice)elif userChoice == "q":print(">>>>>>shoppingCart<<<<<<")for p in shoppingCart:print(p)print("余额，只剩:", salary)exit()else:print("错误选项！")else:print("请输入正确金额！")

 

2.元组 tuple() 只读列表，不能修改；只有两个方法：count,index

 

3.字典： dict{ key:value }

 

#name = "zhouMingGang"#name2 = ["zhou","ming","gang"]names={
     "name":"zmg","name1":"zhou","name2": "ming","name3": "gang",}name4={
     "name":"zhouMingGang",1:2,4:9}#print("+".join(name))#print("+".join(name2))#print("-".join(name3))names.update(name4)names["name4"]="chong"names["name2"]="yun"del names["name2"]names.pop("name4")print(names.get("name1"))print(names)print(names.items())a=dict.fromkeys([9,8,6],[1,{
     "name":"zmg"},99])print(a)

 

 

 循环：

for i in names:

print(i,names[i])

4.集合： set() 去重，无序,关系测试>>>>交集，差集，并集

 

集合：   set()   去重，无序,关系测试>>>>交集，差集，并集list = [1,4,5,6,7,7,8,2,5,6]list1 = set(list)print(list1)list2 = set([1,4,7,8,9])print(list1.intersection(list2))     #交集,共同要素被取出  {8, 1, 4, 7}print(list1 & list2)print(list1.union(list2))            #并集 list1 U list2 {1, 2, 4, 5, 6, 7, 8, 9}print(list1 | list2)print(list1.difference(list2))       #差集,list1独有  {2, 5, 6}print(list1 - list2)print(list2.difference(list1))print(list2 - list1)print(list1.symmetric_difference(list2))   #对称差集 {2, 5, 6, 9}print(list1 ^ list2)list1.issubset(list2)                #子集list1.issuperset(list2)              #父集list1.add("fuck")list1.update(["F","u","C"])list1.pop()