07 set集合,深浅拷⻉以及部分知识点补充
一.while,for 循环知识点补充
二.int, str的相关操作
1.列表变字符串
# lst=["红","橙","黄","绿","青","蓝","紫"]
方法一
# s="_".join(lst)
# print(s)
方法二
# sum=""
# for i in lst:
# sum=sum+i+"_"
# print(sum)
#
字符串连接
# s="金","木","水","火","土"
# s1="_".join(s)
# print(s1)
元组连接
# tu=("男","女","老","少")
# tu1="_".join(tu)
# print(tu1)
2 list删除问题
# lst=["红","橙","黄","绿","青","蓝","紫"]
方法一 clear
# lst.clear()
# print(lst)
方法二 del range(0,len(lst))
# for i in range(0,7):
# del lst[0]
# print(lst)
方法三 pop range(0,len(lst))
# for i in range(0,7):
# lst.pop(0)
# print(lst)
方法四 把要删除的元素保存在新列表中,循环新列表,删除旧列表.
# lst1=[]
# for i in lst:
# lst1.append(i)
# for el in lst1:
# lst.remove(el)
# print(lst)
3 dict 中的fromkey( )
#fromkey : 第一个参数中每一个拿出来和第二个参数组织成键值对.
#注意代码中只是更改了jay那个列表. 但是由于jay和JJ⽤的是同⼀个列表. 所以. 前⾯那个改了. 后⾯那个也会跟着改.
# dic=dict.fromkeys(["jay","jj"],["周杰伦","林俊杰"])
# dic.get("jay").append("蔡依林")
# print(dic)
4. dic中的元素在迭代过程中是不可以删除的.,需要把要删除的内容存在新列表中,循环新列表,删除旧列表
# lst=["红","橙","黄","绿","青","蓝","紫"]
# lst1=["黄","绿"]
# for el in lst1:
# lst.remove(el)
# print(lst)
5. set() 集合 是dict类型的数据,只存key的值,不存value的值.
特点: 元素不重复(应用"去重");无序;内容必须不可变(可哈希);set本身是可变的(增删改).
去重:
# lst=["红","橙","黄","绿","青","蓝","紫","青","绿"]
# s=set(lst)
# print(s)
set 是可变的,里面内容是不可变的
# s={"alex",1,True,{"你好"}} 不可行,因为{"你好"}不可哈希( 可变的)
#s={"麻花",[1,2,3,]} 不可行 因为[1,2,3]是列表,列表不可哈希( 可变的)
#s={"听","说","读","写",(1,3)} 可行 因为(1,3)是元组,元组是不可变的(可哈希)
6.set 增 删 改 查
增一
# s=set()
# s.add("大神")
# print(s)
增二 update迭代更新
# s={"赵","钱","孙","李","张","王"}
# s.update("吉孔孟")
# print(s)
# s.update(["中国"])
# print(s)
# s.update(["中","国"])
# print(s)
删一 pop() 随机删除
# s={"赵","钱","孙","李","张","王"}
# s1=s.pop()
# print(s1) 删除的元素
# print(s) 删除以后的结果
# 删二clear 打印出来是set()
删三remove
# s.remove("王") 删除指定元素
# print(s)
#修改 set无法直接修改,先删掉,再增加.
# s={"赵","钱","孙","李","张","王"}
# s.remove("王")
# s.add("吉")
# print(s)
查询
# s={"赵","钱","孙","李","张","王"}
# for el in s:
# print(el)
7. 其他操作
#交集 print(s1 and s2)
# s1={"赵","钱","孙","李","张","王"}
# s2={"赵","钱","孙","李","张","王","吉","宋"}
# print(s1&s2)
# print(s1.intersection(s2))
并集
# s1={"赵","钱","孙","李","张","王"}
# s2={"赵","钱","孙","李","张","王","吉","宋"}
# print(s1|s2)
# print(s1.union(s2))
差集 print(s1-s2) 得到第一个中单独存在的
# s1={"赵","钱","孙","李","张","王","杜"}
# s2={"赵","钱","孙","李","张","王","吉","宋"}
# print(s1-s2)
# print(s1.difference(s2))
反交集 两个集合中单独存在的数据 (和交集相反)
# s1={"赵","钱","孙","李","张","王","杜"}
# s2={"赵","钱","孙","李","张","王","吉","宋"}
# print(s1^s2)
# print(s1.symmetric_difference(s2))
#子集 print(s1<s2) set1是set2 的子集吗
# s1={"赵","钱","孙","李","张","王",}
# s2={"赵","钱","孙","李","张","王","吉","宋"}
# # print(s1<s2)
# print(s1.issubset(s2))
超集 print(s1<s2) print(s1.issuperset(s2))
# s1={"赵","钱","孙","李","张","王",}
# s2={"赵","钱","孙","李","张","王","吉","宋"}
# print(s1>s2)
# print(s1.issuperset(s2))
8. set集合本身是可以发⽣改变的. 是不可hash的. 我们可以使⽤frozenset来保存数据.frozenset是不可变的. 也就是⼀个可哈希的数据类型
# s1={"赵","钱","孙","李","张","王",}
# s=frozenset(s1)
# dic={s1:"123"}
# print(dic)
#9.深浅拷贝
(1)赋值:对于list, set, dict来说, 直接赋值. 其实是把内存地址交给变量. 并不是复制⼀份内容. 所以.lst1的内存指向和lst2是⼀样的. lst1改变了, lst2也发⽣了改变
list列表
# lst1=["红","橙","黄","绿","青","蓝","紫"]
# lst2=lst1
# print(lst1)
# print(lst2)
# lst1.append("粉")
# print(lst1)
# print(lst2)
ict字典
# dic1={"颜色":"红色","水果":"香蕉"}
# dic2=dic1
# print(dic1)
# print(dic2)
# dic1["蔬菜"]="番茄"
# print(dic1)
# print(dic2)
(2)浅拷贝 只拷贝第一层,第二层的内容不拷贝
# lst1=["风","雨","雪","霜"]
# lst2=lst1.copy()
# lst1.append("冰雹")
# print(lst1)
# print(lst2)
# print(id(lst1),id(lst2))
2
# lst2=["风","雨","雪","霜",["电","雷"]]
# lst3=lst2.copy()
# print(lst3)
# lst2[4].append("冰雹")
# print(lst2)
# print(lst3)
(3)深拷贝 把元素内部的元素完全进⾏拷⻉复制. 不会产⽣⼀个改变另⼀个跟着改变的问题
# import copy
# lst1=["番茄","白菜","胡萝卜",["苹果","荔枝","香蕉"]]
# lst2=copy.deepcopy(lst1)
# print(lst1)
# print(lst2)
# lst1[3].append("芒果")
# print(lst1)
# print(lst2)
.
list的删除问题
四.set 的集合
五深浅拷贝