一.数据基本类型之set集合
set和dict类似，也是一组key的集合，但不存储value。由于key不能重复，所以，在set中，没有重复的key
set集合，是一个无序且不重复的元素集合
1.创建
s = set() #创建空集合s = { 'values1','values2'} #非空集合 2.转换
l = [1,2,5,11]t = (11,22,12)#元组转集合st2 = set(t)#列表转集合st = set(l)print(st)print(st2)
3.常用支持操作
添加元素-->add(key)
s = set()s.add(1)print(s)
删除元素-->remove(key)
s = set([1,2,3])s.remove(1)print(s)
清除元素-->clear()
s = set([1,2,3,4,5])print(s)s.clear()print(s)
比较元素-->difference()
set1 = {1,44,87,23,55}set2 = {1,44,88,23,67}#set1中有而set2中没有的值ret = set1.difference(set2)print(ret) 删除两集合中相同的元素-->difference_update()
set1 = {1,44,87,23,55}set2 = {1,44,88,23,67}#从set1中删除和set2中相同的元素set1.difference_update(set2)print(set1)print(set2) 移除元素-->discard(values)
set1 = {1,44,87,23,55}set2 = {1,44,88,23,67}#移除指定元素,不存在不会报错，remove（）不存在会报错，建议discardset1.discard(44)print(set1) 取交集值-->intersection()
set1 = {1,44,87,23,55}set2 = {1,44,88,23,67}#取两个set集合的交集值ret = set1.intersection(set2)print(ret) 取交集并更新-->intersection_update()
set1 = {1,44,87,23,55}set2 = {1,44,88,23,67}#取交集并更新到set1中set1.intersection_update(set2)print(set1) 判断是否交集-->isdisjoint()
set1 = {1,44,87,23,55}set2 = {1,44,88,23,67}#如果两个集合有交集返回false,反之返回trueprint(set1.isdisjoint(set2)) 子序列-->issubset()
set1 = {1,44,87,23,55}set2 = {1,44,88,23,67}#判断是否是子集，是返回true,反之返回Flaseprint(set1.issubset(set2)) 父序列-->issuperset()
set1 = {1,44,87,23,55}set2 = {1,44,88,23,67}#是否是父序列,是返回True,反之返回Flaseprint(set1.issuperset(set2)) 对称交集-->symmetric_difference()
set1 = {1,44,87,23,55}set2 = {1,44,88,23,67}#对称交集,取两个集合中互不存在的元素，生成一个新的集合ret = set1.symmetric_difference(set2)print(ret) 对称交集并更新-->symmetric_difference_update()
set1 = {1,44,87,23,55}set2 = {1,44,88,23,67}#对称交集,并更新元素到set1中set1.symmetric_difference_update(set2)print(set1) 并集-->union()
set1 = {1,44,87,23,55}set2 = {1,44,88,23,67}#并集并更新到新的集合中ret = set1.union(set2)print(ret) 二.深浅拷贝
1.数字和字符串
对于 数字 和 字符串 而言，赋值、浅拷贝和深拷贝无意义，因为其永远指向同一个内存地址
import copy# ######### 数字、字符串 #########n1 = 123# n1 = "i am alex age 10"print(id(n1))# ## 赋值 ##n2 = n1print(id(n2))# ## 浅拷贝 ##n2 = copy.copy(n1)print(id(n2)) # ## 深拷贝 ##n3 = copy.deepcopy(n1)print(id(n3))
二，其他数据类型
对于list dict,tuple 浅拷贝只拷贝最外一层，深拷贝除了最后一层（因最后一层是字符串）其余的都拷贝
- 赋值
赋值，只是创建一个变量，该变量指向原来内存地址，如
n1 = { "k1": "wu", "k2": 123, "k3": ["alex", 456]} n2 = n1 解析图如下：
- 浅拷贝
浅拷贝，在内存中只额外创建第一层数据
import copy n1 = { "k1": "wu", "k2": 123, "k3": ["alex", 456]} n3 = copy.copy(n1) 解析图如下：
- 深拷贝：
深拷贝，在内存中将所有的数据重新创建一份（排除最后一层，即：python内部对字符串和数字的优化）
import copy n1 = { "k1": "wu", "k2": 123, "k3": ["alex", 456]} n4 = copy.deepcopy(n1) 解析图如下：
三.集合作业
1.寻找差异，并将new_dict的值更新到old_dict中
old_dict = { "#1":11, "#2":22, "#3": 100}new_dict = { "#1":33, "#4":22, "#7": 100} #!/usr/bin/env python# -*- coding:utf-8 -*- old_dict = { "#1":11, "#2":22, "#3": 100}new_dict = { "#1":33, "#4":22, "#7": 100}old_keys = old_dict.keys()new_keys = new_dict.keys()old_set = set(old_keys)new_set = set(new_keys)#old存在new不存在,并删除new中不存在，old 中存在的元素del_set = old_set.difference(new_set)for i in del_set: del old_dict[i]#new存在old不存在add_set = new_set.difference(old_set)for a in add_set: old_dict[a] = new_dict[a]#更新旧数据表update_set = old_set.intersection(new_set)for u in update_set: old_dict[u] = new_dict[u]print(old_dict)