1、作用
1.1 关系运算
friends1 = ["zero","kevin","jason","egon"] friends2 = ["Jy","ricky","jason","egon"] l=[] for x in friends1: if x in friends2: l.append(x) print(l) # 以循环的形式,新列表中存储相同的名字 # ['jason', 'egon']
1.2、去重
2、定义: 在{}内用逗号分隔开多个元素,多个元素满足以下三个条件
1. 集合内元素必须为不可变类型
2. 集合内元素无序
3. 集合内元素没有重复
s = {1, 2} # s=set({1,2}) s = {1, [1, 2]} # 集合内元素必须为不可变类型 s = {1, 'a', 'z', 'b', 4, 7} # 集合内元素无序 s = {1, 1, 1, 1, 1, 1, 'a', 'b'} # 集合内元素没有重复 print(s) # 这三种全都不正确,直接报错了
了解
s={} # 默认是空字典
print(type(s))
定义空集合
s=set()
print(s,type(s))
# <class 'set'>
3、类型转换
# set({1,2,3}) # 错的 res=set('hellolllll') print(res) # {'h', 'e', 'o', 'l'} # print(set([1,1,1,1,1,1])) # 错的 # print(set([1,1,1,1,1,1,[11,222]]) # 报错
4、内置方法
=========================关系运算符=========================
4.1 取交集:两者共同的好友
friends1 = {"zero","kevin","jason","egon"} friends2 = {"Jy","ricky","jason","egon"} res=friends1 & friends2 print(res) print(friends1.intersection(friends2)) # {'jason', 'egon'} # {'jason', 'egon'}
4.2 取并集/合集:两者所有的好友
friends1 = {"zero","kevin","jason","egon"} friends2 = {"Jy","ricky","jason","egon"} print(friends1 | friends2) print(friends1.union(friends2)) # {'Jy', 'egon', 'kevin', 'jason', 'ricky', 'zero'} # {'Jy', 'egon', 'kevin', 'jason', 'ricky', 'zero'}
4.3 取差集:取friends1独有的好友
friends1 = {"zero","kevin","jason","egon"} friends2 = {"Jy","ricky","jason","egon"} print(friends1 - friends2) print(friends1.difference(friends2)) # {'zero', 'kevin'} # {'zero', 'kevin'}
取friends2独有的好友
friends1 = {"zero","kevin","jason","egon"} friends2 = {"Jy","ricky","jason","egon"} print(friends2 - friends1) print(friends2.difference(friends1)) # {'Jy', 'ricky'} # {'Jy', 'ricky'}
4.4 对称差集: 求两个用户独有的好友们(即去掉共有的好友)
friends1 = {"zero","kevin","jason","egon"} friends2 = {"Jy","ricky","jason","egon"} print(friends1 ^ friends2) print(friends1.symmetric_difference(friends2)) # {'zero', 'kevin', 'ricky', 'Jy'} # {'zero', 'kevin', 'ricky', 'Jy'}
4.5 父子集:包含的关系
s1={1,2,3} s2={1,2,4} # 不存在包含关系,下面比较均为False print(s1 > s2) print(s1 < s2) # False # False
s1={1,2,3} s2={1,2} print(s1 > s2) # 当s1大于或等于s2时,才能说是s1是s2他爹 print(s1.issuperset(s2)) print(s2.issubset(s1)) # s2 < s2 =>True # True # True # True
s1={1,2,3} s2={1,2,3} print(s1 == s2) # s1与s2互为父子 print(s1.issuperset(s2)) print(s2.issuperset(s1)) True True True
=========================去重=========================
1、只能针对不可变类型去重
print(set([1,1,1,1,2])) # {1, 2}
2、无法保证原来的顺序
l=[1,'a','b','z',1,1,1,2] l=list(set(l)) print(l) # [1, 'b', 2, 'a', 'z']
手动循环去重
l=[ {'name':'lili','age':18,'sex':'male'}, {'name':'jack','age':73,'sex':'male'}, {'name':'tom','age':20,'sex':'female'}, {'name':'lili','age':18,'sex':'male'}, {'name':'lili','age':18,'sex':'male'}, ] new_l=[] for dic in l: if dic not in new_l: new_l.append(dic) print(new_l) # [{'name': 'lili', 'age': 18, 'sex': 'male'}, # {'name': 'jack', 'age': 73, 'sex': 'male'}, # {'name': 'tom', 'age': 20, 'sex': 'female'}]
其他操作
'''
1.长度
>>> s={'a','b','c'}
>>> len(s)
3
2.成员运算
>>> 'c' in s
True
3.循环
>>> for item in s:
... print(item)
...
c
a
b
'''
# 其他内置方法
s={1,2,3}
# 需要掌握的内置方法1:discard
s.discard(4) # 删除元素不存在do nothing
print(s)
s.remove(4) # 删除元素不存在则报错
# 需要掌握的内置方法2:update # 把集合中本来没有的添加进去
s.update({1,3,5})
print(s)
# 需要掌握的内置方法3:pop # 默认删除最后一个
res=s.pop()
print(res)
# 需要掌握的内置方法4:add # 添加
s.add(4)
print(s)
# 其余方法全为了解
res=s.isdisjoint({3,4,5,6}) # 两个集合完全独立、没有共同部分,返回True
print(res)
# 了解
s.difference_update({3,4,5}) # s=s.difference({3,4,5})跟这个功能一样
print(s)