day10-集合

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

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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

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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'>

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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

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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

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=========================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'}]

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 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
'''

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# 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)