集合类型内置方法

集合类型内置方法

集合基本用不到

1.作用

用于关系运算的集合体，由于集合内的元素无序且集合元素不可重复，因此集合可以去重，但是去重后的集合会打乱原来元素的顺序。

2.定义

#以{}用逗号隔开不可变数据类型
s={1,2,1,'a','b','c'}
print(s)

去重

lis = [1,2,3,1,3]
print(set(lis))
print(list(set(lis)))


print(set())


s = {}  # 空大括号是字典，不是集合，定义空集合必须得用set()
print(type(s))

3.内置方法

pythoners={'jason','nick','tank','sean'}

linuxers={'nick','egon','kevin'}

print(pythoners|linuxers)   #并集
print(pythoners.union(linuxers))

{'sean', 'egon', 'nick', 'jason', 'tank', 'kevin'}
{'sean', 'egon', 'nick', 'jason', 'tank', 'kevin'}

print(pythoners & linuxers)  # 交集
print(pythoners.intersection(linuxers))

{'nick'}
{'nick'}

print(pythoners - linuxers)  # 差集
print(pythoners.difference(linuxers))

{'jason', 'sean', 'tank'}
{'jason', 'sean', 'tank'}

print(pythoners ^ linuxers)  # 交叉补集
print(pythoners.symmetric_difference(linuxers))

{'jason', 'sean', 'kevin', 'egon', 'tank'}
{'jason', 'sean', 'kevin', 'egon', 'tank'}

了解

s = {1,2,3}
s.add(4)
print(s)

{1,2,3,4}

pythoners={'jason','nick','tank','sean'}

linuxers={'nick','egon','kevin'}

pythoners.pop()
print(pythoners)

{'sean', 'tank', 'jason'}

pythoners.update(linuxers)
print(pythoners)

{'nick', 'sean', 'kevin', 'egon', 'jason', 'tank'}

 pythoners.clear()  # 清空
 print(pythoners)

set()

print(pythoners.copy())

{'nick', 'sean', 'tank', 'jason'}

pythoners.remove('nickk')   
print(pythoners)

报错

pythoners.discard('nickk')  
print(pythoners)

{'sean', 'jason', 'nick', 'tank'}

remove和discard区别是remove没有会报错,discard没有不会报错

pythoners = {'jason', 'nick', 'tank', 'sean'}
pythoners2 = {'jason', 'nick', 'tank', 'sean','nick2'}
print(pythoners.issubset(pythoners2))  # 子集
print(pythoners.issuperset(pythoners2))  # 父集

True
False

pythoners = {'jason', 'nick', 'tank', 'sean'}
linuxers = {'nick', 'egon', 'kevin'}  
pythoners.difference_update(linuxers)
print(pythoners)
pythoners.symmetric_difference_update(linuxers)
print(pythoners)

'tank', 'jason', 'sean'}
{'tank', 'jason', 'kevin', 'sean', 'egon', 'nick'}

pythoners = {'jason', 'tank', 'sean'}
linuxers = {'nick', 'egon', 'kevin'}  # 集合没有共同的部分返回True，否则返回False
res = pythoners.isdisjoint(linuxers)
print(res)

4.存一个值还是多个值

多个值

5.有序or无序

无序

6.可变or不可变

可变

s = {1,2}
print(id(s))
s.add(3)
print(id(s))

2175853453096
2175853453096