1 a = {1}
2 print(type(a))
1 a = [1,1,1,2,2,2,22,4,4,44,44]
2 print(set(a))
1 s1 = {1,2,3,4,5}
2 s1.add("hello")
3 print(s1)
4 结果是{1, 2, 3, 4, 5, 'hello'}
删
pop 删除并返回任意set元素
remove 从集合中删除一个元素,它必须是成员
1 s1 = {1,2,3,4,5}
2 print(s1.pop())
3 他的删除是把第一个元素提取出来删掉
4
5 s1 = {1,2,3,4,5}
6 print(s1.remove(5))
7 他的删除是指定一个元素进行删除
1 a = {1,2,3,4,5}
2 a.update({7,8,9})
3 print(a)
4 {"只能放和迭代的对象"}
查
isdisjoint 如果两个集合的焦点为空则返回True
issubset 报告另一个集合石是否包含该集合
issuperset 报告此集合是否包含另一个集合
不是的话False 是的话返回 True,F错T对
1 x = {"a","b","c"}
2 y = {"q","w","e"}
3 print(x.isdisjoint(y))
1 a = {1,2,3,4,5,6,7,8,9}
2 b = {1,2,3,4}
3 print(a.issubset(b))
1 a = {1,2,3,4,5,6,7,8,9}
2 b = {1,2,3,4}
3 print(b.issuperset(a))
a = {}
print(type(a))
b = dict()
1 a = {"name":"foyi","age":"19","sex":"man"}
2 key : value
3 键 对 值
1 a = {"name":"foyi","age":"19","sex":"man"}
2 a.setdefault("name","zz")
3 print(a)
1 a = {"name":"foyi","age":"19","sex":"man"}
2 a.update({"age":100})
3 print(a)
删
pop 查出指定的key的键值对
popitem 返回并删除字典中的最后一对键和值
1 a = {"name":"foyi","age":"19","sex":"man"}
2 a.pop("name") #删除所选的的key
3 a.popitem() #删除最后一个键值对
查
get 获取指定的key对应的value如果key不存在返回none
keys 取出所有的key
values 取出所有的值
items 取出所有键值对a = {"name":"foyi","age":"19","sex":"man"}
a = {"name":"foyi","age":"19","sex":"man"}
print(a.get("zzz"))
a = {"name":"foyi","age":"19","sex":"man"}
print(a.keys())
print(a.values())
print(a.items())