set集合

Posted 2020-07-26 yoyoua

set

set是一个无序且不重复的元素集合

set具有的方法

1.add(self, *args, **kwargs)  添加

s = set([1,2,3,4,5])
s.add(6)
print(s)
{1, 2, 3, 4, 5, 6}

2.clear(self, *args, **kwargs)  清空

s = set([1,2,3,4,5])
s.clear()
print(s)

3.copy(self, *args, **kwargs)  拷贝

s = set([1,2,3,4,5])
a = s.copy()
print(a)
{1, 2, 3, 4, 5}

4.difference(self, *args, **kwargs)  

s = set([1,2,3,4,5,2])
print(s)
a = s.difference([1,2,])
print(a)
{1, 2, 3, 4, 5}
{3, 4, 5}

5.difference_update(self, *args, **kwargs)  删除当前set中的所有包含在 new set 里的元素

s = set([1,2,3,4,5,2])
a = s.difference_update([2])
print(s)
print(a)
{1, 3, 4, 5}
None

6.discard(self, *args, **kwargs)  移除元素

s = set([1,2,3,4,5,2])
s.discard(1)
print(s)

7.intersection(self, *args, **kwargs)  取交集，新创建一个set

s = set([1,2,3,4,5,2])
a = s.intersection([2,3,4])
print(a)
{2, 3, 4}

8.intersection_update(self, *args, **kwargs)  取交集，修改原来set

s = set([1,2,3,4,5,2])
s.intersection_update([2,3,4])
print(s)
{2, 3, 4}

9.isdisjoint(self, *args, **kwargs)  如果没有交集，返回true

s = set([1,2,3,4,5,2])
a = s.isdisjoint([6,7])
print(a)
True

10.issubset(self, *args, **kwargs)  是否是子集

s = set([1,2,3,4,5,])
a = s.issubset([1,2,3,4,5,6])
print(a)
True

11.issuperset(self, *args, **kwargs)  是否是父集

s = set([1,2,3,4,5,])
a = s.issuperset([1,2,3,])
print(a)
True

12. pop(self, *args, **kwargs)  移除

s = set([1,2,3,4,5,])
s.pop()
print(s)
{2, 3, 4, 5}

13.remove(self, *args, **kwargs)  移除

s = set([1,2,3,4,5,])
s.remove(3)
print(s)
{1, 2, 4, 5}

14.symmetric_difference(self, *args, **kwargs)  差集，创建新对象

s = set([1,2,3,4,5,])
a = s.symmetric_difference([1,2,3])
print(a)
{4, 5}

15.symmetric_difference_update(self, *args, **kwargs)  差集，改变原来

s = set([1,2,3,4,5,])
s.symmetric_difference_update([1,2,3])
print(s)

16.union(self, *args, **kwargs)  并集

s = set([1,2,3,4,5,])
a = s.union([1,2,3,6])
print(a)
{1, 2, 3, 4, 5, 6}

17.update(self, *args, **kwargs)  更新

s = set([1,2,3,4,5,])
s.update([1,2,7,8])
print(s)
{1, 2, 3, 4, 5, 7, 8}

 set实例

 1 #练习：寻找差异
 2 # 数据库中原有
 3 old_dict = {
 4     "#1":{ ‘hostname‘:‘c1‘, ‘cpu_count‘: 2, ‘mem_capicity‘: 80 },
 5     "#2":{ ‘hostname‘:‘c1‘, ‘cpu_count‘: 2, ‘mem_capicity‘: 80 },
 6     "#3":{ ‘hostname‘:‘c1‘, ‘cpu_count‘: 2, ‘mem_capicity‘: 80 }
 7 }
 8 
 9 # cmdb 新汇报的数据
10 new_dict = {
11     "#1":{ ‘hostname‘:‘c1‘, ‘cpu_count‘: 2, ‘mem_capicity‘: 800 },
12     "#3":{ ‘hostname‘:‘c1‘, ‘cpu_count‘: 2, ‘mem_capicity‘: 80 },
13     "#4":{ ‘hostname‘:‘c2‘, ‘cpu_count‘: 2, ‘mem_capicity‘: 80 }
14 }
15 old_set = set(old_dict.keys())
16 update_list = list(old_set.intersection(new_dict.keys()))
17 
18 new_list = []
19 del_list = []
20 
21 for i in new_dict.keys():
22     if i not in update_list:
23         new_list.append(i)
24 
25 for i in old_dict.keys():
26     if i not in update_list:
27         del_list.append(i)
28 
29 print(update_list,new_list,del_list)