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)
原文：http://www.cnblogs.com/yoyovip/p/5596596.html