s0 = {1, 2, 3}s1 = set()s2 = set('set')s3 = set([1, 2, 3])s4 = set((1, 2, 3))print(s0, s1, s2, s3, s4, sep='\n')

注意：set与dict一样使用{}来表示，但s={}创建的是空dict，不是创建空set。创建一个空set需要使用s=set()。

s = {1, 2, 'a', 5, 'b'}s.add(7)print(s)s.add('a')print(s)s.add('ab')print(s)

a = {1, 2, 3, 'a', 'b', 'c'}b = 'xyz'c = {'xyz', 'abc'}a.update(b)print(a)a.update(c)print(a)

s1 = {1, 2, 3, 4, 5, 6, 7}s2 = {6, 7, 8, 9, 10, 11, 12}print(s1.union(s2))print(s2.union(s1))print(s1 | s2)

可以看出：A.union(B)==B.union(A)==A|B==B|A

s1 = {1, 2, 3, 4, 5, 6, 7}s2 = {6, 7, 8, 9, 10, 11, 12}print(s1.intersection(s2))print(s2.intersection(s1))print(s1 & s2)

可以看出：A.intersection(B)==B.intersection(A)==A&B==B&A

s1 = {1, 2, 3, 4, 5, 6, 7}s2 = {6, 7, 8, 9, 10, 11, 12}print(s1.difference(s2))print(s2.difference(s1))print(s1 - s2)print(s2 - s1)

可以看出：1）A.difference(B) 不等于 B.difference(A)，从上图中也可以看出来；2）difference可以使用“-”减号代替。

s1 = {1, 2, 3, 4, 5, 6, 7}s2 = {6, 7, 8, 9, 10, 11, 12}print(s1.symmetric_difference(s2))print(s2.symmetric_difference(s1))print(s1 ^ s2)print(s2 ^ s1)

s1 = {1, 2, 3, 4, 5, 6, 7}s2 = {6, 7, 8, 9, 10, 11, 12}s3 = {1, 2, 3}s4 = {1, 2, 3}print(s1.issubset(s2))print(s3.issubset(s1))print(s1.issuperset(s3))print(s4 <= s3)print(s1 > s3)

s = {1, 2, [3, 4, 5]}

s = {1, 2, (3, 4, 5), 'abc'}print(s)

元组、字符串等hashable类型是可以的。

s = {1, 1, 1, 2, 3, 4, 5, 2, 2, 3, 3, 4, 5, 5}s = list(set(s))print(s)

import timest = time.clock()N = 50000arr = [i for i in range(N)]A = [i for i in range(N) if i % 2 == 0]B = [i for i in range(N) if i % 2 == 1]count = 0for x in arr:if x in A:        count += 1elif x in B:        count -= 1print(count)print(time.clock() - st)

耗时近24秒。如果把A、B转换成set，进行同样的运算：

import timest = time.clock()N = 50000arr = [i for i in range(N)]A = set([i for i in range(N) if i % 2 == 0])B = set([i for i in range(N) if i % 2 == 1])count = 0for x in arr:if x in A:        count += 1elif x in B:        count -= 1print(count)print(time.clock() - st)

耗时0.03624秒，list是set耗时的660倍。因此，当需要对一个无重复元素数据集频繁进行查询操作时，使用set是一个很好的选择。