python集合操作

一：python官方文档

　　https://docs.python.org/3.6/library/stdtypes.html#set-types-set-frozenset

二：集合概念

集合是无序的
集合是无重复元素的

一：add()

会改变原有集合

1 >>> s = set('HackerRank')
2 >>> s.add('H')
3 >>> print s
4 set(['a', 'c', 'e', 'H', 'k', 'n', 'r', 'R'])
5 >>> print s.add('HackerRank')#添加后返回none
6 None
7 >>> print s
8 set(['a', 'c', 'e', 'HackerRank', 'H', 'k', 'n', 'r', 'R'])

set.add()

二：pop(),discard(),remove()

　pop()是随机删除元素
　discard(),remove()是删除指定元素的
discard()在删除不存在元素是不会报错，而remvoe()会报keyerror错误
都会改变原有集合

set.pop()

1 >>> s = set([1])
2 >>> print s.pop()
3 1
4 >>> print s
5 set([])
6 >>> print s.pop()
7 KeyError: pop from an empty set

set.pop()

set.remove()

 1 >>> s = set([1, 2, 3, 4, 5, 6, 7, 8, 9])
 2 >>> s.remove(5)
 3 >>> print s
 4 set([1, 2, 3, 4, 6, 7, 8, 9])
 5 >>> print s.remove(4)#默认返回None
 6 None
 7 >>> print s
 8 set([1, 2, 3, 6, 7, 8, 9])
 9 >>> s.remove(0)
10 KeyError: 0

set.remove()

set.discard()

 1 >>> s = set([1, 2, 3, 4, 5, 6, 7, 8, 9])
 2 >>> s.discard(5)
 3 >>> print s
 4 set([1, 2, 3, 4, 6, 7, 8, 9])
 5 >>> print s.discard(4)
 6 None
 7 >>> print s
 8 set([1, 2, 3, 6, 7, 8, 9])
 9 >>> s.discard(0)
10 >>> print s
11 set([1, 2, 3, 6, 7, 8, 9])

set.discard()

三：intersetcion(),difference(),unoin()，symmetric_difference(),交集，差集，并集,对称差集

不改变原有集合

set.intersection()

 1 >>> s = set("Hacker")
 2 >>> print s.intersection("Rank")
 3 set(['a', 'k'])
 4 
 5 
 6 >>> print s.intersection(['R', 'a', 'n', 'k'])
 7 set(['a', 'k'])
 8 
 9 >>> print s.intersection(enumerate(['R', 'a', 'n', 'k']))
10 set([])
11 
12 >>> print s.intersection({"Rank":1})
13 set([])
14 
15 >>> s & set("Rank")
16 set(['a', 'k'])

set.intersection()

set.difference()

 1 #set1有的而set2没有的元素
 2 >>> s = set("Hacker")
 3 >>> print s.difference("Rank")
 4 set(['c', 'r', 'e', 'H'])
 5 
 6 >>> print s.difference(set(['R', 'a', 'n', 'k']))
 7 set(['c', 'r', 'e', 'H'])
 8 
 9 >>> print s.difference(['R', 'a', 'n', 'k'])
10 set(['c', 'r', 'e', 'H'])
11 
12 >>> print s.difference(enumerate(['R', 'a', 'n', 'k']))
13 set(['a', 'c', 'r', 'e', 'H', 'k'])
14 
15 >>> print s.difference({"Rank":1})
16 set(['a', 'c', 'e', 'H', 'k', 'r'])
17 
18 >>> s - set("Rank")
19 set(['H', 'c', 'r', 'e'])

set.difference()

set.unoin()

 1 >>> s = set("Hacker")
 2 >>> print s.union("Rank")
 3 set(['a', 'R', 'c', 'r', 'e', 'H', 'k', 'n'])
 4 
 5 >>> print s.union(set(['R', 'a', 'n', 'k']))
 6 set(['a', 'R', 'c', 'r', 'e', 'H', 'k', 'n'])
 7 
 8 >>> print s.union(['R', 'a', 'n', 'k'])
 9 set(['a', 'R', 'c', 'r', 'e', 'H', 'k', 'n'])
10 
11 >>> print s.union(enumerate(['R', 'a', 'n', 'k']))
12 set(['a', 'c', 'r', 'e', (1, 'a'), (2, 'n'), 'H', 'k', (3, 'k'), (0, 'R')])
13 
14 >>> print s.union({"Rank":1})
15 set(['a', 'c', 'r', 'e', 'H', 'k', 'Rank'])
16 
17 >>> s | set("Rank")
18 set(['a', 'R', 'c', 'r', 'e', 'H', 'k', 'n'])

set.union()

set.symmetric_difference()

 1 >>> s = set("Hacker")
 2 >>> print s.symmetric_difference("Rank")
 3 set(['c', 'e', 'H', 'n', 'R', 'r'])
 4 
 5 >>> print s.symmetric_difference(set(['R', 'a', 'n', 'k']))
 6 set(['c', 'e', 'H', 'n', 'R', 'r'])
 7 
 8 >>> print s.symmetric_difference(['R', 'a', 'n', 'k'])
 9 set(['c', 'e', 'H', 'n', 'R', 'r'])
10 
11 >>> print s.symmetric_difference(enumerate(['R', 'a', 'n', 'k']))
12 set(['a', 'c', 'e', 'H', (0, 'R'), 'r', (2, 'n'), 'k', (1, 'a'), (3, 'k')])
13 
14 >>> print s.symmetric_difference({"Rank":1})
15 set(['a', 'c', 'e', 'H', 'k', 'Rank', 'r'])
16 
17 >>> s ^ set("Rank")
18 set(['c', 'e', 'H', 'n', 'R', 'r'])

set.symmertic_difference

四：update(),difference_update(),set.symmetric_difference_update()

set.update(set2):以集合2还更新集合1，加入到集合1中

1 >> myset.update([1, 2, 3, 4]) # update() only works for iterable objects
2 >> myset
3 {'a', 1, 'c', 'b', 4, 2, (5, 4), 3}
4 >> myset.update({1, 7, 8})
5 >> myset
6 {'a', 1, 'c', 'b', 4, 7, 8, 2, (5, 4), 3}
7 >> myset.update({1, 6}, [5, 13])
8 >> myset
9 {'a', 1, 'c', 'b', 4, 5, 6, 7, 8, 2, (5, 4), 13, 3}

set.update()

set.difference_update(set2):用set1-set2来更新set1，覆盖set1值

1 #用A,B的差集来更新A
2 >>> H = set("Hacker")
3 >>> R = set("Rank")
4 >>> H.difference_update(R)
5 >>> print H
6 set(['c', 'e', 'H', 'r'])

set.difference——update()

set.symmetric_difference_update(set2):用set1和set2的对称差集来更新set1，覆盖set1的值

1 #用A,b互相差集来更新A
2 >>> H = set("Hacker")
3 >>> R = set("Rank")
4 >>> H.symmetric_difference_update(R)
5 >>> print H
6 set(['c', 'e', 'H', 'n', 'r', 'R'])

set.systemeratic_update()

五：isdisjoint(), issubset(), issuperset()

set.isdisjoint():如果两个集合没有交集则返回True。

1s1,s2={1,2,3},{4,5,6}23s1.isdisjoint(s2)4True

set.isdisjoint()

issubset():是否为子集

issubset(other)¶
set <= other: Test whether every element in the set is in other.

set < other：严格子集: Test whether the set is a proper subset of other, that is, set <= other and set != other

 1 >>> set1={1,2,3,4,5}
 2 >>> set2={1,2,3,4,5,6}
 3 >>> set1.issubset(set2)
 4 True
 5 >>> set3={1,2,3,4,5,6}
 6 >>> set3.issubset(set2)
 7 True
 8 >>> set3<set2
 9 False
10 >>> set1<set2
11 True

View Code

issuperset(other)¶
set >= other: Test whether every element in other is in the set.

set > other：严格父集: Test whether the set is a proper superset of other, that is, set >= other and set != other.


https://www.hackerrank.com/challenges/py-check-strict-superset/problem
解决方法

a = set(input().split())
print(all(a > set(input().split()) for _ in range(int(input()))))