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

 >>> s = set('HackerRank')
 >>> s.add('H')
 >>> print s
 set(['a', 'c', 'e', 'H', 'k', 'n', 'r', 'R'])
 >>> print s.add('HackerRank')#添加后返回none
 None
 >>> print s
 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()

 >>> s = set([])
 >>> print s.pop()

 >>> print s
 set([])
 >>> print s.pop()
 KeyError: pop from an empty set

set.pop()

set.remove()

 >>> s = set([, , , , , , , , ])
 >>> s.remove()
 >>> print s
 set([, , , , , , , ])
 >>> print s.remove()#默认返回None
 None
 >>> print s
 set([, , , , , , ])
 >>> s.remove()
 KeyError: 

set.remove()

set.discard()

 >>> s = set([, , , , , , , , ])
 >>> s.discard()
 >>> print s
 set([, , , , , , , ])
 >>> print s.discard()
 None
 >>> print s
 set([, , , , , , ])
 >>> s.discard()
 >>> print s
 set([, , , , , , ])

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()

 >>> s = set("Hacker")
 >>> print s.intersection("Rank")
 set(['a', 'k'])

 >>> print s.intersection(['R', 'a', 'n', 'k'])
 set(['a', 'k'])

 >>> print s.intersection(enumerate(['R', 'a', 'n', 'k']))
 set([])

 >>> print s.intersection({"Rank":})
 set([])

 >>> s & set("Rank")
 set(['a', 'k'])

set.intersection()

set.difference()

 #set1有的而set2没有的元素
 >>> s = set("Hacker")
 >>> print s.difference("Rank")
 set(['c', 'r', 'e', 'H'])

 >>> print s.difference(set(['R', 'a', 'n', 'k']))
 set(['c', 'r', 'e', 'H'])

 >>> print s.difference(['R', 'a', 'n', 'k'])
 set(['c', 'r', 'e', 'H'])

 >>> print s.difference(enumerate(['R', 'a', 'n', 'k']))
 set(['a', 'c', 'r', 'e', 'H', 'k'])

 >>> print s.difference({"Rank":})
 set(['a', 'c', 'e', 'H', 'k', 'r'])

 >>> s - set("Rank")
 set(['H', 'c', 'r', 'e'])

set.difference()

set.unoin()

 >>> s = set("Hacker")
 >>> print s.union("Rank")
 set(['a', 'R', 'c', 'r', 'e', 'H', 'k', 'n'])

 >>> print s.union(set(['R', 'a', 'n', 'k']))
 set(['a', 'R', 'c', 'r', 'e', 'H', 'k', 'n'])

 >>> print s.union(['R', 'a', 'n', 'k'])
 set(['a', 'R', 'c', 'r', 'e', 'H', 'k', 'n'])

 >>> print s.union(enumerate(['R', 'a', 'n', 'k']))
 set(['a', 'c', 'r', 'e', (, 'a'), (, 'n'), 'H', 'k', (, 'k'), (, 'R')])

 >>> print s.union({"Rank":})
 set(['a', 'c', 'r', 'e', 'H', 'k', 'Rank'])

 >>> s | set("Rank")
 set(['a', 'R', 'c', 'r', 'e', 'H', 'k', 'n'])

set.union()

 set.symmetric_difference()

 >>> s = set("Hacker")
 >>> print s.symmetric_difference("Rank")
 set(['c', 'e', 'H', 'n', 'R', 'r'])

 >>> print s.symmetric_difference(set(['R', 'a', 'n', 'k']))
 set(['c', 'e', 'H', 'n', 'R', 'r'])

 >>> print s.symmetric_difference(['R', 'a', 'n', 'k'])
 set(['c', 'e', 'H', 'n', 'R', 'r'])

 >>> print s.symmetric_difference(enumerate(['R', 'a', 'n', 'k']))
 set(['a', 'c', 'e', 'H', (, 'R'), 'r', (, 'n'), 'k', (, 'a'), (, 'k')])

 >>> print s.symmetric_difference({"Rank":})
 set(['a', 'c', 'e', 'H', 'k', 'Rank', 'r'])

 >>> s ^ set("Rank")
 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

 >> myset.update([, , , ]) # update() only works for iterable objects
 >> myset
 {'a', , 'c', 'b', , , (, ), }
 >> myset.update({, , })
 >> myset
 {'a', , 'c', 'b', , , , , (, ), }
 >> myset.update({, }, [, ])
 >> myset
 {'a', , 'c', 'b', , , , , , , (, ), , }

set.update()

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

 #用A,B的差集来更新A
 >>> H = set("Hacker")
 >>> R = set("Rank")
 >>> H.difference_update(R)
 >>> print H
 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

 #用A,b互相差集来更新A
 >>> H = set("Hacker")
 >>> R = set("Rank")
 >>> H.symmetric_difference_update(R)
 >>> print H
 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。

s1,s2={,,},{,,}s1.isdisjoint(s2)True

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

 >>> set1={,,,,}
 >>> set2={,,,,,}
 >>> set1.issubset(set2)
 True
 >>> set3={,,,,,}
 >>> set3.issubset(set2)
 True
 >>> set3<set2
 False
 >>> set1<set2
 True

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()))))