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'])

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([1])
>>> print s.pop()
1
>>> print s
set([])
>>> print s.pop()
KeyError: pop from an empty set

set.remove()

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

set.discard()

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

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":1})
set([])

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

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":1})
set(['a', 'c', 'e', 'H', 'k', 'r'])

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

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', (1, 'a'), (2, 'n'), 'H', 'k', (3, 'k'), (0, 'R')])

>>> print s.union({"Rank":1})
set(['a', 'c', 'r', 'e', 'H', 'k', 'Rank'])

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

 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', (0, 'R'), 'r', (2, 'n'), 'k', (1, 'a'), (3, 'k')])

>>> print s.symmetric_difference({"Rank":1})
set(['a', 'c', 'e', 'H', 'k', 'Rank', 'r'])

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

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

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.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'])

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

set.isdisjoint():如果两个集合没有交集则返回True。

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

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={1,2,3,4,5}
>>> set2={1,2,3,4,5,6}
>>> set1.issubset(set2)
True
>>> set3={1,2,3,4,5,6}
>>> 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()))))