zoukankan      html  css  js  c++  java
  • 集合关系

    list_1 = [1,4,5,7,3,6,7,9]
    list_1 = set(list_1)

    list_2 =set([2,6,0,66,22,8,4])
    print(list_1,list_2)
    '''
    #交集
    print( list_1.intersection(list_2) )

    #并集
    print(list_1.union(list_2))

    #差集 in list_1 but not in list_2
    print(list_1.difference(list_2))
    print(list_2.difference(list_1))

    #子集
    list_3 = set([1,3,7])
    print(list_3.issubset(list_1))
    print(list_1.issuperset(list_3))
    #对称差集
    print(list_1.symmetric_difference(list_2))
    print("-------------")

    #如果两个集合没有交集,则返回true
    list_4 = set([5,6,7,8])
    print(list_3.isdisjoint(list_4)) # Return True if two sets have a null intersection.
    '''


    #交集
    print(list_1 & list_2)
    #union
    print(list_2 | list_1)

    #difference
    print(list_1 - list_2) # in list 1 but not in list 2

    #对称差集
    print(list_1 ^ list_2)

    list_1.add(999)
    list_1.update([888,777,555])
    print(list_1)

    print(list_1.pop())
    print( list_1.discard(888) )
  • 相关阅读:
    [CodeForces
    [CodeChef]RIN(最小割)
    [Bzoj3894]文理分科(最小割)
    [Poj3469]Dual Core CPU(最小割)
    MySQL- 锁(3)
    MySQL- 锁(1)
    MySQL- 锁(2)
    MySQL-中文全文检索
    Solr
    多线程编程-之并发编程:同步容器
  • 原文地址:https://www.cnblogs.com/smlie/p/8099121.html
Copyright © 2011-2022 走看看