• set集合(set是一个无序且不重复的元素集合)


    功能:

    一、可以去掉列表,元组中的重复项

    二、可以求交集,合集,差集等

     def add(self, *args, **kwargs):

          """ 添加 """

    s1 = (1,2,3,4,2,3,1)
    s3 = set(s1)
    s4 = s3.add(5)
    print(s1)
    print(s3)
    print(s4)

    (1, 2, 3, 4, 2, 3, 1)
    {1, 2, 3, 4, 5}
    None

    def clear(self, *args, **kwargs):
            '''清除所有元素'''
    s1 = (1,2,3,4,2,3,1)
    s3 = set(s1)
    s4 = s3.clear()
    print(s1)
    print(s3)
    print(s4)

    (1, 2, 3, 4, 2, 3, 1)
    set()
    Non

    def copy(self, *args, **kwargs):
            '''复制一个set'''
    s1 = (1,2,3,4,2,3,1)
    s3 = set(s1)
    s4 = s3.copy()
    print(s1)
    print(s3)
    print(s4)

    (1, 2, 3, 4, 2, 3, 1)
    {1, 2, 3, 4}
    {1, 2, 3, 4}

    def difference(self, *args, **kwargs):
            '''a.difference(b) a和b求差集,显示a的结果'''
    s1 = (1,2,3,4,2,3,1)
    s2 = (3,4,5,7,1)
    s5 = set(s2)
    s3 = set(s1)
    s4 = s3.difference(s2)
    print(s3)
    print(s5)
    print(s4)

    {1, 2, 3, 4}
    {1, 3, 4, 5, 7}
    {2}

    def difference_update(self, *args, **kwargs):
          '''删除当前a中的所有包含在 b 里的元素,在原有集合中直接删除'''
    s1 = (1,2,3,4,2,3,1)
    s2 = (3,4,5,7,1)
    s5 = set(s2)
    s3 = set(s1)
    print(s3)
    s4 = s3.difference_update(s2)
    print(s3)
    print(s5)
    print(s4)

    {1, 2, 3, 4}
    {2}
    {1, 3, 4, 5, 7}
    None

    def discard(self, *args, **kwargs):
          '''删除集合中的元素,直接在原有集合中删除'''
    s1 = (1,2,3,4,2,3,1)
    s2 = (3,4,5,7,1)
    s5 = set(s2)
    s3 = set(s1)
    print(s3)
    s4 = s3.discard(2)
    print(s3)
    print(s5)
    print(s4)
    {1, 2, 3, 4}
    {1, 3, 4}
    {1, 3, 4, 5, 7}
    None
    def intersection(self, *args, **kwargs):
            """ 取交集,新创建一个set """
    s1 = (1,2,3,4,2,3,1)
    s2 = (3,4,5,7,1)
    s5 = set(s2)
    s3 = set(s1)
    print(s3)
    print(s5)
    s4 = s3.intersection(s5)
    print(s3)
    print(s4)

    {1, 2, 3, 4}
    {1, 3, 4, 5, 7}
    {1, 2, 3, 4}
    {1, 3, 4}


    def intersection_update(self, *args, **kwargs):
          '''取交集,修改原来set'''
    s1 = (1,2,3,4,2,3,1)
    s2 = (3,4,5,7,1)
    s5 = set(s2)
    s3 = set(s1)
    print(s3)
    print(s5)
    s4 = s3.intersection_update(s5)
    print(s3)
    print(s4)

    {1, 2, 3, 4}
    {1, 3, 4, 5, 7}
    {1, 3, 4}
    None


    def isdisjoint(self, *args, **kwargs):
            '''如果没有交集,返回true'''
    s1 = (1,2,3,4,2,3,1)
    s2 = (3,4,5,7,1)
    s5 = set(s2)
    s3 = set(s1)
    print(s3)
    print(s5)
    s4 = s3.isdisjoint(s5)
    print(s3)
    print(s4)
    {1, 2, 3, 4}
    {1, 3, 4, 5, 7}
    {1, 2, 3, 4}
    False

    def issubset(self, *args, **kwargs):
              '''是否是子集'''
    def issuperset(self, *args, **kwargs):
               """ 是否是父集 """
      
    
    
    def pop(self, *args, **kwargs):
              '''移除'''
    def remove(self, *args, **kwargs):

               """ 移除 """
       
    def symmetric_difference(self, *args, **kwargs): 
                  """ 差集,创建新对象"""
    def symmetric_difference_update(self, *args, **kwargs):
               """ 差集,改变原来 """
       
    def union(self, *args, **kwargs): 
                    """ 并集 """
    def update(self, *args, **kwargs): 
                    """ 更新 """
     
     

          

      

      
     

      

        

          

          

          
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  • 原文地址:https://www.cnblogs.com/xianyin/p/5645282.html
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