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  • set集合

    set

    set是一个无序且不重复的元素集合

    set具有的方法

    1.add(self, *args, **kwargs)  添加

    s = set([1,2,3,4,5])
    s.add(6)
    print(s)
    {1, 2, 3, 4, 5, 6}

    2.clear(self, *args, **kwargs)  清空

    s = set([1,2,3,4,5])
    s.clear()
    print(s)

    3.copy(self, *args, **kwargs)  拷贝

    s = set([1,2,3,4,5])
    a = s.copy()
    print(a)
    {1, 2, 3, 4, 5}

    4.difference(self, *args, **kwargs)  

    s = set([1,2,3,4,5,2])
    print(s)
    a = s.difference([1,2,])
    print(a)
    {1, 2, 3, 4, 5}
    {3, 4, 5}

    5.difference_update(self, *args, **kwargs)  删除当前set中的所有包含在 new set 里的元素

    s = set([1,2,3,4,5,2])
    a = s.difference_update([2])
    print(s)
    print(a)
    {1, 3, 4, 5}
    None

    6.discard(self, *args, **kwargs)  移除元素

    s = set([1,2,3,4,5,2])
    s.discard(1)
    print(s)

    7.intersection(self, *args, **kwargs)  取交集,新创建一个set

    s = set([1,2,3,4,5,2])
    a = s.intersection([2,3,4])
    print(a)
    {2, 3, 4}

    8.intersection_update(self, *args, **kwargs)  取交集,修改原来set

    s = set([1,2,3,4,5,2])
    s.intersection_update([2,3,4])
    print(s)
    {2, 3, 4}

    9.isdisjoint(self, *args, **kwargs)  如果没有交集,返回true

    s = set([1,2,3,4,5,2])
    a = s.isdisjoint([6,7])
    print(a)
    True

    10.issubset(self, *args, **kwargs)  是否是子集

    s = set([1,2,3,4,5,])
    a = s.issubset([1,2,3,4,5,6])
    print(a)
    True

    11.issuperset(self, *args, **kwargs)  是否是父集

    s = set([1,2,3,4,5,])
    a = s.issuperset([1,2,3,])
    print(a)
    True

    12. pop(self, *args, **kwargs)  移除

    s = set([1,2,3,4,5,])
    s.pop()
    print(s)
    {2, 3, 4, 5}

    13.remove(self, *args, **kwargs)  移除

    s = set([1,2,3,4,5,])
    s.remove(3)
    print(s)
    {1, 2, 4, 5}

    14.symmetric_difference(self, *args, **kwargs)  差集,创建新对象

    s = set([1,2,3,4,5,])
    a = s.symmetric_difference([1,2,3])
    print(a)
    {4, 5}

    15.symmetric_difference_update(self, *args, **kwargs)  差集,改变原来

    s = set([1,2,3,4,5,])
    s.symmetric_difference_update([1,2,3])
    print(s)

    16.union(self, *args, **kwargs)  并集

    s = set([1,2,3,4,5,])
    a = s.union([1,2,3,6])
    print(a)
    {1, 2, 3, 4, 5, 6}

    17.update(self, *args, **kwargs)  更新

    s = set([1,2,3,4,5,])
    s.update([1,2,7,8])
    print(s)
    {1, 2, 3, 4, 5, 7, 8}

     set实例

     1 #练习:寻找差异
     2 # 数据库中原有
     3 old_dict = {
     4     "#1":{ 'hostname':'c1', 'cpu_count': 2, 'mem_capicity': 80 },
     5     "#2":{ 'hostname':'c1', 'cpu_count': 2, 'mem_capicity': 80 },
     6     "#3":{ 'hostname':'c1', 'cpu_count': 2, 'mem_capicity': 80 }
     7 }
     8 
     9 # cmdb 新汇报的数据
    10 new_dict = {
    11     "#1":{ 'hostname':'c1', 'cpu_count': 2, 'mem_capicity': 800 },
    12     "#3":{ 'hostname':'c1', 'cpu_count': 2, 'mem_capicity': 80 },
    13     "#4":{ 'hostname':'c2', 'cpu_count': 2, 'mem_capicity': 80 }
    14 }
    15 old_set = set(old_dict.keys())
    16 update_list = list(old_set.intersection(new_dict.keys()))
    17 
    18 new_list = []
    19 del_list = []
    20 
    21 for i in new_dict.keys():
    22     if i not in update_list:
    23         new_list.append(i)
    24 
    25 for i in old_dict.keys():
    26     if i not in update_list:
    27         del_list.append(i)
    28 
    29 print(update_list,new_list,del_list)
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  • 原文地址:https://www.cnblogs.com/yoyovip/p/5596596.html
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