Python入门之set集合

set集合

set是一个无序且不重复的元素集合，有以下优点：

• 1、访问速度快
• 2、解决重复问题

1、set创建：

#字符串拆解形成set集合；其中字符串中重复字符，咋set集合中只会显示一个
>>> s1 = set("chengdd") #2个字符d
>>> s1
{'c', 'g', 'e', 'd', 'n', 'h'}    #只包含一个字符d
#list元素形成set集合，list如果有多个相同元素在集合中只显示一个；其中list元素可以为int/str
>>> s2 = set(["chengd","xrd",99,101,99])
>>> s2
{101, 99, 'chengd', 'xrd'}
#直接{}创建set集合
>>> s3 = {"chengd",999}
>>> s3
{'chengd', 999}

　　注：

• 在list元素创建set集合时，list中不能包list/dict元素；{}创建set集合时不能包含list/dict

  #list元素不能包含list/dict
  >>> s4 = set(["chengd","xrd",99,101,99,[1,2,3]])
  Traceback (most recent call last):
    File "<pyshell#14>", line 1, in <module>
      s4 = set(["chengd","xrd",99,101,99,[1,2,3]])
  TypeError: unhashable type: 'list'
  #{}创建元素不能包含list/dict
  >>> s5 = {1,2,"chengd",[1,2,3]}
  Traceback (most recent call last):
    File "<pyshell#15>", line 1, in <module>
      s5 = {1,2,"chengd",[1,2,3]}
  TypeError: unhashable type: 'list'
  >>> s6 = {1,2,"chengd",{"name":"xrd"}}
  Traceback (most recent call last):
    File "<pyshell#16>", line 1, in <module>
      s6 = {1,2,"chengd",{"name":"xrd"}}
  TypeError: unhashable type: 'dict'

• 字典创建set集合时，集合元素为key

  >>> old_dict = {
      "#1":{ 'hostname':c1, 'cpu_count': 2, 'mem_capicity': 80 },
      "#2":{ 'hostname':c1, 'cpu_count': 2, 'mem_capicity': 80 },
      "#3":{ 'hostname':c1, 'cpu_count': 2, 'mem_capicity': 80 },
  }
  >>> new_dict = {
      "#1":{ 'hostname':c1, 'cpu_count': 2, 'mem_capicity': 800 },
      "#3":{ 'hostname':c1, 'cpu_count': 2, 'mem_capicity': 80 },
      "#4":{ 'hostname':c2, 'cpu_count': 2, 'mem_capicity': 80 },
  }
  >>> print(new_dict.keys())
  dict_keys(['#3', '#4', '#1'])
  >>> old_set = set(old_dict)
  >>> new_set = set(new_dict)
  >>> old_set
  {'#3', '#2', '#1'}
  >>> new_set
  {'#3', '#4', '#1'}

   

2、set常用方法：

• set.add(value)：添加单个元素；添加多个元素报错

  >>> s2
  {101, 99, 'chengd', 'xrd'}
  >>> s2.add("age")
  >>> s2
  {101, 99, 'chengd', 'xrd', 'age'}
  >>> s2.add("ages",123)
  Traceback (most recent call last):
    File "<pyshell#20>", line 1, in <module>
      s2.add("ages",123)
  TypeError: add() takes exactly one argument (2 given)

   

• set.clear()：清空集合元素

  >>> s2
  {101, 99, 'chengd', 'xrd', 'age'}
  >>> s2.clear()
  >>> s2
  set()

• set.copy()：浅拷贝set集合元素

  >>> s1
  {'c', 'g', 'e', 'd', 'n', 'h'}
  >>> s2 = s1.copy()
  >>> s2
  {'h', 'c', 'e', 'g', 'n', 'd'}
  #深浅拷贝和直接赋值区别：http://www.cnblogs.com/chengd/articles/7020282.html

• set1.difference(set2)：对比set1和set2，set1与set2不同的元素将会输出到成为新的set集合。(注意比较顺序)

  >>> s1 = {1,2,3,4,5}
  >>> s2 = {3,4,5,6,7,8}
  >>> s1
  {1, 2, 3, 4, 5}
  >>> s2
  {3, 4, 5, 6, 7, 8}
  >>> s1.difference(s2)
  {1, 2}    #s1在前面则表示s1包含但s2不包含元素
  >>> s2.difference(s1)
  {8, 6, 7}   #s2在前面则表示s2包含但s1不包含元素

  注：set1.difference(set2)的结果可以赋给新的set集合

  >>> s3 = s1.difference(s2)
  >>> s3
  {1, 2}
  >>> s4 = s2.difference(s1)
  >>> s4
  {8, 6, 7}

• set1.difference_update(set2)：对比set1和set2；set1和set2相同的选项将被去除，set1被改变，set2不变

  >>> s1
  {2, 3, 4}
  >>> s2
  {3, 4, 5, 6, 7, 8}
  >>> s1.difference_update(s2)
  >>> s1
  {2}
  >>> s2
  {3, 4, 5, 6, 7, 8}

• set.discard()：删除单个元素

  >>> s1
  {1, 2, 3, 4, 5}
  >>> s1.discard(1)
  >>> s1
  {2, 3, 4, 5}
  >>> s1.discard(5)
  >>> s1
  {2, 3, 4}

• set1.intersection(set2)：取set1和set2交集；结果可以赋给新的集合

  >>> s1
  {2, 3, 4}
  >>> s2
  {3, 4, 5, 6, 7, 8}
  >>> s1.intersection(s2)
  {3, 4}
  >>> s3 = s1.intersection(s2)
  >>> s3
  {3, 4}

• set1.intersection_update(set2)：取set1和set2交集；set1为交集结果

  >>> s1
  {1, 3, 5, 7}
  >>> s2
  {3, 4, 5, 6, 7, 8}
  >>> s1.intersection_update(s2)
  >>> s1
  {3, 5, 7}
  >>> s2
  {3, 4, 5, 6, 7, 8}

• set1.isdisjoint(set2)：判断set1和set2是否有交集，没有返回True

  #有交集返回Flase
  >>> s1
  {3, 5, 7}
  >>> s2
  {3, 4, 5, 6, 7, 8}
  >>> s1.isdisjoint(s2)
  False
  #没有交集返回True
  >>> s1 = {11,12,13}
  >>> s2
  {3, 4, 5, 6, 7, 8}
  >>> s1.isdisjoint(s2)
  True

• set1.issuperset(s2)：判断set1是否是set2父集，是返回真

  >>> s1 = {1,2,3,4,5}
  >>> s2 = {1,2,3}
  >>> s3 = {11,12,13}
  >>> s1.issuperset(s2)
  True
  >>> s1.issuperset(s3)
  False

• set1.issubset()：判断set1是否是set2子集，是返回真

  >>> s1 = {1,2,3}
  >>> s2 = {11,12,13}
  >>> s3 = {1,2,3,4,5,6}
  >>> s1.issubset(s3)
  True
  >>> s2.issubset(s3)
  False

• set1.pop()：从左到右删除元素，并输出结果；也可以直接赋值

  >>> s2
  {11, 12, 13}
  >>> s3
  {1, 2, 3, 4, 5, 6}
  >>> s3.pop()
  1
  >>> s3.pop()
  2
  >>> s3
  {3, 4, 5, 6}
  >>> i = s3.pop()    #赋值
  >>> i
  3
  >>> s2
  {11, 12, 13}

• set.remove(value)：删除指定元素

  >>> s = {1,2,3,4,5,6}
  >>> s.remove(1,3,4)    #一次不能删除多个元素
  Traceback (most recent call last):
    File "<pyshell#117>", line 1, in <module>
      s.remove(1,3,4)
  TypeError: remove() takes exactly one argument (3 given)
  >>> s.remove(14)    #删除不存在元素报错
  Traceback (most recent call last):
    File "<pyshell#118>", line 1, in <module>
      s.remove(14)
  KeyError: 14
  >>> s.remove(4)
  >>> s
  {1, 2, 3, 5, 6}

• set1.symmetric_difference(set2)：取set1和set2差集，可用于赋值新集合

  >>> s1
  {1, 2, 3}
  >>> s2
  {1, 2, 3, 4, 5}
  >>> s1.symmetric_difference(s2)
  {4, 5}
  >>> s2.symmetric_difference(s1)
  {4, 5}

• set1.symmetric_difference_update(set2)：取set1和set2差集，set1为差集结果

  >>> s1
  {1, 2, 3}
  >>> s2
  {1, 2, 3, 4, 5}
  >>> s1.symmetric_difference_update(s2)
  >>> s1
  {4, 5}
  >>> s2
  {1, 2, 3, 4, 5}

• set1.union(set2)：取set1和set2并集，结果可赋值新集合

  >>> s1
  {11, 12, 13}
  >>> s2
  {1, 2, 3, 4, 5}
  >>> s1.union(s2)
  {1, 2, 3, 4, 5, 11, 12, 13}
  >>> s1
  {11, 12, 13}
  >>> s2
  {1, 2, 3, 4, 5}
  >>> s3 = s1.union(s2)
  >>> s3
  {1, 2, 3, 4, 5, 11, 12, 13}

• set1.update(set2)：取set1和set2并集；set1为并集结果

  >>> s1
  {11, 12, 13}
  >>> s2
  {1, 2, 3, 4, 5}
  >>> s1.update(s2)
  >>> s1
  {1, 2, 3, 4, 5, 11, 12, 13}
  >>> s2
  {1, 2, 3, 4, 5}

   

3、常用内置函数

len(set)：查看set元素个数