跟老齐学Python之集合的关系

冻结的集合

前面一节讲述了集合的基本概念，注意，那里所涉及到的集合都是可原处修改的集合。还有一种集合，不能在原处修改。这种集合的创建方法是：

>>>f_set=frozenset("qiwsir")#看这个名字就知道了frozen，冻结的set
>>>f_set
frozenset(["q","i","s","r","w"])
>>>f_set.add("python")#报错
Traceback(mostrecentcalllast):
File"<stdin>",line1,in<module>
AttributeError:"frozenset"objecthasnoattribute"add"

>>>a_set=set("github")#对比看一看，这是一个可以原处修改的set
>>>a_set
set(["b","g","i","h","u","t"])
>>>a_set.add("python")
>>>a_set
set(["b","g","i","h","python","u","t"])

集合运算

先复习一下中学数学（准确说是高中数学中的一点知识）中关于集合的一点知识，主要是唤起那痛苦而青涩美丽的回忆吧，至少对我是。

元素与集合的关系

元素是否属于某个集合。

>>>aset
set(["h","o","n","p","t","y"])
>>>"a"inaset
False
>>>"h"inaset
True

集合与集合的纠结

假设两个集合A、B

A是否等于B，即两个集合的元素完全一样
在交互模式下实验

>>>a
set(["q","i","s","r","w"])
>>>b
set(["a","q","i","l","o"])
>>>a==b
False
>>>a!=b
True

A是否是B的子集，或者反过来，B是否是A的超集。即A的元素也都是B的元素，但是B的元素比A的元素数量多。
实验一下

>>>a
set(["q","i","s","r","w"])
>>>c
set(["q","i"])
>>>c<a#c是a的子集
True
>>>c.issubset(a)#或者用这种方法，判断c是否是a的子集
True
>>>a.issuperset(c)#判断a是否是c的超集
True

>>>b
set(["a","q","i","l","o"])
>>>a<b#a不是b的子集
False
>>>a.issubset(b)#或者这样做
False

A、B的并集，即A、B所有元素，如下图所示

>>>a
set(["q","i","s","r","w"])
>>>b
set(["a","q","i","l","o"])
>>>a|b#可以有两种方式，结果一样
set(["a","i","l","o","q","s","r","w"])
>>>a.union(b)
set(["a","i","l","o","q","s","r","w"])

A、B的交集，即A、B所公有的元素，如下图所示

>>>a
set(["q","i","s","r","w"])
>>>b
set(["a","q","i","l","o"])
>>>a&b#两种方式，等价
set(["q","i"])
>>>a.intersection(b)
set(["q","i"])

我在实验的时候，顺手敲了下面的代码，出现的结果如下，看官能解释一下吗？（思考题）

>>>aandb
set(["a","q","i","l","o"])

A相对B的差（补），即A相对B不同的部分元素，如下图所示

>>>a
set(["q","i","s","r","w"])
>>>b
set(["a","q","i","l","o"])
>>>a-b
set(["s","r","w"])
>>>a.difference(b)
set(["s","r","w"])

-A、B的对称差集，如下图所示

>>>a
set(["q","i","s","r","w"])
>>>b
set(["a","q","i","l","o"])
>>>a.symmetric_difference(b)
set(["a","l","o","s","r","w"])

以上是集合的基本运算。在编程中，如果用到，可以用前面说的方法查找。不用死记硬背。