一、集合的定义
set集合，是一个无序且不重复的元素集合。
集合对象是一组无序排列的可哈希的值，集合成员可以做字典中的键。集合支持用in和not in操作符检查成员，由len()内建函数得到集合的基数(大小)， 用 for 循环迭代集合的成员。但是因为集合本身是无序的，不可以为集合创建索引或执行切片(slice)操作，也没有键(keys)可用来获取集合中元素的值。
二、集合的创建
s = set()s = {11,22,33,44}*注：创建空集合时，只能用set(),如果用第二种方法s={}，创建的实际上是一个空字典。s = {}print(type(s))<class "dict">
a=set("boy")b=set(["y", "b", "o","o"])c=set({"k1":"v1","k2":"v2"})d={"k1","k2","k2"}e={("k1", "k2","k2")}print(a,type(a))print(b,type(b))print(c,type(c))print(d,type(d))print(e,type(e))执行结果如下：{"o", "b", "y"} <class "set">{"o", "b", "y"} <class "set">{"k1", "k2"} <class "set">{"k1", "k2"} <class "set">{("k1", "k2", "k2")} <class "set">
三、集合的功能
源码
基本功能：
增加
a=set("python")a.add("tina")print(a)b=set("python")b.update("tina")print(b)执行结果如下：{"tina", "o", "p", "n", "t", "y", "h"}{"o", "i", "p", "a", "n", "t", "y", "h"}##################由以上代码可以看出，add是单个元素的添加，而update是批量的添加。输出结果是无序的，并非添加到尾部。
删除（remove，discard，pop）
c={"p", "i", "h", "n", "o", "y", "t"}c.remove("p")print(c)
c={"p", "i", "h", "n", "o", "y", "t"}
c.discard("p")
print(c)c={"p", "i", "h", "n", "o", "y", "t"}
c.pop()
print(c)执行结果如下： {"i", "h", "t", "o", "y", "n"}
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当执行c.remove("p","i")和c.discard("p","i")时，报错：TypeError: remove() takes exactly one argument (2 given)，说明remove和discard删除元素时都只能一个一个的删，同add对应。
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remove，pop和discard的区别：
discard删除指定元素，当指定元素不存在时，不报错；
remove删除指定元素，但当指定元素不存在时，报错：KeyError。
pop删除任意元素，并可将移除的元素赋值给一个变量，不能指定元素移除。
清空
c={"p", "i", "h", "n", "o", "y", "t"}c.clear()print(c)执行结果如下：set()
set的特有功能：
s1 = {0}s2 = {i % 2 for i in range(10)}s = set("hi")t = set(["h", "e", "l", "l", "o"])print(s.intersection(t), s & t) # 交集print(s.union(t), s | t) # 并集print(s.difference(t), s - t) # 差集print(s.symmetric_difference(t), s ^ t) # 对称差集print(s1.issubset(s2), s1 <= s2) # 子集（被包含）print(s1.issuperset(s2), s1 >= s2) # 父集（包含）执行结果如下：{"h"} {"h"}{"i", "e", "h", "l", "o"} {"i", "e", "h", "l", "o"}{"i"} {"i"}{"e", "l", "o", "i"} {"e", "l", "o", "i"}True TrueFalse False
s = {11,22,33}t = {22,44}print(s.isdisjoint(t))#（disjoint脱节的，）即如果没有交集，返回True，否则返回Falses.difference_update(t)#将差集覆盖到源集合，即从当前集合中删除和B中相同的元素print(s)执行结果如下：False{33, 11}s = {11,22,33}t = {22,44}s.intersection_update(t)#将交集覆盖到源集合print(s)执行结果如下：{22}s = {11,22,33}t = {22,44}s.symmetric_difference_update(t)#将对称差集覆盖到源集合print(s)执行结果如下：{33, 11, 44}
四、集合的转换
se = set(range(4))li = list(se)tu = tuple(se)st = str(se)print(li,type(li))print(tu,type(tu))print(st,type(st))执行结果如下：[0, 1, 2, 3] <class "list">(0, 1, 2, 3) <class "tuple">{0, 1, 2, 3} <class "str">
五、练习题
寻找差异：哪些需要删除？哪些需要新建？哪些需要更新？
# 数据库中原有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 }} # cmdb 新汇报的数据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 }}
注意：无需考虑内部元素是否改变，只要原来存在，新汇报也存在，就是需要更新
del_dict = set(old_dict).difference(set(new_dict))add_dict = set(new_dict).difference(set(old_dict))update_dict = set(new_dict).intersection(set(old_dict))print(del_dict)print(add_dict)print(update_dict)执行结果如下：{"#2"}{"#4"}{"#3", "#1"}
#!/usr/bin/python# -*- coding:utf-8 -*-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},}new_set = set()old_set = set()for i in new_dict: new_set.add(i)for j in old_dict: old_set.add(j)new_add = new_set.difference(old_set) #new_dict中有，old_dict中没有old_del = old_set.difference(new_set) #old_dict中有，new_dict中没有update = new_set.intersection(old_set) #old_dict和new_dict共同有的，需要把new_dict更新到old_dict中for k in new_add: old_dict[k] = new_dict[k] #将new_dict中新增的内容添加到old_dict中for v in old_del: del old_dict[v] #将old_dict中失效的内容删除for m in update: old_dict[m] = new_dict[m] #把new_dict更新到old_dict中print(old_dict)
