标签：一个 ror call card ddr 不重复 diff 移除 error
set集合，是一个无序的，且不重复的元素集合
定义方式使用"{}",也可以使用set(iterable)内置函数定义，但iterable参数只能是可迭代对象的对象
>>> set1 = {1,2,3,4,1,3} #定义集合，默认会将重复的去掉
>>> set1
{1, 2, 3, 4}
>>> aa = ‘abcde‘
>>> set2 = set(aa) #使用set函数定义集合
>>> set2
{‘a‘, ‘b‘, ‘d‘, ‘c‘, ‘e‘}
>>> set3 = set(‘python‘)
>>> set3
{‘y‘, ‘p‘, ‘o‘, ‘h‘, ‘n‘, ‘t‘}
>>> set2 = set(range(5)) #如果使用数字必须使用迭代序列
>>> set2
{0, 1, 2, 3, 4}
>>> set2 {1, 2, 3, 4, 5} >>> set2.add(6) >>> set2 {1, 2, 3, 4, 5, 6} >>> set2.add(‘addr‘) >>> set2 {1, 2, 3, 4, 5, 6, ‘addr‘} 2、s.clear()情况所有元素 >>> set2 {1, 2, 3, 4, 5, 6, ‘addr‘} >>> set2.clear() >>> set2 set()
>>> set1 {1, 2, 3, 4, 6, ‘addr‘} >>> set3=set1.copy() >>> set3 {1, 2, 3, 4, 6, ‘addr‘}
返回两个集合差集中s的元素
>>> set1 = {1,2,3,4}
>>> set2 = {3,4,5,6}
>>> set1.difference(set2) #用set2比较set1时，返回set1和set2交集中set1剩余的元素
{1, 2}
>>> set2.difference(set1) #理解同上，返回自身集合中对方集合没有的新集合
{5, 6}
从s集合中移除与b集合交集的元素，并更新到s集合
>>> set1 = {1,2,3,4}
>>> set2 = {3,4,5,6}
>>> set1.difference_update(set2)
>>> set1
{1, 2}
>>> set2.difference_update(set1) #没有交集元素，所以不变
>>> set2
{3, 4, 5, 6}
删除指定元素，如不存在不会报错
>>> set2 {3, 4, 5, 6} >>> set2.discard(6) >>> set2 {3, 4, 5} >>> set2.discard(7) #元素不存在，没有报错 >>> set2 {3, 4, 5}
返回两个或多个集合的交集，并返回
>>> set1 = {1,2,3,4}
>>> set2 = {3,4,5,6}
>>> set1.intersection(set2) #返回交集值
{3, 4}
>>> set3 = {3,6,8,9}
>>> set1.intersection(set2,set3) #必须三个set都存在才返回
{3}
根据当前s与比较b的交集值更新到s集合中
>>> set1 = {1,2,3,4,5,6}
>>> set2 = {3,4,5,7,8,9}
>>> set1.intersection_update(set2) #取交集值更新到自身set
>>> set1
{3, 4, 5}
>>> set2
{3, 4, 5, 7, 8, 9}
如果两个集合没有交集返回True否则返回False
>>> set1 = {1,2,3,4,5}
>>> set2 = {4,5,6,7,8}
>>> set1.isdisjoint(set2)
False
>>> set3 = {22,33}
>>> set1.isdisjoint(set3)
True
s集合元素是否被b集合包含，是则返回True否则返回False
>>> s1 = {1,2}
>>> s2 = {1,2,3,4}
>>> s3 = {11,22}
>>> s1.issubset(s2)
True
>>> s1.issubset(s3)
False
判断s集合元素是否包含了所以b的集合元素，是返回True否则返回False
>>> s1 = {1,2}
>>> s2 = {1,2,3,4}
>>> s2.issuperset(s1)
True
>>> s1.issuperset(s2)
False
集合默认排序后，随机删除一个元素，并返回该元素，默认删除第一个元素
>>> s2 = {‘name‘,‘age‘,‘salary‘,‘shool‘}
>>> s2
{‘salary‘, ‘name‘, ‘shool‘, ‘age‘}
>>> s2.pop()
‘salary‘
>>> s2.pop()
‘name‘
>>> s2.pop()
‘shool‘
移除集合中的指定值，和discard一样，只是remove在值不存在时会报错。
>>> s1 = {1,2,3,4}
>>> s1.remove(2)
>>> s1
{1, 3, 4}
>>> s1.remove(5)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
KeyError: 5
返回两个集合的差集
>>> s1 = {1,2,3,4}
>>> s2 = {3,4,5,6}
>>> s1.symmetric_difference(s2) #返回两个集合不交集的元素
{1, 2, 5, 6}
>>> s2.symmetric_difference(s1)
{1, 2, 5, 6}
将两个集合不交集的元素返回给s
>>> s1 = {1,2,3,4}
>>> s2 = {3,4,5,6}
>>> s1.symmetric_difference_update(s2)
>>> s1
{1, 2, 5, 6}
返回两个集合的并集
>>> s1 = {1,2,3,4}
>>> s2 = {3,4,5,6}
>>> s1.union(s2)
{1, 2, 3, 4, 5, 6}
以集合b扩展s集合
>>> s1 = {1,2,3}
>>> s2 = {4,5,6}
>>> s1.update(s2)
>>> s1
{1, 2, 3, 4, 5, 6}
标签：一个 ror call card ddr 不重复 diff 移除 error
原文地址：http://www.cnblogs.com/zhangxinqi/p/7603021.html
