# 创建集合AA={1,2,3}# 创建集合BB={2,3,4}# 输出集合A和Bprint("集合A:",A)print("集合B:",B)# 使用intersection()方法判断交集intersectionAB=A.intersection(B)# 使用&运算符判断交集intersectionAB_operator=A&B# 输出判断结果print("集合A和集合B的交集(使用intersection方法):",intersectionAB)pri...
set2 = {3,4,5}# 使用 | 运算符计算并集union_set_operator = set1 | set2print(union_set_operator)# 输出: {1, 2, 3, 4, 5}# 使用 |= 运算符更新集合为并集set1 |= set2print(set1)# 输出: {1, 2, 3, 4, 5}# 使用update()方法set1.update(set2)print(set1)# 输出: {1, ...
返回set “s”的一个浅复制 请注意:union(), intersection(), difference() 和 symmetric_difference() 的非运算符(non-operator,就是形如 s.union()这样的)版本将会接受任何 iterable 作为参数。相反,它们的运算符版本(operator based counterparts)要求参数必须是 sets。这样可以避免潜在的错误,如:为了更可读而...
D = {100,200,300}print(A.intersection(D))print(B.intersection(D))print(C.intersection(D)) print(A.intersection(B, C, D)) Run Code Output {100} {200} {300} set() Example 3: Set Intersection Using & operator You can also find the intersection of sets using&operator. A = {100,7...
# 计算交集使用运算符 &intersection_result=set_a&set_b&set_c# 打印结果print("The intersection using '&' operator is:",intersection_result) 1. 2. 3. 4. 5. 执行此代码得到的输出也将是相同的结果: The intersection using '&' operator is: {3} ...
1、set:集合/设置 2、add:添加 3、update:更新 4、discard:丢弃 5、intersection:相交 6、union:联合 7、difference:差数 8、symmetric:对称 9、in:在…里面 10、not:不/不是 11、disjoint:不相交 12、subset:子集 13、superset:父集/超集 14、copy:复制 ...
& operatorprint (A.intersection(B))#Output:{2,4}print (A&B)#Output:{2,4}示例2：找到两个以上的交集A={1,2,3,4,5}B={2,4,6,8,10}C={2,4}print (A&B&C)#Output:{2,4}print (A.intersection(B,C))#Output:{2,4}intersection()方法和&运算符之间的区别：· intersection()：接受...
相反,它们的运算符版本(operator based counterparts)要求参数必须是 sets。这样可以避免潜在的错误,如:为了更可读而使用 set('abc') & 'cbs' 来替代 set('abc').intersection('cbs')。从 2.3.1 版本中做的更改:以前所有参数都必须是 sets。 另外,Set 和 ImmutableSet 两者都支持 set 与 set 之间的比较。
5. What if you want to check for combinations of set values? Suppose that you want to find any drink that has orange juice or vermouth? Let’s use the set intersection operator, which is an ampersand (&): >>> for name, contents in drinks.items(): ... if contents & {'vermouth',...
Join 3 sets, and return a set with items that is present in all 3 sets: x ={"a","b","c"} y = {"c","d","e"} z = {"f","g","c"} result = x.intersection(y, z) print(result) Try it Yourself » Example Join 3 sets with the&operator: ...