set1={1,2,3,4,5}set2={4,5,6,7,8}intersection_set=set1.intersection(set2)print(intersection_set) 1. 2. 3. 4. Output: {4, 5} 1. In the above code, we have two setsset1andset2. Theintersectionfunction is called onset1and passedset2as an argument. The resulting setintersectio...
| | (i.e. all elements that are in this set but not the others.) | | intersection(...) | Return the intersection of two sets as a new set. | | (i.e. all elements that are in both sets.) | | isdisjoint(...) | Return True if two sets have a null intersection. | | ...
52defintersection_update(self, *args, **kwargs):#real signature unknown53"""Update a set with the intersection of itself and another."""54pass55 56defisdisjoint(self, *args, **kwargs):#real signature unknown57"""Return True if two sets have a null intersection."""58pass59 60defissubse...
p_s= set(python_1)#把列表转换为集合l_s = set(linux_1)#把列表转换为集合print(p_s,l_s)print(p_s.intersection(l_s))#可以用集合的intersection方法print(p_s&l_s)#用&也是一样的效果 这是求交集的符号 &C:\python35\python3.exe D:/pyproject/day12列表/set-集合.py {'ggq','mr','ytj...
因为set 是一个无序不重复元素集,因此,两个 set 可以做数学意义上的 union(并集), intersection(交集), difference(差集) 等操作。 例子: 代码语言:javascript 代码运行次数:0 运行 AI代码解释 set1=set('hello') set2=set(['p','y','y','h','o','n']) print(set1) print(set2) # 交集 (求...
Return the intersection of two sets as a new set. 交集 (i.e. all elements that are in both sets.) """ pass defintersection_update(self, *args, **kwargs): # real signature unknown """ Update a set with the intersection of itself and another. 取交集并更更新到A中 """ ...
使用集合的交集函数。a_set.intersection(b_set)如果至少有一个元素是公共的,则返回正整数,否则返回0。因此,我们在set_a中插入a,在set_b中插入b,然后检查a_set.intersection(b_set),并根据值返回。 defcommon_member(a,b):a_set=set(a)b_set=set(b)iflen(a_set.intersection(b_set))>0:return(True...
Return a set that contains the items that exist in both setx, and sety: x ={"apple","banana","cherry"} y = {"google","microsoft","apple"} z = x.intersection(y) print(z) Try it Yourself » Definition and Usage Theintersection()method returns a set that contains the similarity be...
array1=[1,2,3,4,5]array2=[4,5,6,7,8]set1=set(array1)set2=set(array2)common_elements=set1.intersection(set2)print("Common elements:",list(common_elements)) 1. 2. 3. 4. 5. 6. 7. 8. 3. 列表推导式 列表推导式是Python中一种简洁而强大的构建列表的方式,可以快速对比两个数组并...
intersection() & Returns a set, that is the intersection of two other sets intersection_update() &= Removes the items in this set that are not present in other, specified set(s) isdisjoint() Returns whether two sets have a intersection or not issubset() <= Returns whether another set con...