find两个列表的交集？
我知道如何获得两个平面列表的交集：
b1 = [1,2,3,4,5,9,11,15] b2 = [4,5,6,7,8] b3 = [val for val in b1 if val in b2] 要么
def intersect(a, b): return list(set(a) & set(b)) print intersect(b1, b2)
但是，当我必须find嵌套列表的交集，然后我的问题开始：
c1 = [1, 6, 7, 10, 13, 28, 32, 41, 58, 63] c2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]]
最后我想收到：
c3 = [[13,32],[7,13,28],[1,6]]
你们能帮我一把吗？
有关
- 在python中展平浅表
如果你想：
c1 = [1, 6, 7, 10, 13, 28, 32, 41, 58, 63] c2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]] c3 = [[13, 32], [7, 13, 28], [1,6]]
那么这里是你的Python 2的解决scheme：
c3 = [filter(lambda x: x in c1, sublist) for sublist in c2]
在Python 3中， filter返回一个iterable而不是list ，所以你需要用list()包装filter调用：
c3 = [list(filter(lambda x: x in c1, sublist)) for sublist in c2]
说明：
filter部分获取每个子列表项并检查它是否在源列表c1中。 列表理解是针对c2中的每个子列表执行的。
你不需要定义交集。 这已经是一套一stream的一部分了。
>>> b1 = [1,2,3,4,5,9,11,15] >>> b2 = [4,5,6,7,8] >>> set(b1).intersection(b2) set([4, 5])
对于只想查找两个列表交集的人来说，Asker提供了两种方法：
b1 = [1,2,3,4,5,9,11,15] b2 = [4,5,6,7,8] b3 = [val for val in b1 if val in b2]和
def intersect(a, b): return list(set(a) & set(b)) print intersect(b1, b2)
但是有一种更高效的混合方法，因为你只需要在list / set之间进行一次转换，而不是三次转换：
b1 = [1,2,3,4,5] b2 = [3,4,5,6] s2 = set(b2) b3 = [val for val in b1 if val in s2]
这将在O（n）中运行，而他涉及列表理解的原始方法将运行在O（n ^ 2）
纯粹的列表理解版本
>>> c1 = [1, 6, 7, 10, 13, 28, 32, 41, 58, 63] >>> c2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]] >>> c1set = frozenset(c1)
展平变体：
>>> [n for lst in c2 for n in lst if n in c1set] [13, 32, 7, 13, 28, 1, 6]
嵌套变体：
>>> [[n for n in lst if n in c1set] for lst in c2] [[13, 32], [7, 13, 28], [1, 6]]
function方法：
input_list = [[1, 2, 3, 4, 5], [2, 3, 4, 5, 6], [3, 4, 5, 6, 7]] result = reduce(set.intersection, map(set, input_list))
它可以应用于更一般的1+列表的情况
＆运算符取两个交集。
{1,2,3}＆{2,3,4}出[1]：{2,3}
你应该使用这个代码扁平化（取自http://kogs-www.informatik.uni-hamburg.de/~meine/python_tricks ），代码是未经testing的，但我很确定它的工作原理：
def flatten(x): """flatten(sequence) -> list Returns a single, flat list which contains all elements retrieved from the sequence and all recursively contained sub-sequences (iterables). Examples: >>> [1, 2, [3,4], (5,6)] [1, 2, [3, 4], (5, 6)] >>> flatten([[[1,2,3], (42,None)], [4,5], [6], 7, MyVector(8,9,10)]) [1, 2, 3, 42, None, 4, 5, 6, 7, 8, 9, 10]""" result = [] for el in x: #if isinstance(el, (list, tuple)): if hasattr(el, "__iter__") and not isinstance(el, basestring): result.extend(flatten(el)) else: result.append(el) return result
在列表完成之后，以通常的方式执行交叉点：
c1 = [1, 6, 7, 10, 13, 28, 32, 41, 58, 63] c2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]] def intersect(a, b): return list(set(a) & set(b)) print intersect(flatten(c1), flatten(c2))
由于intersect被定义，基本的列表理解就足够了：
>>> c3 = [intersect(c1, i) for i in c2] >>> c3 [[32, 13], [28, 13, 7], [1, 6]]
S. Lott的评论和TM的相关评论得到了改进：
>>> c3 = [list(set(c1).intersection(i)) for i in c2] >>> c3 [[32, 13], [28, 13, 7], [1, 6]]
2列表交叉的pythonic方式是：
[x for x in list1 if x in list2]
你认为[1,2]与[1, [2]]相交吗？ 也就是说，它只是你关心的数字，还是列表结构呢？
如果只有数字，调查如何“扁平化”列表，然后使用set()方法。
我不知道我是否迟到回答你的问题。 读完你的问题后，我想出了一个函数intersect（），可以在列表和嵌套列表上工作。 我用recursion来定义这个函数，这很直观。 希望这是你在找什么：
def intersect(a, b): result=[] for i in b: if isinstance(i,list): result.append(intersect(a,i)) else: if i in a: result.append(i) return result
例：
>>> c1 = [1, 6, 7, 10, 13, 28, 32, 41, 58, 63] >>> c2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]] >>> print intersect(c1,c2) [[13, 32], [7, 13, 28], [1, 6]] >>> b1 = [1,2,3,4,5,9,11,15] >>> b2 = [4,5,6,7,8] >>> print intersect(b1,b2) [4, 5]
鉴于：
> c1 = [1, 6, 7, 10, 13, 28, 32, 41, 58, 63] > c2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]]
我发现下面的代码运行良好，如果使用set操作可能更简洁：
> c3 = [list(set(f)&set(c1)) for f in c2]
它得到：
> [[32, 13], [28, 13, 7], [1, 6]]
如果需要订购：
> c3 = [sorted(list(set(f)&set(c1))) for f in c2]
我们有：
> [[13, 32], [7, 13, 28], [1, 6]]
顺便说一句，对于更多的python风格，这个也很好：
> c3 = [ [i for i in set(f) if i in c1] for f in c2]
我也正在寻找一种方法来做到这一点，最终结果如下：
def compareLists(a,b): removed = [x for x in a if x not in b] added = [x for x in b if x not in a] overlap = [x for x in a if x in b] return [removed,added,overlap]
要定义正确考虑元素的基数的交集，请使用Counter ：
from collections import Counter >>> c1 = [1, 2, 2, 3, 4, 4, 4] >>> c2 = [1, 2, 4, 4, 4, 4, 5] >>> list((Counter(c1) & Counter(c2)).elements()) [1, 2, 4, 4, 4]
c1 = [1, 6, 7, 10, 13, 28, 32, 41, 58, 63] c2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]] c3 = [list(set(c2[i]).intersection(set(c1))) for i in xrange(len(c2))] c3 ->[[32, 13], [28, 13, 7], [1, 6]]
我们可以使用这个设置方法：
c1 = [1, 6, 7, 10, 13, 28, 32, 41, 58, 63] c2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]] result = [] for li in c2: res = set(li) & set(c1) result.append(list(res)) print result
# Problem: Given c1 and c2: c1 = [1, 6, 7, 10, 13, 28, 32, 41, 58, 63] c2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]] # how do you get c3 to be [[13, 32], [7, 13, 28], [1, 6]] ?
这里有一个设置c3 ，不涉及集合：
c3 = [] for sublist in c2: c3.append([val for val in c1 if val in sublist])
但是，如果您只想使用一条线，则可以这样做：
c3 = [[val for val in c1 if val in sublist] for sublist in c2]
这是列表理解中的列表理解，这有点不同寻常，但我认为你不应该有太多的麻烦。
