我知道如何得到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]
or
def intersect(a, b):
return list(set(a) & set(b))
print intersect(b1, b2)
但当我必须为嵌套列表找到交集时，我的问题就开始了:
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]]
最后我希望收到:
c3 = [[13,32],[7,13,28],[1,6]]
你们能帮我一下吗?
相关的
在python中扁平一个浅列表
当前回答
# 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]] ?
这里有一种不涉及集合的方法来设置c3:
c3 = []
for sublist in c2:
c3.append([val for val in c1 if val in sublist])
但如果你喜欢只用一行，你可以这样做:
c3 = [[val for val in c1 if val in sublist] for sublist in c2]
它是列表推导式中的列表推导式，这有点不寻常，但我认为你应该不会有太多的问题。
其他回答
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(i) & set(c1)) for i in c2]
c3
[[32, 13], [28, 13, 7], [1, 6]]
对我来说，这是非常优雅和快速的方法。
考虑到:
> 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]]
我发现下面的代码工作得很好，如果使用set操作可能会更简洁:
> c3 = [list(set(f)&set(c1)) for f in c2]
它有:
> [[32, 13], [28, 13, 7], [1, 6]]
如需订购:
> c3 = [sorted(list(set(f)&set(c1))) for f in c2]
我们有:
> [[13, 32], [7, 13, 28], [1, 6]]
顺便说一下，对于更python的风格，这个也很好:
> c3 = [ [i for i in set(f) if i in c1] for f in c2]
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]]
下面是Python 2的解决方案:
c3 = [filter(lambda x: x in c1, sublist) for sublist in c2]
在Python 3中，filter返回一个可迭代对象而不是list，所以你需要用list()来包装filter调用:
c3 = [list(filter(lambda x: x in c1, sublist)) for sublist in c2]
解释:
过滤器部分获取每个子列表的项并检查它是否在源列表c1中。 对c2中的每个子列表执行列表推导式。
你认为[1,2]与[1，[2]]相交吗?也就是说，你只关心数字，还是也关心列表结构?
如果只有数字，研究如何“扁平化”列表，然后使用set()方法。
你应该使用这段代码(来自http://kogs-www.informatik.uni-hamburg.de/~meine/python_tricks)，这段代码未经测试，但我很确定它是有效的:
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
在你平摊了列表之后，你用通常的方式执行交叉:
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))
