Python列表过滤:从列表列表中删除子集

Question

使用Python如何通过有序子集匹配减少列表列表[[..],[..],..]？

在此问题的上下文中,列表L是列表的子集,M如果M包含所有成员L,并且顺序相同.例如,列表[1,2]是列表[1,2,3]的子集,但不是列表[2,1,3]的子集.

输入示例:

a. [[1, 2, 4, 8], [1, 2, 4, 5, 6], [1, 2, 3], [2, 3, 21], [1, 2, 3, 4], [1, 2, 3, 4, 5, 6, 7]]
b. [[2, 16, 17], [1, 2, 3, 4, 5, 6, 7], [1], [1, 2, 3, 4], [1, 2], [17, 18, 19, 22, 41, 48], [2, 3], [1, 2, 3], [50, 69], [1, 2, 3], [2, 3, 21], [1, 2, 3], [1, 2, 4, 8], [1, 2, 4, 5, 6]]

预期结果:

a. [[1, 2, 4, 8], [2, 3, 21], [1, 2, 3, 4, 5, 6, 7]]
b. [[2, 16, 17], [1, 2, 3, 4, 5, 6, 7], [17, 18, 19, 22, 41, 48], [50, 69],  [2, 3, 21], [1, 2, 4, 8], [1, 2, 4, 5, 6]]

更多例子:

L = [[1, 2, 3, 4, 5, 6, 7], [1, 2, 5, 6]] - 没有减少

L = [[1, 2, 3, 4, 5, 6, 7], [1, 2, 3],[1, 2, 4, 8]]- 是减少

L = [[1, 2, 3, 4, 5, 6, 7], [7, 6, 5, 4, 3, 2, 1]] - 没有减少

(很抱歉导致与不正确的数据集混淆.)

Answer 1

这可以简化,但是:

l = [[1, 2, 4, 8], [1, 2, 4, 5, 6], [1, 2, 3], [2, 3, 21], [1, 2, 3, 4], [1, 2, 3, 4, 5, 6, 7]]
l2 = l[:]

for m in l:
    for n in l:
        if set(m).issubset(set(n)) and m != n:
            l2.remove(m)
            break

print l2
[[1, 2, 4, 8], [2, 3, 21], [1, 2, 3, 4, 5, 6, 7]]

Answer 2

这段代码应该具有相当的内存效率。除了存储您的初始列表列表之外，此代码使用的额外内存可忽略不计（不会创建列表的临时集或副本）。

def is_subset(needle,haystack):
   """ Check if needle is ordered subset of haystack in O(n)  """

   if len(haystack) < len(needle): return False

   index = 0
   for element in needle:
      try:
         index = haystack.index(element, index) + 1
      except ValueError:
         return False
   else:
      return True

def filter_subsets(lists):
   """ Given list of lists, return new list of lists without subsets  """

   for needle in lists:
      if not any(is_subset(needle, haystack) for haystack in lists
         if needle is not haystack):
         yield needle

my_lists = [[1, 2, 4, 8], [1, 2, 4, 5, 6], [1, 2, 3], 
            [2, 3, 21], [1, 2, 3, 4], [1, 2, 3, 4, 5, 6, 7]]    
print list(filter_subsets(my_lists))

>>> [[1, 2, 4, 8], [2, 3, 21], [1, 2, 3, 4, 5, 6, 7]]

而且，只是为了好玩，一个单线：

def filter_list(L):
    return [x for x in L if not any(set(x)<=set(y) for y in L if x is not y)]

Answer 3

感谢所有提出解决方案并处理我有时错误的数据集的人。使用@hughdbrown解决方案我将其修改为我想要的：

修改是在目标上使用滑动窗口以确保找到子集序列。我认为我应该使用比“设置”更合适的词来描述我的问题。

def is_sublist_of_any_list(cand, lists):
    # Compare candidate to a single list
    def is_sublist_of_list(cand, target):
        try:
            i = 0            
            try:
                start = target.index(cand[0])
            except:
                return False

            while start < (len(target) + len(cand)) - start:
                if cand == target[start:len(cand)]:
                    return True
                else:
                    start = target.index(cand[0], start + 1)
        except ValueError:
            return False

    # See if candidate matches any other list
    return any(is_sublist_of_list(cand, target) for target in lists if len(cand) <= len(target))

# Compare candidates to all other lists
def super_lists(lists):
    a = [cand for i, cand in enumerate(lists) if not is_sublist_of_any_list(cand, lists[:i] + lists[i+1:])]
    return a

lists = [[2, 16, 17], [1, 2, 3, 4, 5, 6, 7], [1], [1, 2, 3, 4], [1, 2], [17, 18, 19, 22, 41, 48], [2, 3], [1, 2, 3], [50, 69], [1, 2, 3], [2, 3, 21], [1, 2, 3], [1, 2, 4, 8], [1, 2, 4, 5, 6]]
expect = [[2, 16, 17], [1, 2, 3, 4, 5, 6, 7], [17, 18, 19, 22, 41, 48], [50, 69],  [2, 3, 21], [1, 2, 4, 8], [1, 2, 4, 5, 6]]

def test():
    out = super_lists(list(lists))

    print "In  : ", lists
    print "Out : ", out

    assert (out == expect)

结果：

In  :  [[2, 16, 17], [1, 2, 3, 4, 5, 6, 7], [1], [1, 2, 3, 4], [1, 2], [17, 18, 19, 22, 41, 48], [2, 3], [1, 2, 3], [50, 69], [1, 2, 3], [2, 3, 21], [1, 2, 3], [1, 2, 4, 8], [1, 2, 4, 5, 6]]
Out :  [[2, 16, 17], [1, 2, 3, 4, 5, 6, 7], [17, 18, 19, 22, 41, 48], [50, 69], [2, 3, 21], [1, 2, 4, 8], [1, 2, 4, 5, 6]]