如何从列表中删除所有元素，该列表是 python 中同一列表中另一个更大元素的子序列？

Question

我有一个列表列表，如下所示：

[['rest'],['look'],['rest','look'],['resting','look],['apple','mango'],['apple','man' ],['apple','banana','mango'],['rest','resting','look','looked','it','spit']]

必须删除属于另一个元素的子串/子序列的所有元素。 例如元素['rest']和['look']已经存在于列表元素['rest','look']和['rest','resting','look','looked','it','spit'] ，因此必须将它们从最终列表中删除。 此外，元素['rest','look']是 ['rest','resting','look','looked','it','spit'] 的子序列, so it should be removed. Similarly, , so it should be removed. Similarly, ['resting','look'] is a substring of ，因此它也必须被删除。元素 ['apple','mango'] 因为它是 ['apple','banana','mango'] 的子序列，所以应该删除，但是 ['apple','man'] 不应该删除，因为它不是一个公共子序列。output 必须是列表而不是集合。

我试过这个：

x = [['rest'],['look'],['rest','look'],['resting','look'],['apple','mango'],['apple','man'],['apple','banana','mango'],['rest', 'resting', 'look', 'looked', 'it', 'spit']]

res=[]

found=0

for i in x:
   for item in i:
     for j in x:
          for item1 in j:
            if item == item1:
               found=1
     if found==0:
          res.append(item)

print res

我得到的 output 是一个空列表。 所需的 output 是：

[['apple','man'],['apple','banana','mango'],['rest','resting','look','looked','it','spit']]

Answer 1

您可以通过将列表展平为一个集合来使用集合推导式。

x = [['rest'],['look'],['rest','look'],['resting','look'],['rest', 'resting', 'look', 'looked', 'it', 'spit']]
In [2]: results = {s_ for s in x for s_ in s}
In [2]: results
Out[3]: {'it', 'look', 'looked', 'rest', 'resting', 'spit'}

Answer 2

我认为这就是您尝试实施的内容。 我不认为你需要found变量，你可以只检查子列表中的每个项目（例如['rest', 'look']已经append编辑到res 。

x = [['rest'], ['look'], ['rest', 'look'], ['resting', 'look'], ['rest', 'resting', 'look', 'looked', 'it', 'spit']]
res = []
for L in x:
    for item in L:
        if item not in res:
            res.append(item)


print(res)
# ['rest', 'look', 'resting', 'looked', 'it', 'spit']