Python - 比较两个嵌套列表并编写第三个嵌套列表

Question

some_list = [[app_num, product, prod_size, prod_amt]]
other_list = [[app_num, product, prod_size, prod_amt]]

I have two lists of lists. There are no matches between some_list and other_list for app_num. However, for a given app_num in other_list, the product, prod_size_ and prod_amt may be the same as in a given app_num in some_list. I am trying to create a final_some_list that is the same as some_list, except that it has deleted any list element from some_list if that list element has the same product, prod_size, and prod_amt as an element in other_list.

I have tried nested for loops, list comprehension (below are some examples of many failed attempts).

final_some_list = [app for app in some_list for other_app in other_list 
if not app[1] == other_app[1] and app[2] == other_app[2] and app[3] == 
other_app[3]

final_some_list = [app for app in some_list for other_app in other_list 
if not app[1] and app[2] and app[3] == in other_app

Answer 1

final_some_list = [x for x in some_list if all(x[1:] != y[1:] for y in other_list)]

Edit :

Solution above is O ( nm ) where n == len(some_list) and m == len(other_list) . You can make it O ( n log m ) if you want. Put all the tails of all the elements of other_list in a binary search tree (assuming those elements are ordable), now querying that tree is O (log m ). In case the elements are hashable you can even have an O ( n ) solution because querying a hash-set is O (1).

Btw: The elements of both lists look like tuples to me, this will get you more close to a better solution.

Answer 2

Only the first comparison contains the not attribute, add != to all of the comparisons and this should work:

final_some_list = [app for app in some_list for other_app in other_list 
if app[1] != other_app[1] and app[2] != other_app[2] and app[3] != 
other_app[3]]