Recursive Function - Compare 2 lists, return the elements that don't exist in both

Recursive Function - Compare 2 lists, return the elements that don't exist in both - Printable Version

+- Python Forum (https://python-forum.io)
+-- Forum: Python Coding (https://python-forum.io/forum-7.html)
+--- Forum: Homework (https://python-forum.io/forum-9.html)
+--- Thread: Recursive Function - Compare 2 lists, return the elements that don't exist in both (/thread-14894.html)

Recursive Function - Compare 2 lists, return the elements that don't exist in both - KellyBaptist - Dec-22-2018

I have to do a RECURSIVE FUNCTION, comparing two lists and returning the elements that they don't have in common.

This is what I have so far:

def compare(list1, list2):
    if list2[0] in list1:
        list1.remove(list2[0])
        return compare(list1[1:], list2[1:])

    return list1  

#Example 
 >>>>compare([2, 3, 4, 5], [2, 3])
 [4, 5]

I can compare if the first element of the list (list[0]) is the same but I am having trouble on how to compare the elements if they are not in the first position...I've tried many options but I'm a begginer in programming and don't really know how to do it. It must be a RECURSIVE FUNCTION, I can't use FOR or WHILE. And is there any way that I can do it without using remove()? Thank you so much

RE: Recursive Function - Compare 2 lists, return the elements that don't exist in both - Gribouillis - Dec-23-2018

Recursion works by supposing that you have the solution for a shorter copy of the problem. In your case, assuming that list2 is not empty, the list

result = compare(list1, list2[1:])

contains all the items that are in list1 but not in list2[1:] or in list2[1:] but not in list1. If list2[0] is in list1, it needs to be removed from the result. If it is not in list1, it needs to be added to the result.

In the case where list2 is empty you can return a copy of list1 (don't return list1 itself because the result is modified by the recursive calls)