Comparing two lists and only printing the differences? (XORing two lists)

Question

I'm trying to create a function that takes in 2 lists and returns the list that only has the differences of the two lists.

Example:

a = [1,2,5,7,9]
b = [1,2,4,8,9]

The result should print [4,5,7,8]

The function so far:

def xor(list1, list2):
    list3=list1+list2
    for i in range(0, len(list3)):
        x=list3[i]
        y=i
        while y>0 and x<list3[y-1]:
            list3[y]=list3[y-1]
            y=y-1
        list3[y]=x

        last=list3[-1]
    for i in range(len(list3) -2, -1, -1):
        if last==list3[i]:
            del list3[i]
        else:
            last=list3[i]

    return list3 
print xor([1,2,5,7,8],[1,2,4,8,9])

The first for loop sorts it, second one removes the duplicates. Problem is the result is [1,2,4,5,7,8,9] not [4,5,7,8], so it doesn't completely remove the duplicates? What can I add to do this. I can't use any special modules, .sort, set or anything, just loops basically.

Answer 1

You basically want to add an element to your new list if it is present in one and not present in another. Here is a compact loop which can do it. For each element in the two lists (concatenate them with list1+list2), we add element if it is not present in one of them:

[a for a in list1+list2 if (a not in list1) or (a not in list2)]

You can easily transform it into a more unPythonic code with explicit looping through elements as you have now, but honestly I don't see a point (not that it matters):

def xor(list1, list2):
    outputlist = []
    list3 = list1 + list2
    for i in range(0, len(list3)):
        if ((list3[i] not in list1) or (list3[i] not in list2)) and (list3[i] not in outputlist):
             outputlist[len(outputlist):] = [list3[i]]
    return outputlist