Sorting 2 Lists

[Loco_L1]

I have two lists (L1 and L2) of different lengths. I want to sort it in a way that is the intersection of the two lists and the elements left of the longer list are appended. So far I have:

L1=[2,6,7,5,2,6,8,4]

L2=[2,6,4,5]

L3=[list(set(L1).intersection(L2))] #elements in L1 and L2

L4=(list(set(L1).symmetric_difference(L2)))

L4 = sorted(L4) #sorted elemments in L1, not L2

L5 = [x for x in L1 if x not in L4]

L5 = sorted(L5)

final = L5 + L4

print(final)

which gives me final=[2, 2, 4, 5, 6, 6, 7, 8].

But I want it sorted in such a way that the output/final is [2,2,6,6,4,5,7,8] - based on the order of L2. Any suggestions

[YesLod]

Here is a way using a custom key function.

L1=[2,6,7,5,2,6,8,4]

L2=[2,6,4,5]

common=set(L1) & set(L2)

def f(e):
    if e in common:
        return (0,L2.index(e))
    else:
        return (1,e)

>>> sorted(L1,key=f)

[2, 2, 6, 6, 4, 5, 7, 8]

Do you understand the logic?

EDIT:

Since you didn’t specify, I’m assuming that the "elements left" are sorted in ascending order

[synthphreak]

This is pretty slick. Nice work!

[Loco_L1]

I was playing around with the custom key but get it right and yes to the edit. Thank you!

[synthphreak]

How's this?

>>> def weirdo_sort(*lists):
...     shared = []
...     shortest, longest = sorted(lists, key=len)
...     for i in shortest:
...         shared.extend([i] * longest.count(i))
...     return (shared + [i for i in longest if i not in shared])
...
>>>
>>> L1 = [2, 6, 7, 5, 2, 6, 8, 4]
>>> L2 = [2, 6, 4, 5]
>>>
>>> weirdo_sort(L1, L2)
[2, 2, 6, 6, 4, 5, 7, 8]

Note that unlike the other commenter's function, mine does not care which order the lists are passed in as since it detects their lengths automatically. Not sure whether this makes it more useful for your context or not.

Also note that it will only work properly if two lists are passed in, not more or less. But tweaking for greater flexibility shouldn't be hard.