Interleave different length lists, elimating duplicates, and preserve order

Question

I have two lists, let\'s say:

keys1 = [\'A\', \'B\', \'C\', \'D\', \'E\',           \'H\', \'I\']
keys2 = [\'A\', \'B\',           \'E\', \'F\', \'G\', \'H\',            


        

Answer 1

I recently had stumbled upon a similar issue while implementing a feature. I tried to clearly define the problem statement first. If I understand right, here is the problem statement

Problem Statement

Write a function merge_lists which will merge a list of lists with overlapping items, while preserving the order of items.

Constraints

1. If item A comes before item B in all the lists where they occur together, then item A must precede item B in the final list also

2. If item A and item B interchange order in different lists, ie in some lists A precedes B and in some others B precedes A, then the order of A and B in the final list should be the same as their order in the first list where they occur together. That is, if A precedes B in l1 and B precedes A in l2, then A should precede B in final list

3. If Item A and Item B do not occur together in any list, then their order must be decided by the position of the list in which each one occurs first. That is, if item A is in l1 and l3, item B is in l2 and l6, then the order in the final list must be A then B

Test case 1:

Input:

l1 = ["Type and Size", "Orientation", "Material", "Locations", "Front Print Type", "Back Print Type"]

l2 = ["Type and Size", "Material", "Locations", "Front Print Type", "Front Print Size", "Back Print Type", "Back Print Size"]

l3 = ["Orientation", "Material", "Locations", "Color", "Front Print Type"]

merge_lists([l1,l2,l3])

Output:

['Type and Size', 'Orientation', 'Material', 'Locations', 'Color', 'Front Print Type', 'Front Print Size', 'Back Print Type', 'Back Print Size']

Test case 2:

Input:

l1 = ["T", "V", "U", "B", "C", "I", "N"]

l2 = ["Y", "V", "U", "G", "B", "I"]

l3 = ["X", "T", "V", "M", "B", "C", "I"]

l4 = ["U", "P", "G"]

merge_lists([l1,l2,l3, l4])

Output:

['Y', 'X', 'T', 'V', 'U', 'M', 'P', 'G', 'B', 'C', 'I', 'N']

Test case 3:

Input:

l1 = ["T", "V", "U", "B", "C", "I", "N"]

l2 = ["Y", "U", "V", "G", "B", "I"]

l3 = ["X", "T", "V", "M", "I", "C", "B"]

l4 = ["U", "P", "G"]

merge_lists([l1,l2,l3, l4])

Output:

['Y', 'X', 'T', 'V', 'U', 'M', 'P', 'G', 'B', 'C', 'I', 'N']

Solution

I arrived at a reasonable solution which solved it correctly for all the data I had. (It might be wrong for some other data set. Will leave it for others to comment that). Here is the solution

def remove_duplicates(l):
    return list(set(l))

def flatten(list_of_lists):
    return [item for sublist in list_of_lists for item in sublist]

def difference(list1, list2):
    result = []
    for item in list1:
        if item not in list2:
            result.append(item)
    return result

def preceding_items_list(l, item):
    if item not in l:
        return []
    return l[:l.index(item)]

def merge_lists(list_of_lists):
    final_list = []
    item_predecessors = {}

    unique_items = remove_duplicates(flatten(list_of_lists))
    item_priorities = {}

    for item in unique_items:
        preceding_items = remove_duplicates(flatten([preceding_items_list(l, item) for l in list_of_lists]))
        for p_item in preceding_items:
            if p_item in item_predecessors and item in item_predecessors[p_item]:
                preceding_items.remove(p_item)
        item_predecessors[item] = preceding_items
    print "Item predecessors ", item_predecessors

    items_to_be_checked = difference(unique_items, item_priorities.keys())
    loop_ctr = -1
    while len(items_to_be_checked) > 0:
        loop_ctr += 1
        print "Starting loop {0}".format(loop_ctr)
        print "items to be checked ", items_to_be_checked
        for item in items_to_be_checked:
            predecessors = item_predecessors[item]
            if len(predecessors) == 0:
                item_priorities[item] = 0
            else:
                if all(pred in item_priorities for pred in predecessors):
                    item_priorities[item] = max([item_priorities[p] for p in predecessors]) + 1
        print "item_priorities at end of loop ", item_priorities
        items_to_be_checked = difference(unique_items, item_priorities.keys())
        print "items to be checked at end of loop ", items_to_be_checked
        print

    final_list = sorted(unique_items, key=lambda item: item_priorities[item])
    return final_list

I've also open sourced the code as a part of the library named toolspy. So you can just do this

pip install toolspy

from toolspy import merge_lists
lls=[['a', 'x', 'g'], ['x', 'v', 'g'], ['b', 'a', 'c', 'x']]
merge_lists(lls)