How to the find the longest match from a set of lists in Python?

Question

If I have a list like this

my_list = [1, 7, 3, 4, 2, 9, 6, 5, 8]

Also I have other lists of length less than the length of my_list. For instance, let us say I have two other lists as follows:

my_other_list_1 = [1, 7, 3, 4, 2]
my_other_list_2 = [1, 7, 4, 9, 8]

All lists have distinct elements and the other lists have elements from the original list.

My problem is to find the list (either my_other_list_1 or my_other_list_2) that has the longest match in my_list. First, I would like to ask what is this problem called? Then, I prefer to get the length of the longest match of each list. How do I do this?

In the example, I would return my_other_list_1 because it has a match of length 5 since [1, 7, 3, 4, 2] is already in my_list. Also, I would return 2 for my_other_list_2 because there is a match of length 2 in it, i.e., [1, 7].

Summarizing, if I have an algorithm A and the inputs are my_list, my_other_list_1 and my_other_list_2, the algorithm should return

my_other_list_1 has match 5
my_other_list_2 has match 2

NB. I think this is a problem called longest common subsequences (LCS) problem but as I understand in LCS problem the subsequences do not need to be consecutive.

If you don't have the requirement of coding this yourself, you could use a difflib.SequenceMatcher -- they even have a find_longest_match method :-). — mgilson
– mgilson, Commented Jun 8, 2016 at 15:32
This problem is called the longest common substring problem. I've also heard it referred to as the longest contiguous subsequence problem. You can find more information here, and the code for this problem is the second part of this answer — Greg
– Greg, Commented Jun 8, 2016 at 15:45

TheLazyScripter · Accepted Answer · 2016-06-08 17:29:49Z

1

def longest_match(main_list, *other_lists, **options):
    try:
        if options:
            ordered = options['ordered']
        else:
            raise Exception
    except:
        ordered = False

    best_match = 0
    longest_match = 0
    for index, list in enumerate(other_lists):
        current_match = 0
        if not ordered:
            while True:
                if list[:current_match] == main_list[:current_match]:
                    current_match += 1
                else:
                    if current_match > longest_match:
                        longest_match = current_match
                        best_match = index
                    break
        else:
            for i, letter in enumerate(list):
                current_match = i
                if letter == main_list[0]:
                    current_match += 1
                    while True:
                        if list[i:current_match] == main_list[:current_match]:
                            current_match += 1
                        else:
                            if current_match > longest_match:
                                longest_match = current_match
                                best_match = index
                            break
    return other_lists[best_match]

print longest_match(my_list, my_other_list_1, my_other_list_2, ordered=True) #kwarg odered=True will allow for mid list comprehension

I'm on my phone and havnt tested the following, but it should do the trick!

edited Jun 8, 2016 at 17:29

answered Jun 8, 2016 at 15:54

TheLazyScripter

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2 Comments

Jared Over a year ago

what if the best match starts in the middle of the main_list?

TheLazyScripter Over a year ago

@Jared You are correct so I have adapted the code to work even if the match does not begin at the start of the list, only if the ordered option is true!

Philip Kalinda · Accepted Answer · 2016-06-09 08:34:51Z

1

I'd be more than happy to help bud! Here you go!

def matcher_algorithm_large(original_list, **kwargs):
match_dict = {}
for i in kwargs:
    counter=0
    for j in kwargs[i]:
        if j == original_list[kwargs[i].index(j)]:
            counter += 1
        else:
            break
    match_dict[i]=counter
for key,value in match_dict.items():
    print('{} has match {}'.format(key,value))

This so basically you can plug them in like so:

my_list = [1, 7, 3, 4, 2, 9, 6, 5, 8]
my_other_list_1 = [1, 7, 3, 4, 2]
my_other_list_2 = [1, 7, 4, 9, 8]

matcher_algorithm_large(my_list,my_other_list_1=my_other_list_1,my_other_list_2=my_other_list_2)

This will give you the following output:

my_other_list_1 has match 5
my_other_list_2 has match 2

I hope this helps you out. I've done it so you can pass in as many lists as you would like. The inputs could be made a lot cleaner if you had a dict of lists, but that up to you if you'd like to have it that way. I'd be more than happy to script that. :)

edited Jun 9, 2016 at 8:34

answered Jun 8, 2016 at 16:23

Philip Kalinda

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3 Comments

Philip Kalinda Over a year ago

this is assuming that you wish to maintain the names of the lists in the outputs as opposed to that other answer

Greg Over a year ago

this doesn't find matches that begin in the middle of the original_list.

Philip Kalinda Over a year ago

Oh, I didn't think that was a requirement based on the question description. Slightly more challenging problem but I can give it a go should it be required :)

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