Question

i know this question might look like a duplicate. but i had a hard time trying to solve this and i couldn’t find a helpful solution for my case

i’am implementing a genetic algorithm using python for the traveling salesman problem

assume we have those lists ( tours)

a = [1,0,2,5,4,3,1]
b = [1,2,5,4,3,0,1]
c = [1,3,5,4,2,0,1]

as you can see, the [5,4] is repeated in the whole 3 lists
and a regular intersection would return all the elements in the list.

i want some function like intersect_list(a,b)

that returns [5,4]

is there a python built-in way to find this? or do you have any suggestion?.

Note : i know i can loop it to solve this , but please put in mind that in my case i have around 400 lists. and at the length of 401 each.

in other words : i want to see the common path between those lists.

please let me know if anything was unclear
thanks in advance.

Answer 1

After taking a look at the links posted by @pyfunc, I came up with the following:

def shortest_of(lists):
    return min(lists, key=len)

def contains_sublist(lst, sublst):
    n = len(sublst)
    return any((sublst == lst[i:i+n]) for i in xrange(len(lst)-n+1)) 

def longest_common(lists):
    if not lists:
        return ()
    res = set()    
    base = shortest_of(lists)
    length = len(base)

    for i in xrange(length, 0, -1):
        for j in xrange(length - i + 1):
            candidate = ', ' + str(base[j:i+j]).strip('[]') + ','
            #candidate = base[j:i+j]  

            for alist in lists:
                if not candidate in ', ' + str(alist).strip('[]') + ',':
                #if not contains_sublist(alist, candidate):   
                    break
            else:
                res.add(tuple([int(a) for a in candidate[2:-1].split(',')]))
                #res.add(tuple(candidate))

        if res:
            return tuple(res)    

    return ()

if __name__ == '__main__':
    a = [1,0,2,5,4,3,1]
    b = [1,2,5,4,3,0,1]
    c = [1,3,5,4,2,0,1]

    print longest_common([a,b,c])
    print longest_common([b,c])

output:

((5, 4),)
((0, 1), (5, 4))

EDIT:

Updated solution to use string conversions and matching as it happened to be way faster. Previous solution parts are commented out. Also, it now gives all possibilities.