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Python - Generic Lists Among Lists in a List

I need to find the first shared list (which is a list of coordinates in this case) between a variable number of lists.

i.e. this list

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

should return

>>> [3,4]

To put it simply, I can work with a list of all common lists (coordinates) between lists containing coordinates.

I cannot use sets or dictionaries because lists are not hashed (I guess?).

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4 answers


Correct, list

objects are not hashed because they are mutable. tuple

objects are hashed (assuming all their elements are hashed). Since your innermost lists are just integers, this is a great way to get around the inappropriateness of lists:

>>> lists = [[[1,2],[3,4],[6,7]],[[3,4],[5,9],[8,3],[4,2]],[[3,4],[9,9]]]
>>> sets = [set(tuple(x) for x in y) for y in lists]
>>> set.intersection(*sets)
set([(3, 4)])

Here I am giving you a set that contains tuples of coordinates that are present in all sublists. To get a list of lists like you started with:

[list(x) for x in set.intersection(*sets)]


does the trick.

To fix the problem with @wim, if you really want to reference the first element in the intersection (where the first

first one is defined in lists[0]

), the easiest way is probably like this:

#... Stuff as before
intersection = set.intersection(*sets)
reference_to_first = next( (x for x in lists[0] if tuple(x) in intersection), None )

This returns None

if the intersection is empty.

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If you are looking for the first child list that is common among all parent lists, the following will work.



def first_common(lst):
    first = lst[0]
    rest = lst[1:]
    for x in first:
        if all(x in r for r in rest):
            return x
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Solution with recursive function. :)

This gets the first duplicated item.

def get_duplicated_element(array):
    global result, checked_elements
    checked_elements = []
    result = -1
    def array_recursive_check(array):
        global result, checked_elements
        if result != -1: return
        for i in array:
            if type(i) == list:
                if i in checked_elements:
                    result = i
                    return
                checked_elements.append(i)
                array_recursive_check(i)
    array_recursive_check(array)
    return result

get_duplicated_element([[[1,2],[3,4],[6,7]],[[3,4],[5,9],[8,3],[4,2]],[[3,4],[9,9]]])
[3, 4]
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you can achieve this with a list comprehension:

>>> l = [[[1,2],[3,4],[6,7]],[[3,4],[5,9],[8,3],[4,2]],[[3,4],[9,9]]]
>>> lcombined =  sum(l, [])
>>> [k[0] for k in [(i,lcombined.count(i)) for i in lcombined] if k[1] > 1][0]
[3, 4]
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