python - 列表中列表之间的公共列表

Question

I need to be able to find the first common list (which is a list of coordinates in this case) between a variable amount of lists.

ie this list

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

should return

>>> [3,4]

If easier, I can work with a list of all common lists(coordinates) between the lists that contain the coordinates.

I can't use sets or dictionaries because lists are not hashable(i think?).

Answer 1

Correct, list objects are not hashable because they are mutable. tuple objects are hashable (provided that all their elements are hashable). Since your innermost lists are all just integers, that provides a wonderful opportunity to work around the non-hashableness 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 give you a set which contains tuples of the coordinates which are present in all the sublists. To get a list of list like you started with:

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

does the trick.

To address the concern by @wim, if you really want a reference to the first element in the intersection (where first is defined by being first 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 will return None if the intersection is empty.

Answer 2

If you are looking for the first child list that is common amongst 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

Answer 3

Solution with recursive function. :)

This gets first duplicated element.

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]

Answer 4

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]