获取列表之间第一个和最后一个共同元素索引的最快方法

Question

I have two sorted lists

x = [-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10]
y = [3,4,5,6]

Between these x and y I would like to return imin = (6,0) and imax = (9,3) . If these lists do not share any elements, I would like to return imin = None and imax = None .

A solution is

def inds(x,y):
    arr = [(i,j) for i,xx in enumerate(x) for j,yy in enumerate(y) if xx==yy ]
    if arr!=[]: # to obtain proper None output
        imin = (min(i for i,_ in arr), min(j for _,j in arr))
        imax = (max(i for i,_ in arr), max(j for _,j in arr))
    else: 
        imin = None
        imax = None 
    return (imin,imax)

This does a lot of unnecessary computation (O(n**2)) and is a bottleneck of one of my programs. Can anyone suggest something faster?

ADDITIONAL (NON-MINIMAL EXAMPLE) INFO

If it helps, I actually have a list of objects.

objects = [(A1,B1),(A2,B2)] 

x and y would be attributes of each element of this objects list like so:

x = objects[0][0].attrib
y = objects[0][1].attrib

and I actually want to generate

[(imin1,imax1),(imin2,imax2)]

Which could come, for example, from

def attribs(A,B):
    return (A.attrib,B.attrib)

[inds(*attribs(*v)) for v in objects]

note: I added a numpy tag just because I'm open to using numpy for this if it's faster.

Answer 1

This should be what you are after

c = set(x).intersection(y)  # O(n) time
def get_first(l):
    return next((idx for idx, elm in enumerate(l) if elm in c), None)  # O(n) time
imin = (get_first(x), get_first(y))
imax = (len(x) - get_first(x[::-1]) - 1, len(y) - get_first(y[::-1]) - 1)

From here onwards you can do a few tweaks but it will still run O(n)

Answer 2

Using np.intersect1d and returning the indices, you can do the following

idxes = np.stack(np.intersect1d(x,y, return_indices=True)[1:])
ix = tuple(idxes[:,0])
iy = tuple(idxes[:,-1])

>>> ix
(6, 0)
>>> iy
(9, 3)

Explanation

idxes is a 2d array of the indices where there are intersections between your two arrays:

>>> idxes
array([[6, 7, 8, 9],
       [0, 1, 2, 3]])

So you can just take the first and last using

ix = tuple(idxes[:,0])
iy = tuple(idxes[:,-1])

Answer 3

You can also sort the intersection list and use .index() to find out the indices.

z = list(set(x).intersection(set(y))) # O(n)
z.sort() # O(nlogn)

imin = (x.index(z[0]), y.index(z[0])) # O(n)
imax = (x.index(z[-1]), y.index(z[-1])) # O(n)