Python-сообщество

     # [
     # [elem1, ... comment11]
     # [elem1, ... comment12]
     # ...
     # [elem2, ... comment21]
     # ...
     # ]
     # [
     # [elem1, ... [comment11, comment12]]
     # [elem2, ... [comment21, ...]]
     # ...
     # ]

set().intersection()

>>> a = [1,2,3,4,6]
>>> b = [1,2,3,4,8]
>>> c = [0,1,2,3,5]
>>> d = [0,1,2,3,8]
>>> res = []
>>> res.append(a)
>>> res.append(b)
>>> res.append(c)
>>> res.append(d)

from collections import defaultdict
d = defaultdict(set)
for (i, *x) in res:
    d[i] = d[i].union(set(x))
print(d)

defaultdict(<class 'set'>,{
  0: {1, 2, 3, 5, 8},
  1: {2, 3, 4, 6, 8}
})

from itertools import groupby
from operator import itemgetter
a = [1,2,3,4,6]
b = [1,2,3,4,8]
c = [0,1,2,3,5]
d = [0,1,2,3,8]
tmp = ((x[:-1], x[-1]) for x in (a, b, c, d))
for k, g in groupby(tmp, itemgetter(0)):
    k.append([x[-1] for x in g])
    print(k)

from itertools import groupby
data = [
    [1,2,3,4,8],
    [1,2,3,4,6],
    [0,1,2,3,5],
    [0,1,2,3,8]
]
def keyfunc(x):
    return x[:-1]
data.sort()
res = []
for k, g in groupby(data, keyfunc):
    k.append([x[-1] for x in g])
    res.append(k)
print res