Returning duplicate tuples that match at a number of different indices

Question

I have tuples nested in a list i.e. [(0,1,2,3,4,5,6,7), (etc)].

I'm trying to return the tuples that are matching at index 0 & 1, and 3 to 5 (and to keep the ordering of the data inside the tuple)

So with the code below I'm trying to remove the duplicates, then with the result I'm comparing this to the original list to identify the duplicates removed:

seen = set()
seen_add = seen.add
newL = []
for a in myList:
     if a[:2] not in seen and not seen_add(a[:2]):
        if a[3:6] not in seen and not seen_add(a[3:6]):
            newL.append(a)



result = list(set(myList) - set(newL))
for i in result: print i

But the first part removes a tuple that doesn't even have a duplicate.

N.B. The code for removing the first two elements came from here (by Martijn Pieters): Removing Duplicates from Nested List Based on First 2 Elements; but removing additional elements resulted in the aforementioned 'error'.

To return the duplicates (this code follows the answer by @unutbu)

for i in result:
        for e in newL:
            if i[:2]==e[:2]:
                if i[3:6]==e[3:6]:
                    print e, i

Answer 1

If you want to regard elements as matching when a[:2]+a[3:6] is the same, then you need to add a[:2]+a[3:6] to seen, rather than a[:2] and a[3:6] separately:

seen = set()
seen_add = seen.add
newL = [a  for a in myList
        if a[:2]+a[3:6] not in seen
        and not seen_add(a[:2]+a[3:6])]

newL will then contain "unique" elements from myList, with the order preserved. Note that calling set(myList) will destroy the order of the items in myList, so

result = list(set(myList) - set(newL))

will contain the duplicate elements, but the order will not be preserved.