n-largest elements in a sequence (need to retain duplicates)

I need to find the n largest elements in a list of tuples. Here is an example for top 3 elements.

# I have a list of tuples of the form (category-1, category-2, value)
# For each category-1, ***values are already sorted descending by default***
# The list can potentially be approximately a million elements long.
lot = [('a', 'x1', 10), ('a', 'x2', 9), ('a', 'x3', 9), 
       ('a', 'x4',  8), ('a', 'x5', 8), ('a', 'x6', 7),
       ('b', 'x1', 10), ('b', 'x2', 9), ('b', 'x3', 8), 
       ('b', 'x4',  7), ('b', 'x5', 6), ('b', 'x6', 5)]

# This is what I need. 
# A list of tuple with top-3 largest values for each category-1
ans = [('a', 'x1', 10), ('a', 'x2', 9), ('a', 'x3', 9), 
       ('a', 'x4', 8), ('a', 'x5', 8),
       ('b', 'x1', 10), ('b', 'x2', 9), ('b', 'x3', 8)]

I tried using heapq.nlargest. However it only returns the first 3 largest elements and doesn't return duplicates. For example,

heapq.nlargest(3, [10, 10, 10, 9, 8, 8, 7, 6])
# returns
[10, 10, 10]
# I need
[10, 10, 10, 9, 8, 8]

I can only think of a brute force approach. This is what I have and it works.

res, prev_t, count = [lot[0]], lot[0], 1
for t in lot[1:]:
    if t[0] == prev_t[0]:
        count = count + 1 if t[2] != prev_t[2] else count
        if count <= 3:
            res.append(t)   
    else:
        count = 1
        res.append(t)
    prev_t = t

print res

Any other ideas on how I can implement this?

EDIT: timeit results for a list of 1 million elements show that mhyfritz's solution runs in 1/3rd the time of brute force. Didn't want to make the question too long. So added more details in my answer.