I'm trying to remove the duplicate from the list and count the list after removing the duplicates

seq = [[1,2,3], [1,2,3], [2,3,4], [4,5,6]]
new_seq = [[1,2,3], [2,3,4], [4,5,6]]
count = 3 

My code takes around 23 seconds for around 66,000 lists in a list

How can I make my code faster?

def unique(seq):
    new_seq = []
    count = 0
    for i in seq:
        if i not in new_seq:
            count += 1
    return count
  • 2
    \$\begingroup\$ What are you really trying to accomplish? Is this function part of a larger program? Tell us about the context. \$\endgroup\$ May 6, 2016 at 19:41
  • \$\begingroup\$ The lists comes from another function which calculates an algorithm \$\endgroup\$
    – jack
    May 6, 2016 at 19:45

1 Answer 1


Your function is slow because it is O(n2): each element being added to new_seq has to be compared against every previously added element.

To deduplicate a sequence, use a set. Constructing the set is only O(n) because it uses hashing.

Then, to obtain the size of the set, use len().

def unique(seq):
    return len(set(tuple(element) for element in seq))

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