Question

How do I find the duplicates in a list of integers and create another list of the duplicates?

Answer 1

To remove duplicates use set(a). To print duplicates, something like:

a = [1,2,3,2,1,5,6,5,5,5]

import collections
print([item for item, count in collections.Counter(a).items() if count > 1])

## [1, 2, 5]

Note that Counter is not particularly efficient (timings) and probably overkill here. set will perform better. This code computes a list of unique elements in the source order:

seen = set()
uniq = []
for x in a:
    if x not in seen:
        uniq.append(x)
        seen.add(x)

or, more concisely:

seen = set()
uniq = [x for x in a if x not in seen and not seen.add(x)]    

I don’t recommend the latter style, because it is not obvious what not seen.add(x) is doing (the set add() method always returns None, hence the need for not).

To compute the list of duplicated elements without libraries:

seen = set()
dupes = []

for x in a:
    if x in seen:
        dupes.append(x)
    else:
        seen.add(x)

or, more concisely:

seen = set()
dupes = [x for x in a if x in seen or seen.add(x)]    

If list elements are not hashable, you cannot use sets/dicts and have to resort to a quadratic time solution (compare each with each). For example:

a = [[1], [2], [3], [1], [5], [3]]

no_dupes = [x for n, x in enumerate(a) if x not in a[:n]]
print no_dupes # [[1], [2], [3], [5]]

dupes = [x for n, x in enumerate(a) if x in a[:n]]
print dupes # [[1], [3]]