# How do I find the duplicates in a list and create another list with them?

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

## Solution

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)
``````

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:
``````

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]]
``````

Source: StackOverflow.com