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)
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]]
```

Source: StackOverflow.com