How to remove duplicates from a list then sort by most frequent

Question

I have a list with assorted keywords that may repeat. I need to generate a list with distinct keywords but sorted by the frequency of which they appeared on the original list. How would be the idiomatic Scala for that? Here is a working but ugly implementation: ``` val keys = List("c","a","b","b","a","a") keys.groupBy(p => p).toList.sortWith( (a,b) => a._2.size > b._2.size ).map(_._1) // List("a","b","c") ```

Accepted Answer

Shorter version:

```
keys.distinct.sortBy(keys count _.==).reverse
```

That is not particular efficient, however. The `groupBy` version ought to perform better, though it can be improved:

```
keys.groupBy(identity).toSeq.sortBy(_._2.size).map(_._1)
```

One can also get rid of the `reverse` in the first version by declaring an `Ordering`:

```
val ord = Ordering by (keys count (_: String).==)
keys.distinct.sorted(ord.reverse)
```

Note that `reverse` in this version just produces a new `Ordering` that works in the opposite manner of the original. This version also suggests a way to get better performance:

```
val freq = collection.mutable.Map.empty[String, Int] withDefaultValue 0
keys foreach (k => freq(k) += 1)
val ord = Ordering by freq
keys.distinct.sorted(ord.reverse)
```

How to remove duplicates from a list then sort by most frequent

Solution

Problem