Scala: groupBy (identity) of List Elements - scala

Question

I develop an application that builds pairs of words in (tokenised) text and produces the number of times each pair occurs (even when same-word pairs occur multiple times, it's OK as it'll be evened out later in the algorithm).
When I use
elements groupBy()
I want to group by the elements' content itself, so I wrote the following:
def self(x: (String, String)) = x
/**
* Maps a collection of words to a map where key is a pair of words and the
* value is number of
* times this pair
* occurs in the passed array
*/
def producePairs(words: Array[String]): Map[(String,String), Double] = {
var table = List[(String, String)]()
words.foreach(w1 =>
words.foreach(w2 =>
table = table ::: List((w1, w2))))
val grouppedPairs = table.groupBy(self)
val size = int2double(grouppedPairs.size)
return grouppedPairs.mapValues(_.length / size)
}
Now, I fully realise that this self() trick is a dirty hack. So I thought a little a came out with a:
grouppedPairs = table groupBy (x => x)
This way it produced what I want. However, I still feel that I clearly miss something and there should be easier way of doing it. Any ideas at all, dear all?
Also, if you'd help me to improve the pairs extraction part, it'll also help a lot – it looks very imperative, C++ - ish right now. Many thanks in advance!

Answer 1

I'd suggest this:
def producePairs(words: Array[String]): Map[(String,String), Double] = {
val table = for(w1 <- words; w2 <- words) yield (w1,w2)
val grouppedPairs = table.groupBy(identity)
val size = grouppedPairs.size.toDouble
grouppedPairs.mapValues(_.length / size)
}
The for comprehension is much easier to read, and there is already a predifined function identity, with is a generalized version of your self.

Answer 2

you are creating a list of pairs of all words against all words by iterating over words twice, where i guess you just want the neighbouring pairs. the easiest is to use a sliding view instead.
def producePairs(words: Array[String]): Map[(String, String), Int] = {
val pairs = words.sliding(2, 1).map(arr => arr(0) -> arr(1)).toList
val grouped = pairs.groupBy(t => t)
grouped.mapValues(_.size)
}
another approach would be to fold the list of pairs by summing them up. not sure though that this is more efficient:
def producePairs(words: Array[String]): Map[(String, String), Int] = {
val pairs = words.sliding(2, 1).map(arr => arr(0) -> arr(1))
pairs.foldLeft(Map.empty[(String, String), Int]) { (m, p) =>
m + (p -> (m.getOrElse(p, 0) + 1))
}
}
i see you are return a relative number (Double). for simplicity i have just counted the occurances, so you need to do the final division. i think you want to divide by the number of total pairs (words.size - 1) and not by the number of unique pairs (grouped.size)..., so the relative frequencies sum up to 1.0

Answer 3

Alternative approach which is not of order O(num_words * num_words) but of order O(num_unique_words * num_unique_words) (or something like that):
def producePairs[T <% Traversable[String]](words: T): Map[(String,String), Double] = {
val counts = words.groupBy(identity).map{case (w, ws) => (w -> ws.size)}
val size = (counts.size * counts.size).toDouble
for(w1 <- counts; w2 <- counts) yield {
((w1._1, w2._1) -> ((w1._2 * w2._2) / size))
}
}