Multiple yields in sequence comprehension?

I'm trying to learn Scala and tried to write a sequence comprehension that extracts unigrams, bigrams and trigrams from a sequence. E.g., [1,2,3,4] should be transformed to (not Scala syntax)

[1; _,1; _,_,1; 2; 1,2; _,1,2; 3; 2,3; 1,2,3; 4; 3,4; 2,3,4]

In Scala 2.8, I tried the following:

def trigrams(tokens : Seq[T]) = {
  var t1 : Option[T] = None
  var t2 : Option[T] = None
  for (t3 <- tokens) {
    yield t3
    yield (t2,t3)
    yield (t1,t2,Some(t3))
    t1 = t2
    t2 = t3
  }
}

But this doesn't compile as, apparently, only one yield is allowed in a for-comprehension (no block statements either). Is there any other elegant way to get the same behavior, with only one pass over the d开发者_Python百科ata?

You can't have multiple yields in a for loop because for loops are syntactic sugar for the map (or flatMap) operations:

for (i <- collection) yield( func(i) )

translates into

collection map {i => func(i)}

Without a yield at all

for (i <- collection) func(i)

translates into

collection foreach {i => func(i)}

So the entire body of the for loop is turned into a single closure, and the presence of the yield keyword determines whether the function called on the collection is map or foreach (or flatMap). Because of this translation, the following are forbidden:

1. Using imperative statements next to a yield to determine what will be yielded.
2. Using multiple yields

(Not to mention that your proposed verison will return a List[Any] because the tuples and the 1-gram are all of different types. You probably want to get a List[List[Int]] instead)

Try the following instead (which put the n-grams in the order they appear):

val basis = List(1,2,3,4)
val slidingIterators = 1 to 4 map (basis sliding _)

for {onegram <- basis
     ngram <- slidingIterators if ngram.hasNext}
     yield (ngram.next)

or

val basis = List(1,2,3,4)
val slidingIterators = 1 to 4 map (basis sliding _)
val first=slidingIterators head
val buf=new ListBuffer[List[Int]]

while (first.hasNext)
   for (i <- slidingIterators)
      if (i.hasNext)
         buf += i.next

If you prefer the n-grams to be in length order, try:

val basis = List(1,2,3,4)
1 to 4 flatMap { basis sliding _ toList }

scala> val basis = List(1, 2, 3, 4)
basis: List[Int] = List(1, 2, 3, 4)

scala> val nGrams = (basis sliding 1).toList ::: (basis sliding 2).toList ::: (basis sliding 3).toList
nGrams: List[List[Int]] = ...

scala> nGrams foreach (println _)
List(1)
List(2)
List(3)
List(4)
List(1, 2)
List(2, 3)
List(3, 4)
List(1, 2, 3)
List(2, 3, 4)

I guess I should have given this more thought.

def trigrams(tokens : Seq[T]) : Seq[(Option[T],Option[T],T)] = {
  var t1 : Option[T] = None
  var t2 : Option[T] = None
  for (t3 <- tokens)
    yield {
      val tri = (t1,t2,t3)
      t1 = t2
      t2 = Some(t3)
      tri
    }
}

Then extract the unigrams and bigrams from the trigrams. But can anyone explain to me why 'multi-yields' are not permitted, and if there's any other way to achieve their effect?

val basis = List(1, 2, 3, 4)
val nGrams = basis.map(x => (x)) ::: (for (a <- basis; b <- basis) yield (a, b)) ::: (for (a <- basis; b <- basis; c <- basis) yield (a, b, c))
nGrams: List[Any] = ...
nGrams foreach (println(_))
1
2
3
4
(1,1)
(1,2)
(1,3)
(1,4)
(2,1)
(2,2)
(2,3)
(2,4)
(3,1)
(3,2)
(3,3)
(3,4)
(4,1)
(4,2)
(4,3)
(4,4)
(1,1,1)
(1,1,2)
(1,1,3)
(1,1,4)
(1,2,1)
(1,2,2)
(1,2,3)
(1,2,4)
(1,3,1)
(1,3,2)
(1,3,3)
(1,3,4)
(1,4,1)
(1,4,2)
(1,4,3)
(1,4,4)
(2,1,1)
(2,1,2)
(2,1,3)
(2,1,4)
(2,2,1)
(2,2,2)
(2,2,3)
(2,2,4)
(2,3,1)
(2,3,2)
(2,3,3)
(2,3,4)
(2,4,1)
(2,4,2)
(2,4,3)
(2,4,4)
(3,1,1)
(3,1,2)
(3,1,3)
(3,1,4)
(3,2,1)
(3,2,2)
(3,2,3)
(3,2,4)
(3,3,1)
(3,3,2)
(3,3,3)
(3,3,4)
(3,4,1)
(3,4,2)
(3,4,3)
(3,4,4)
(4,1,1)
(4,1,2)
(4,1,3)
(4,1,4)
(4,2,1)
(4,2,2)
(4,2,3)
(4,2,4)
(4,3,1)
(4,3,2)
(4,3,3)
(4,3,4)
(4,4,1)
(4,4,2)
(4,4,3)
(4,4,4)

You could try a functional version without assignments:

def trigrams[T](tokens : Seq[T]) = {
  val s1 = tokens.map { Some(_) }
  val s2 = None +: s1
  val s3 = None +: s2
  s1 zip s2 zip s3 map {
    case ((t1, t2), t3) => (List(t1), List(t1, t2), List(t1, t2, t3))
  }
}