Scala for-loop. Getting index in consice way

问题:

In this code I want to increment index to put it to each yielding result.

var index=0

for(str <- splitToStrings(text) ) yield  {

  if (index != 0) index += 1               // but index is equal to `0` all the time

  new Word(str, UNKNOWN_FORM, index )
}

Why I can not change index ? And what the best way to implement this logic then, trying to be concise?

回答1:

The zipWithIndex method on most sequence-like collections will give you a zero-based index, incrementing with each element:

for ((str, index) <- splitToStrings(text).zipWithIndex)
  yield new Word(str, UNKNOWN_FORM, index)

回答2:

Because initially index is set to 0, thus your condition index != 0 is never executes to true and index is never got incremented. Maybe you don't need this condition? Maybe you can count results afterwards? Now I see that index is used within loop. Then you have to either use @BenJames answer or go recursive.

回答3:

zipWithIndex will copy and create a new collection, so better make it lazy when the collection is potentially large

for ((str, index) <- splitToStrings(text).view.zipWithIndex)
  yield new Word(str, UNKNOWN_FORM, index)

In fact, if you are working with an indexed sequence, then a more efficient way is to use indices, which produces the range of all indices of this sequence.

val strs = splitToStrings(text)

for(i <- strs.indices) yield  {
  new Word(strs(i), UNKNOWN_FORM, i )
}

回答4:

splitToStrings(text).foldLeft(0,List[Word]){(a,b) => {
   if(a._1!=0) (a._1+1,new Word(str, UNKNOWN_FORM, index) :: b)
   else (a._1,new Word(str, UNKNOWN_FORM, index) :: b)
}}

I am using foldLeft here with a tuple as: starting base with index = 0 and an empty List. I then iterate over each element.

Above a is this tuple. I check the index value and increment it. Else I dont add the index. And I add the new Word to the list.

Ultimately in the end you get a tuple containing the index value and the total List containing all Words.