Scala for-loop. Getting index in consice way

Question

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?

score 25 · Accepted Answer · answered Jun 02 '13 at 14:59

25

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)

answered Jun 02 '13 at 14:59

Ben James

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That's nice. I've already seen this kind of transformation. But it still in passive knowledge. That's concise indeed. – ses Jun 02 '13 at 15:05
wouldn't that be inefficient? – Yan King Yin Mar 21 '15 at 10:23

score 6 · Answer 2 · edited May 23 '17 at 12:17

6

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.

edited May 23 '17 at 12:17

Community

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answered Jun 02 '13 at 14:51

om-nom-nom

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yep. that was stupid fro my side. at least I'm glad that no surprise from yield. (I just put it because I want to start from 0-index). I tent to think that something goes not right way because of scala but not me :). but this differennt case %) there are many surprises I've got recently.. that's why – ses Jun 02 '13 at 14:54

Max · Answer 3 · 2013-06-05T03:46:45.837

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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 )
}

edited Jun 05 '13 at 03:46

answered Jun 03 '13 at 01:08

Max

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this will go go though whole list all the time to get the item by index: strs(i). might be slow. – ses Jun 03 '13 at 18:13
updated answer. Sequence need to be indexed in order to provide a good random access performance. – Max Jun 05 '13 at 03:47

Jatin · Answer 4 · 2013-06-02T15:41:32.723

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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.

edited Jun 02 '13 at 15:41

answered Jun 02 '13 at 15:01

Jatin

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Code like this could easily use some explanatory words to go with it. – likeitlikeit Jun 02 '13 at 15:23

Scala for-loop. Getting index in consice way

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