scala,diff,sequence,seq , Scala - finding first position in which two Seq differ

## Question:

Tag: scala,diff,sequence,seq

Scala comes with the nice `corresponds` method:

``````val a = scala.io.Source.fromFile("fileA").getLines().toSeq()
val b = scala.io.Source.fromFile("fileB").getLines().toSeq()

val areEqual = a.corresponds(b){_.equals(_)}

if(areEqual) ...
``````

And I quite like the brevity of that.

Is there a similar method already defined that will also report to me the first position in which the two sequences differ?

I.e. is there a more idiomatic way to write something like this:

``````val result = ((seqA zip seqB).zipWithIndex).find{case ((a,b),i) => !a.equals(b)} match{
case Some(((a,b),i)) => s"seqA and seqB differ in pos \$i: \$a <> \$b"
case _ => "no difference"
}
``````

Because as you can see, that's a bloody pain in the neck to read. And it gets even worse if I want to use triplets instead of tuples of tuples:

``````val result = (((seqA zip seqB).zipWithIndex) map {case (t,i) => (t._1,t._2,i)}).find{case (a,b,i) => !a.equals(b)} match{
case Some((a,b,i)) => s"seqA and seqB differ in pos \$i: \$a <> \$b"
case _ => "no difference"
}
``````

I am aware of the `diff` method. Unfortunately, that one disregards the order of the elements.

You can use `indexWhere` (see ScalaDoc) as follows:

``````(as zip bs).indexWhere{case (x,y) => x != y}
``````

Example:

``````scala> val as = List(1,2,3,4)
scala> val bs = List(1,2,4,4)

scala> (as zip bs).indexWhere{case (x,y) => x != y}

res0: Int = 2
``````

However, note that all solutions based on `zip` may report no differences if one Seq is longer than the other (`zip` truncates the longer Seq ) - this might or might not be what you need...

Update: For Seqs of equal length, a different approach is as follows:

``````as.indices.find(i => as(i) != bs(i))
``````

This is nice as it returns an `Option[Int]`, so it returns `None` rather than a magical -1 if there is no difference between the Seqs.

It behaves the same as the other solution if `as` is shorter than `bs`, but fails if `as` is longer (you could take the minimum length, of course).

However, because it addresses both Seqs by index, it will only perform well for `IndexedSeq`s.

Update 2: We can deal with different Seq lengths by using `lift`, so that we get an Option when retrieving elements by index:

``````bs.indices.find(i => as.lift(i) != bs.lift(i))
``````

so if `as = [1,2]` and `bs = [1,2,3]`, the first index by which they differ is 2 (because this element is missing in `as`). However, in this case we need to call `indices` on the longest Seq rather than the shortest - or explicitly check which is longest using `max`, e.g.

``````(0 until (as.length max bs.length)).find(i => as.lift(i) != bs.lift(i))
``````

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