how do I initialize an immutable map programmatically?

The easiest way is probably,
val mymap = ndx zip vals toMap

On Thu, Dec 15, 2011 at 9:23 PM, Kevin Murphy <murphyk2@gmail.com> wrote:

Suppose I have lists of indices (keys) and values
and I want to make an immutable k->v map.
How can I do this?

Here's my first try:

 val mymap =  for ( (i,v) <- ndx.zip(vals) ) yield { i -> v}

But this fails, as shown in this example:

 val ndx = List(1,3)
 val vals = List(0.5, 0.6)
val mymap =  for ( (i,v) <- ndx.zip(vals) ) yield { i -> v}
//  List[(Int, Double)] = List((1,0.5), (3,0.6))

So I ended up using this ugly (but correct) imperative code:

 val mapping = mutable.Map[Int, Double]()
 for ( (i,v) <- ndx.zip(vals)) { mapping(i) = v }
 val mymap = mapping.toMap
// scala.collection.immutable.Map[Int,Double] = Map(3 -> 0.6, 1 ->
0.5)

Is there a better way?

Tx
Kevin

Also, if you specify the return type, you can also skip the intermediate creation of a list of 2-tuples by using collection.breakOut:
val mymap: Map[Int,Double] = ndx.zip(vals)(collection.breakOut)

On Thu, Dec 15, 2011 at 9:23 PM, Kevin Murphy <murphyk2@gmail.com> wrote:

Suppose I have lists of indices (keys) and values
and I want to make an immutable k->v map.
How can I do this?

Here's my first try:

 val mymap =  for ( (i,v) <- ndx.zip(vals) ) yield { i -> v}

But this fails, as shown in this example:

 val ndx = List(1,3)
 val vals = List(0.5, 0.6)
val mymap =  for ( (i,v) <- ndx.zip(vals) ) yield { i -> v}
//  List[(Int, Double)] = List((1,0.5), (3,0.6))

So I ended up using this ugly (but correct) imperative code:

 val mapping = mutable.Map[Int, Double]()
 for ( (i,v) <- ndx.zip(vals)) { mapping(i) = v }
 val mymap = mapping.toMap
// scala.collection.immutable.Map[Int,Double] = Map(3 -> 0.6, 1 ->
0.5)

Is there a better way?

Tx
Kevin

Another option: Map(ndx zip vals: _*)

On Thu, Dec 15, 2011 at 9:52 PM, Tom Switzer <thomas.switzer@gmail.com> wrote:

Also, if you specify the return type, you can also skip the intermediate creation of a list of 2-tuples by using collection.breakOut:
val mymap: Map[Int,Double] = ndx.zip(vals)(collection.breakOut)

On Thu, Dec 15, 2011 at 9:23 PM, Kevin Murphy <murphyk2@gmail.com> wrote:

Suppose I have lists of indices (keys) and values
and I want to make an immutable k->v map.
How can I do this?

Here's my first try:

 val mymap =  for ( (i,v) <- ndx.zip(vals) ) yield { i -> v}

But this fails, as shown in this example:

 val ndx = List(1,3)
 val vals = List(0.5, 0.6)
val mymap =  for ( (i,v) <- ndx.zip(vals) ) yield { i -> v}
//  List[(Int, Double)] = List((1,0.5), (3,0.6))

So I ended up using this ugly (but correct) imperative code:

 val mapping = mutable.Map[Int, Double]()
 for ( (i,v) <- ndx.zip(vals)) { mapping(i) = v }
 val mymap = mapping.toMap
// scala.collection.immutable.Map[Int,Double] = Map(3 -> 0.6, 1 ->
0.5)

Is there a better way?

Tx
Kevin

Thanks for all these helpful replies.
I found the same question (and answer) here
http://stackoverflow.com/questions/2763584/how-would-i-yield-an-immutabl...

Using the factor Map constructor (rather than toMap method)
has the advantage that we can also create mutable maps,
eg

scala> mutable.Map((ndx zip vals):_*)
res11: scala.collection.mutable.Map[Int,Double] = Map(3 -> 0.6, 1 ->
0.5)

Apparently "the :_* is needed to give the compiler a hint to use the
list as a varargs argument"
(which I interpret to be like using m{:} instead of m(:) in Matlab,
FWIW)

A pretty complex discussion of breakOut is given here:
http://stackoverflow.com/questions/1715681/scala-2-8-breakout

Kevin

On Dec 15, 11:24 pm, Naftoli Gugenheim wrote:
> Another option: Map(ndx zip vals: _*)
>
> On Thu, Dec 15, 2011 at 9:52 PM, Tom Switzer wrote:
>
>
>
>
>
>
>
> > Also, if you specify the return type, you can also skip the intermediate
> > creation of a list of 2-tuples by using collection.breakOut:
>
> > val mymap: Map[Int,Double] = ndx.zip(vals)(collection.breakOut)
>
> > On Thu, Dec 15, 2011 at 9:23 PM, Kevin Murphy wrote:
>
> >> Suppose I have lists of indices (keys) and values
> >> and I want to make an immutable k->v map.
> >> How can I do this?
>
> >> Here's my first try:
>
> >>  val mymap =  for ( (i,v) <- ndx.zip(vals) ) yield { i -> v}
>
> >> But this fails, as shown in this example:
>
> >>  val ndx = List(1,3)
> >>  val vals = List(0.5, 0.6)
> >> val mymap =  for ( (i,v) <- ndx.zip(vals) ) yield { i -> v}
> >> //  List[(Int, Double)] = List((1,0.5), (3,0.6))
>
> >> So I ended up using this ugly (but correct) imperative code:
>
> >>  val mapping = mutable.Map[Int, Double]()
> >>  for ( (i,v) <- ndx.zip(vals)) { mapping(i) = v }
> >>  val mymap = mapping.toMap
> >> // scala.collection.immutable.Map[Int,Double] = Map(3 -> 0.6, 1 ->
> >> 0.5)
>
> >> Is there a better way?
>
> >> Tx
> >> Kevin