Scala filter by set - scala

Question

Say I have a map that looks like this
val map = Map("Shoes" -> 1, "heels" -> 2, "sneakers" -> 3, "dress" -> 4, "jeans" -> 5, "boyfriend jeans" -> 6)
And also I have a set or collection that looks like this:
val set = Array(Array("Shoes", "heels", "sneakers"), Array("dress", "maxi dress"), Array("jeans", "boyfriend jeans", "destroyed jeans"))
I would like to perform a filter operation on my map so that only one element in each of my set retains. Expected output should be something like this:
map = Map("Shoes" -> 1, "dress" -> 4 ,"jeans" -> 5)
The purpose of doing this is so that if I have multiple sets that indicate different categories of outfits, my output map doesn't "repeat" itself on technically the same objects.
Any help is appreciated, thanks!

Answer 1

So first get rid of the confusion that your sets are actually arrays. For the rest of the example I will use this definition instead:
val arrays = Array(Array("Shoes", "heels", "sneakers"), Array("dress", "maxi dress"), Array("jeans", "boyfriend jeans", "destroyed jeans"))
So in a sense you have an array of arrays of equivalent objects and want to remove all but one of them?
Well first you have to find which of the elements in an array are actually used as keys in the mep. So we just filter out all elements that are not used as keys:
array.filter(map.keySet)
Now, we have to chose one element. As you said, we just take the first one:
array.filter(map.keySet).head
As your "sets" are actually arrays, this is really the first element in your array that is also used as a key. If you would actually use sets this code would still work as sets actually have a "first element". It is just highly implementations specific and it might not even be deterministic over various executions of the same program. At least for immutable sets it should however be deterministic over several calls to head, i.e., you should always get the same element.
Instead of the first element we are actually interested in all other elements, as we want to remove them from the map:
array.filter(map.keySet).tail
Now, we just have to remove those from the map:
map -- array.filter(map.keySet).tail
And to do it for all arrays:
map -- arrays.flatMap(_.filter(map.keySet).tail)
This works fine as long as the arrays are disjoined. If they are not, we can not take the initial map to filter the array in every step. Instead, we have to use one array to compute a new map, then take the next starting with the result from the last and so on. Luckily, we do not have to do much:
arrays.foldLeft(map){(m,a) => m -- a.filter(m.keySet).tail}
Note: Sets are also functions from elements to Boolean, this is, why this solution works.

Answer 2

This code solves the problem:
var newMap = map
set.foreach { list =>
var remove = false
list.foreach { _key =>
if (remove) {
newMap -= _key
}
if (newMap.contains(_key)) {
remove = true
}
}
}
I'm completely new at Scala. I have taken this as my first Scala
example, please any hints from Scala's Gurus is welcome.

Answer 3

The basic idea is to use groupBy. Something like
map.groupBy{ case (k,v) => g(k) }.
map{ case (_, kvs) => kvs.head }
This is the general way to group similar things (using some function g). Now the question is just how to make the g that you need. One way is
val g = set.zipWithIndex.
flatMap{ case (a, i) => a.map(x => x -> i) }.
toMap
which labels each set with a number, and then forms a map so you can look it up. Maps have an apply function, so you can use it as above.

Answer 4

A slightly simpler version
set.flatMap(_.find(map.contains).map(y => y -> map(y)))