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I'm trying to pass reverse ordering to the scala.util.Sorting.quickSort with this code:
val a = Array(3, 5, 1, 2)
scala.util.Sorting.quickSort(a)(Ordering.Int.reverse)
It doesn't work saying that:
Error: Unit does not take parameters
However, the code like that works:
val a = Array(3, 5, 1, 2)
a.sorted(Ordering.Int.reverse)
I don't understand why the quickSort example doesn't work? The Ordering.Int.reverse produce Ordering[Int] and according to documentation, the quickSort accepts it implicitly.
I'm running with Scala 2.12.7.
Expanding on Yuval Itzchakov's comment, here is the source code for scala.util.Sorting.quickSort:
def quickSort(a: Array[Int]): Unit = java.util.Arrays.sort(a)
A few lines down, another overloaded quicksort is defined as:
def quickSort[K: Ordering](a: Array[K]): Unit = {...
You mean to use this latter one. But when you call quickSort, the compiler is picking the first one which does not accept an Ordering in its parameter list so it complains about extra parameters. You will need to specify type parameter if you are using Int, Double, or Float as explained here. To summarize:
val a : Array[Int] = Array(3, 5, 1, 2)
val b = Array(3L, 5L, 1L, 2L) //Longs do not have this problem!
scala.util.Sorting.quickSort[Int](a)(Ordering.Int.reverse)
scala.util.Sorting.quickSort(b)(Ordering.Long.reverse)
println(a.toList)
println(b.toList)
List(5, 3, 2, 1)
List(5, 3, 2, 1)
According to documentation the signature of method that uses Ordering is:
def quickSort[K](a: Array[K])(implicit arg0: math.Ordering[K]): Unit
Sort array a with quicksort, using the Ordering on its elements. This
algorithm sorts in place, so no additional memory is used aside from
what might be required to box individual elements during comparison.
try this:
scala> val a = Array(3, 5, 1, 2)
a: Array[Int] = Array(3, 5, 1, 2)
scala> scala.util.Sorting.quickSort(a)(Ordering.Int.reverse)
<console>:13: error: Unit does not take parameters
scala.util.Sorting.quickSort(a)(Ordering.Int.reverse)
^
scala> scala.util.Sorting.quickSort[Int](a)(Ordering[Int].reverse)
scala> a
res2: Array[Int] = Array(5, 3, 2, 1)
Why is the below statement valid for .map() but not for .flatMap()?
val tupled = input.map(x => (x*2, x*3))
//Compilation error: cannot resolve reference flatMap with such signature
val tupled = input.flatMap(x => (x*2, x*3))
This statement has no problem, though:
val tupled = input.flatMap(x => List(x*2, x*3))
Assuming input if of type List[Int], map takes a function from Int to A, whereas flatMap takes a function from Int to List[A].
Depending on your use case you can choose either one or the other, but they're definitely not interchangeable.
For instance, if you are merely transforming the elements of a List you typically want to use map:
List(1, 2, 3).map(x => x * 2) // List(2, 4, 6)
but you want to change the structure of the List and - for example - "explode" each element into another list then flattening them, flatMap is your friend:
List(1, 2, 3).flatMap(x => List.fill(x)(x)) // List(1, 2, 2, 3, 3, 3)
Using map you would have had List(List(1), List(2, 2), List(3, 3, 3)) instead.
To understand how this is working, it can be useful to explicitly unpack the functions you're sending to map and flatMap and examine their signatures. I've rewritten them here, so you can see that f is a function mapping from Int to an (Int, Int) tuple, and g is a function that maps from Int to a List[Int].
val f: (Int) => (Int, Int) = x => (x*2, x*3)
val g: (Int) => List[Int] = x => List(x*2, x*3)
List(1,2,3).map(f)
//res0: List[(Int, Int)] = List((2,3), (4,6), (6,9))
List(1,2,3).map(g)
//res1: List[List[Int]] = List(List(2, 3), List(4, 6), List(6, 9))
//List(1,2,3).flatMap(f) // This won't compile
List(1,2,3).flatMap(g)
//res2: List[Int] = List(2, 3, 4, 6, 6, 9)
So why won't flatMap(f) compile? Let's look at the signature for flatMap, in this case pulled from the List implementation:
final override def flatMap[B, That](f : scala.Function1[A, scala.collection.GenTraversableOnce[B]])(...)
This is a little difficult to unpack, and I've elided some of it, but the key is the GenTraversableOnce type. List, if you follow it's chain of inheritance up, has this as a trait it is built with, and thus a function that maps from some type A to a List (or any object with the GenTraversableOnce trait) will be a valid function. Notably, tuples do not have this trait.
That is the in-the-weeds explanation why the typing is wrong, and is worth explaining because any error that says 'cannot resolve reference with such a signature' means that it can't find a function that takes the explicit type you're offering. Types are very often inferred in Scala, and so you're well-served to make sure that the type you're giving is the type expected by the method you're calling.
Note that flatMap has a standard meaning in functional programming, which is, roughly speaking, any mapping function that consumes a single element and produces n elements, but your final result is the concatenation of all of those lists. Therefore, the function you pass to flatMap will always expect to produce a list, and no flatMap function would be expected to know how to act on single elements.
The following code snippet
val keys=List(3,2,1,0)
val unsorted=List(1, 2, 3, 4)
val sorted =keys map unsorted
does sorting based on the keys.
Normally, map method takes a lambda and apply it to every element in the source. Instead the above code takes a list and does a sorting based on the index.
What is happening in this particular situation?
List is a partial function that goes from its indices to its values. You could look at it like this:
scala> List(3,2,1,0) map (i => List(1,2,3,4)(i))
res0: List[Int] = List(4, 3, 2, 1)
to verify
scala> val f: Int => Int = List(1, 2, 3, 4)
f: Int => Int = List(1, 2, 3, 4)
It does not sorting them, map takes a function, lets see:
that statement translates to:
keys.map(unsorted.apply)
Scala's List.apply(n: Int) function return nth element of list, in this case map makes it:
unsorted(3), unsorted(2), unsorted(1), unsorted(4)
which is
4,3,2,1 which is sorted only by chance in this case.
Test this:
scala> val keys = List(3,0,1,2)
keys: List[Int] = List(3, 0, 1, 2)
scala> val unsorted = List(10,20,30,40)
unsorted: List[Int] = List(10, 20, 30, 40)
scala> keys map unsorted
res0: List[Int] = List(40, 10, 20, 30)
In keys map unsorted for each value in keys we call the apply method in list unsorted which yields the element at a key position.
Swap elements in keys and note the outcome, for instance note the last key,
val keys=List(2,1,0,3)
val sorted = keys map unsorted
sorted: List[Int] = List(3, 2, 1, 4)
The function .map has not specific relation with sorting.
Your snipet is a specific case that happens to sort. If desugaring, it does as follows.
/* keys */List(3,2,1,0).map { i: Int =>
unsorted.apply(i)
}
It applies unsorted.apply with each value of keys (used as indexes).
It's not sorting at all. It is normal situation: map method takes a lambda and apply it to every element in the source. Just change your sorted list and you'll see it.
Using scalactic, one can convert an Every to a List like this:
scala> Every(1,2,3).toList
res6: List[Int] = List(1, 2, 3)
How can I perform the reverse operation, though, i.e. try to convert a List or an Iterable to an Every? Is there a built-in method that does this?
Simply:
val everyOpt: Option[Every[Int]] = Every.from(List(1, 2, 3))
If the list is known to be non empty:
val every: Every[Int] = Every.from(List(1, 2, 3)).get
This was quite an unplesant surprise:
scala> Set(1, 2, 3, 4, 5)
res18: scala.collection.immutable.Set[Int] = Set(4, 5, 1, 2, 3)
scala> Set(1, 2, 3, 4, 5).toList
res25: List[Int] = List(5, 1, 2, 3, 4)
The example by itself suggest a "no" answer to my question. Then what about ListSet?
scala> import scala.collection.immutable.ListSet
scala> ListSet(1, 2, 3, 4, 5)
res21: scala.collection.immutable.ListSet[Int] = Set(1, 2, 3, 4, 5)
This one seems to work, but should I rely on this behavior?
What other data structure is suitable for an immutable collection of unique items, where the original order must be preserved?
By the way, I do know about distict method in List. The problem is, I want to enforce uniqueness of items (while preserving the order) at interface level, so using distinct would mess up my neat design..
EDIT
ListSet doesn't seem very reliable either:
scala> ListSet(1, 2, 3, 4, 5).toList
res28: List[Int] = List(5, 4, 3, 2, 1)
EDIT2
In my search for a perfect design I tried this:
scala> class MyList[A](list: List[A]) { val values = list.distinct }
scala> implicit def toMyList[A](l: List[A]) = new MyList(l)
scala> implicit def fromMyList[A](l: MyList[A]) = l.values
Which actually works:
scala> val l1: MyList[Int] = List(1, 2, 3)
scala> l1.values
res0: List[Int] = List(1, 2, 3)
scala> val l2: List[Int] = new MyList(List(1, 2, 3))
l2: List[Int] = List(1, 2, 3)
The problem, however, is that I do not want to expose MyList outside the library. Is there any way to have the implicit conversion when overriding? For example:
trait T { def l: MyList[_] }
object O extends T { val l: MyList[_] = List(1, 2, 3) }
scala> O.l mkString(" ") // Let's test the implicit conversion
res7: String = 1 2 3
I'd like to do it like this:
object O extends T { val l = List(1, 2, 3) } // Doesn't work
That depends on the Set you are using. If you do not know which Set implementation you have, then the answer is simply, no you cannot be sure. In practice I usually encounter the following three cases:
I need the items in the Set to be ordered. For this I use classes mixing in the SortedSet trait which when you use only the Standard Scala API is always a TreeSet. It guarantees the elements are ordered according to their compareTo method (see the Ordered trat). You get a (very) small performance penalty for the sorting as the runtime of inserts/retrievals is now logarithmic, not (almost) constant like with the HashSet (assuming a good hash function).
You need to preserve the order in which the items are inserted. Then you use the LinkedHashSet. Practically as fast as the normal HashSet, needs a little more storage space for the additional links between elements.
You do not care about order in the Set. So you use a HashSet. (That is the default when using the Set.apply method like in your first example)
All this applies to Java as well, Java has a TreeSet, LinkedHashSet and HashSet and the corresponding interfaces SortedSet, Comparable and plain Set.
It is my belief that you should never rely on the order in a set. In no language.
Apart from that, have a look at this question which talks about this in depth.
ListSet will always return elements in the reverse order of insertion because it is backed by a List, and the optimal way of adding elements to a List is by prepending them.
Immutable data structures are problematic if you want first in, first out (a queue). You can get O(logn) or amortized O(1). Given the apparent need to build the set and then produce an iterator out of it (ie, you'll first put all elements, then you'll remove all elements), I don't see any way to amortize it.
You can rely that a ListSet will always return elements in last in, first out order (a stack). If that suffices, then go for it.