Ok, I wouldn't be coming to You for help if I knew what to do, anyways, still having problems with my "program".
class Mark(val name: String, val style_mark: Double, val other_mark: Double) {}
object Test extends Application
{
val m1 = new Mark("Smith", 18, 16);
val m2 = new Mark("Cole", 14, 7);
val m3 = new Mark("James", 13, 15);
val m4 = new Mark("Jones", 14, 16);
val m5 = new Mark("Richardson", 20, 19);
val m6 = new Mark("James", 4, 18);
val marks = List[Mark](m1, m2, m3, m4, m5, m6);
def avg(xs: List[Double]) = xs.sum / xs.length
val marksGrouped = marks.groupBy(_.name).map { kv => new Mark(kv._1, avg(kv._2.map(_.style_mark)), avg(kv._2.map(_.other_mark))) }
val marksSorted = marksGrouped.sortWith((m1, m2) => m1._style_mark < m2._style_mark)
}
And this is the error I get: error: value sortWith is not a member of scala.collection.immutable.Iterable[Mark]
You'll have to call toList on marksGrouped first. Iterable does not have a sortWith method, but List does.
Basic collection hierarchy:
TraversableOnce: might only be traversable once, like iterators.
Traversable: May be traversed, with foreach, but no other guarantees provided.
Iterable: Can produce iterator, which enables lazy traversal.
Seq: contents have a fixed traversal order.
IndexedSeq: contents can be efficiently retrieved by position number.
Set: contents only contain one element of each type.
Map: contents can be efficiently retrieved by a key.
So the problem you face is that Iterable does not provide support to define the traversal order, which is what sortWith does. Only collections derived from Seq can -- List, Vector, ArrayBuffer, etc.
The method toSeq will return a Seq out of an Iterable. Or you may choose a more specific collection, with characteristics well matched to your algorithm.
For extra higher-order programming goodness, use sortBy, rather than sortWith
val marksSorted = marksGrouped.toList.sortBy(_.style_mark)
Related
I have a scala list and a funcion:
val l = List((1,2), (3,4), (5,6))
def f(x:Int, y:Int) = x+y
I want to build a dictionary that associated to each tuple its sum.
The result f(1,2) should be kept in memory and not recomputed each time that it is called. How can I do that? In python I would use a dictionary.
l.map(x => x -> (f _).tupled(x)).toMap
An analog to Python dictionaries in Scala is Map:
λ val resultsMap = Map( // Create a map from a Seq of (key, value) pairs
l.zip( // List(((1, 2), SOMETHING), ((3, 4), SOMETHING), etc.)
l.map((f _).tupled) // Invoke `f` on each entry in `l`
).toSeq : _*)
resultsMap: Map[(Int, Int), Int] = Map((1, 2) -> 3, (3, 4) -> 7, (5, 6) -> 11)
Usage:
λ resultsMap((1, 2))
res0: Int = 3
Using a for comprehension we can extract the tuple values as well as bind each tuple for constructing the associations, like this,
for ( t#(a,b) <- l ) yield t -> f(a,b)
Thus applying toMap on the collection from the comprehension we get the desired map. In this approach though we must declare as many items in the tuple as we wish to be applied to the function.
Let's say that I have a Map of strings -> List of Integers. I would like to create a function which takes in as a parameter a List of strings and returns all the integers correlating to all the string in that list. I.e. if the Map X contains the following mappings:
database = [("Bob",[1,2,3]),("John",[1,5,6]),("Trevor",[4,5,7])]
If this function takes in ["Bob","John"] as the list of names, it should return,
[1,2,3,5,6]
Since Bob correlates to 1,2,3 and John correlates to 1,5,6 (same entries for both names aren't duplicated). I also would like to not introduce a mutable variable if I don't have to, thus leading me to believe a for comprehension that yields this list of number values would be the best way to achieve this, but I'm not sure how.
If you want to use a for-comprehension you can so this:
val result = for {
key <- keys
nums <- map.get(key).toSeq
num <- nums
} yield num
result.distinct
Explanation:
for each key in the list try to get an entry and convert it to a Seq (necessary because flatMap expects a Seq in this case) and add every number in the list to the result. If the key is not present in the map the collection will be empty and therefore not yield any results. At the end call distinct to remove the duplicates.
val myMap = Map("Bob" -> List(1,2,3), "John" -> List(1,5,6), "Trevor" -> List(4,5,7))
val names = List("Bob", "John")
You can add default value to Map using method withDefaultValue:
val mapWithDefaul = myMap withDefaultValue Nil
Then you could use Map as function in flatMap:
names.flatMap(mapWithDefaul).distinct
// List(1, 2, 3, 5, 6)
Let
val db = Map("Bob" -> List(1,2,3), "John" -> List(1,5,6), "Trevor" -> List(4,5,7))
val names = List("Bob", "John")
Then a similar approach to #senia's using flatMap,
implicit class mapCorr[A,B](val db: Map[A,List[B]]) extends AnyVal {
def corr(keys: List[A]): List[B] = {
keys.flatMap{ k => db get k }.flatten.distinct
}
}
and
scala> db.corr(keys)
res0: List[Int] = List(1, 2, 3, 5, 6)
Here we allow for key lists of type A and maps from type A to type List[B] .
val myset = Set("Bob","John")
val database = Map(("Bob"->List(1,2,3)),("John"->List(1,5,6)),("Trevor"->List(4,5,7)))
val ids = database.filter(m => myset.contains(m._1)).map(_._2).flatten.toList.distinct
outputs:
ids: List[Int] = List(1, 2, 3, 5, 6)
Something like:
val result = database.filter(elem => list.contains(elem._1)).foldLeft(List())((res,elem) => res ++ elem._2)
where list is the input list of names.
Let's say I have List(1,2,3,4,5) and I want to get
List(3,5,7,9), that is, the sums of the element and the previous (1+2, 2+3,3+4,4+5)
I tried to do this by making two lists:
val list1 = List(1,2,3,4)
val list2 = (list1.tail ::: List(0)) // 2,3,4,5,0
for (n0_ <- list1; n1th_ <- list2) yield (n0_ + n1_)
But that combines all the elements with each other like a cross product, and I only want to combine the elements pairwise. I'm new to functional programming and I thought I'd use map() but can't seem to do so.
List(1, 2, 3, 4, 5).sliding(2).map(_.sum).to[List] does the job.
Docs:
def sliding(size: Int): Iterator[Seq[A]]
Groups elements in fixed size blocks by passing a "sliding window" over them (as opposed to partitioning them, as is done in grouped.)
You can combine the lists with zip and use map to add the pairs.
val list1 = List(1,2,3,4,5)
list1.zip(list1.tail).map(x => x._1 + x._2)
res0: List[Int] = List(3, 5, 7, 9)
Personally I think using sliding as Infinity has is the clearest, but if you want to use a zip-based solution then you might want to use the zipped method:
( list1, list1.tail ).zipped map (_+_)
In addition to being arguably clearer than using zip, it is more efficient in that the intermediate data structure (the list of tuples) created by zip is not created with zipped. However, don't use it with infinite streams, or it will eat all of your memory.
This describes the problem pretty well:
scala> var l2 = List(1,2,3)
l2: List[Int] = List(1, 2, 3)
scala> l2(2) = 55
<console>:10: error: value update is not a member of List[Int]
l2(2) = 55
^
scala.List is immutable, meaning you cannot update it in place. If you want to create a copy of your List which contains the updated mapping, you can do the following:
val updated = l2.updated( 2, 55 )
There are mutable ordered sequence types as well, in scala.collection.mutable, such as Buffer types which seem more like what you want. If you try the following you should have more success:
scala> import scala.collection._
import scala.collection._
scala> val b = mutable.Buffer(1,2,3)
b: scala.collection.mutable.Buffer[Int] = ArrayBuffer(1, 2, 3)
scala> b(2) = 55
scala> b
res1: scala.collection.mutable.Buffer[Int] = ArrayBuffer(1, 2, 55)
Edit: Just to note that some other answers have mentioned that you should use a "mutable List type" - this is true, but "List" in Scala just refers to the single-linked list, whereas in Java it's generally used for any ordered, random-access collection. There is also a DoubleLinkedList, which is more like a Java LinkedList, and a MutableList, which is a type used for the internals of some other types.
Generally speaking what you probably want in Scala is a Buffer for this job; especially since the default implementation is an ArrayBuffer, which is pretty close to being the same as ArrayList, most peoples' default, in Java.
If you ever want to find out what the closest "mapping" of a Java collections interface to the Scala world is, though, the easiest thing to do is probably just check what JavaConversions does. In this case you can see the mapping is to Buffer:
scala.collection.mutable.Buffer <=> java.util.List
The List you are creating is immutable.
scala> val l = List(1,2,3)
l: List[Int] = List(1, 2, 3)
scala> l.getClass
res3: java.lang.Class[_] = class scala.collection.immutable.$colon$colon
Use a mutable List instead and you should be fine.
But you can wrap the assignment in an implicit if you don't like updated
implicit class RichList[A](l: List[A]) {
def update(which: Int, what: A): List[A] = {
l.updated(which, what)
}
}
val l = List(1, 2, 3)
val l2 = l(2) = 55
List(1, 2, 55)
Your problem is that Lists in scala are immutable, you need to use a mutable list instead.
Import scala.collection.mutable.Queue and use that instead. Queue is a MutableList, so it should do what you want.
Lists are immutable in scala. But if you want to replace an element in a List you can use "updated" method like this
val num:List[Int]=List(1,5,4,7,8,2,10)
num=num.updated(0,22)
Since Lists are immutable,this creates a copy of the first List called num(as we assigned the new list to 'num') by replacing 0th element to 22.So the genaral updated method is
listname.updated(index,value)
I would like a List, Seq, or even an Iterable that is a read-only view of a part of a List, in my specific case, the view will always start at the first element.
List.slice, is O(n) as is filter. Is there anyway of doing better than this - I don't need any operations like +, - etc. Just apply, map, flatMap, etc to provide for list comprehension syntax on the sub list.
Is the answer to write my own class whose iterators keep a count to know where the end is?
How about Stream? Stream is Scala's way to laziness. Because of Stream's laziness, Stream.take(), which is what you need in this case, is O(1). The only caveat is that if you want to get back a List after doing a list comprehension on a Stream, you need to convert it back to a List. List.projection gets you a Stream which has most of the opeations of a List.
scala> val l = List(1, 2, 3, 4, 5)
l: List[Int] = List(1, 2, 3, 4, 5)
scala> val s = l.projection.take(3)
s: Stream[Int] = Stream(1, ?)
scala> s.map(_ * 2).toList
res0: List[Int] = List(2, 4, 6)
scala> (for (i <- s) yield i * 2).toList
res1: List[Int] = List(2, 4, 6)
List.slice and List.filter both return Lists -- which are by definition immutable.The + and - methods return a different List, they do not change the original List. Also, it is hard to do better than O(N). A List is not random access, it is a linked list. So imagine if the sublist that you want is the last element of the List. The only way to access that element is to iterate over the entire List.
Well, you can't get better than O(n) for drop on a List. As for the rest:
def constantSlice[T](l: List[T], start: Int, end: Int): Iterator[T] =
l.drop(start).elements.take(end - start)
Both elements and take (on Iterator) are O(1).
Of course, an Iterator is not an Iterable, as it is not reusable. On Scala 2.8 a preference is given to returning Iterable instead of Iterator. If you need reuse on Scala 2.7, then Stream is probably the way to go.