val A = 3
val (A) = (3)
Both correct. But:
val (A,B) = (2,3)
can't be compiled:
scala> val (A,B) = (2,3)
<console>:7: error: not found: value A
val (A,B) = (2,3)
^
<console>:7: error: not found: value B
val (A,B) = (2,3)
^
Why?
In the second code snippet, it using pattern matching to do assessment.
It is translated to the follow code:
val Tuple(A, B) = Tuple2(2,3)
When Scala is doing pattern matching, variable starts with a upper case in the pattern is considered as an constant value (or singleton Object), so val (a, b) = (2, 3) works but not val (A, B) = (2, 3).
BTW, your first code snippet does not using pattern matching, it's just an ordinary variable assignment.
If you using Tuple1 explicitly, it will have same error.
scala> val Tuple1(Z) = Tuple1(3)
<console>:7: error: not found: value Z
val Tuple1(Z) = Tuple1(3)
Here is some interesting example:
scala> val A = 10
A: Int = 10
scala> val B = 20
B: Int = 20
scala> val (A, x) = (10, 20)
x: Int = 20
scala> val (A, x) = (10, 30)
x: Int = 30
scala> val (A, x) = (20, 20)
scala.MatchError: (20,20) (of class scala.Tuple2$mcII$sp)
at .<init>(<console>:9)
at .<clinit>(<console>)
Related
I have several methods that operate on Vector sequences and the following idiom is common when combining data from multiple vectors into a single one with the use of a for comprehension / yield:
(for (i <- 0 until y.length) yield y(i) + 0.5*dy1(i)) toVector
Notice the closing toVector and the enclosing parentheses around the for comprehension. I want to get rid of it because it's ugly, but removing it produces the following error:
type mismatch;
found : scala.collection.immutable.IndexedSeq[Double]
required: Vector[Double]
Is there a better way of achieving what I want that avoids explicitly calling toVector many times to essentially achieve a non-operation (converting and indexed sequence...to an indexed sequence)?
One way to avoid collection casting, e.g. toVector, is to invoke, if possible, only those methods that return the same collection type.
y.zipWithIndex.map{case (yv,idx) => yv + 0.5*dy1(idx)}
for yield on Range which you are using in your example yields a Vector[T] by default.
example,
scala> val squares= for (x <- Range(1, 3)) yield x * x
squares: scala.collection.immutable.IndexedSeq[Int] = Vector(1, 4)
check the type,
scala> squares.isInstanceOf[Vector[Int]]
res14: Boolean = true
Note that Vector[T] also extends IndexedSeq[T].
#SerialVersionUID(-1334388273712300479L)
final class Vector[+A] private[immutable] (private[collection] val startIndex: Int, private[collection] val endIndex: Int, focus: Int)
extends AbstractSeq[A]
with IndexedSeq[A]
with GenericTraversableTemplate[A, Vector]
with IndexedSeqLike[A, Vector[A]]
with VectorPointer[A #uncheckedVariance]
with Serializable
with CustomParallelizable[A, ParVector[A]]
That's why above result is also an instance of IndexedSeq[T],
scala> squares.isInstanceOf[IndexedSeq[Int]]
res15: Boolean = true
You can define the type of your result as IndexedSeq[T] and still achieve what you want with Vector without explicitly calling .toVector
scala> val squares: IndexedSeq[Int] = for (x <- Range(1, 3)) yield x * x
squares: IndexedSeq[Int] = Vector(1, 4)
scala> squares == Vector(1, 4)
res16: Boolean = true
But for yield on Seq[T] gives List[T] by default.
scala> val squares = for (x <- Seq(1, 3)) yield x * x
squares: Seq[Int] = List(1, 9)
Only in that case if you want vector you must .toVector the result.
scala> squares.isInstanceOf[Vector[Int]]
res21: Boolean = false
scala> val squares = (for (x <- Seq(1, 3)) yield x * x).toVector
squares: Vector[Int] = Vector(1, 9)
The following code is supposed to append to a Seq, but it prints an empty list, what's wrong here?
object AppendToSeq extends App{
val x = Seq[Int]()
x :+ 1
x :+ 2
println(x)
}
the value x created is an immutable Sequence and the method :+ defined on the immutable sequence return a new Seq object.
so your code should have x has a var (a mutable variable) and it should have its value re-assigned after each append (:+) operation as shown below.
scala> var x = Seq[Int]()
x: Seq[Int] = List()
scala> x = x :+ 1
x: Seq[Int] = List(1)
scala> x = x :+ 2
x: Seq[Int] = List(1, 2)
scala> x
res2: Seq[Int] = List(1, 2)
x :+ 1 creates a new Seq by appending 1 to the existing Seq, x, but the new Seq isn't saved anywhere, i.e. it isn't assigned to any variable, so it's just thrown away.
If you want to modify an existing Seq you can make the variable a var instead of a val. Then when you create a new Seq you can save it under the same name.
scala> var x = Seq[Int]()
x: Seq[Int] = List()
scala> x = x :+ 7
x: Seq[Int] = List(7)
In ML, one can assign names for each element of a matched pattern:
fun findPair n nil = NONE
| findPair n (head as (n1, _))::rest =
if n = n1 then (SOME head) else (findPair n rest)
In this code, I defined an alias for the first pair of the list and matched the contents of the pair. Is there an equivalent construct in Scala?
You can do variable binding with the # symbol, e.g.:
scala> val wholeList # List(x, _*) = List(1,2,3)
wholeList: List[Int] = List(1, 2, 3)
x: Int = 1
I'm sure you'll get a more complete answer later as I'm not sure how to write it recursively like your example, but maybe this variation would work for you:
scala> val pairs = List((1, "a"), (2, "b"), (3, "c"))
pairs: List[(Int, String)] = List((1,a), (2,b), (3,c))
scala> val n = 2
n: Int = 2
scala> pairs find {e => e._1 == n}
res0: Option[(Int, String)] = Some((2,b))
OK, next attempt at direct translation. How about this?
scala> def findPair[A, B](n: A, p: List[Tuple2[A, B]]): Option[Tuple2[A, B]] = p match {
| case Nil => None
| case head::rest if head._1 == n => Some(head)
| case _::rest => findPair(n, rest)
| }
findPair: [A, B](n: A, p: List[(A, B)])Option[(A, B)]
I'm trying to figure out how to use StateT to combine two State state transformers based on a comment on my Scalaz state monad examples answer.
It seems I'm very close but I got an issue when trying to apply sequence.
import scalaz._
import Scalaz._
import java.util.Random
val die = state[Random, Int](r => (r, r.nextInt(6) + 1))
val twoDice = for (d1 <- die; d2 <- die) yield (d1, d2)
def freqSum(dice: (Int, Int)) = state[Map[Int,Int], Int]{ freq =>
val s = dice._1 + dice._2
val tuple = s -> (freq.getOrElse(s, 0) + 1)
(freq + tuple, s)
}
type StateMap[x] = State[Map[Int,Int], x]
val diceAndFreqSum = stateT[StateMap, Random, Int]{ random =>
val (newRandom, dice) = twoDice apply random
for (sum <- freqSum(dice)) yield (newRandom, sum)
}
So I got as far as having a StateT[StateMap, Random, Int] that I can unwrap with initial random and empty map states:
val (freq, sum) = diceAndFreqSum ! new Random(1L) apply Map[Int,Int]()
// freq: Map[Int,Int] = Map(9 -> 1)
// sum: Int = 9
Now I'd like to generate a list of those StateT and use sequence so that I can call list.sequence ! new Random(1L) apply Map[Int,Int](). But when trying this I get:
type StT[x] = StateT[StateMap, Random, x]
val data: List[StT[Int]] = List.fill(10)(diceAndFreqSum)
data.sequence[StT, Int]
//error: could not find implicit value for parameter n: scalaz.Applicative[StT]
data.sequence[StT, Int]
^
Any idea? I can use some help for the last stretch - assuming it's possible.
Ah looking at the scalaz Monad source, I noticed there was an implicit def StateTMonad that confirms that StateT[M, A, x] is a monad for type parameter x. Also monads are applicatives, which was confirmed by looking at the definition of the Monad trait and by poking in the REPL:
scala> implicitly[Monad[StT] <:< Applicative[StT]]
res1: <:<[scalaz.Monad[StT],scalaz.Applicative[StT]] = <function1>
scala> implicitly[Monad[StT]]
res2: scalaz.Monad[StT] = scalaz.MonadLow$$anon$1#1cce278
So this gave me the idea of defining an implicit Applicative[StT] to help the compiler:
type StT[x] = StateT[StateMap, Random, x]
implicit val applicativeStT: Applicative[StT] = implicitly[Monad[StT]]
That did the trick:
val data: List[StT[Int]] = List.fill(10)(diceAndFreqSum)
val (frequencies, sums) =
data.sequence[StT, Int] ! new Random(1L) apply Map[Int,Int]()
// frequencies: Map[Int,Int] = Map(10 -> 1, 6 -> 3, 9 -> 1, 7 -> 1, 8 -> 2, 4 -> 2)
// sums: List[Int] = List(9, 6, 8, 8, 10, 4, 6, 6, 4, 7)
I'm new to scala, and what I'm learning is tuple.
I can define a tuple as following, and get the items:
val tuple = ("Mike", 40, "New York")
println("Name: " + tuple._1)
println("Age: " + tuple._2)
println("City: " + tuple._3)
My question is:
How to get the length of a tuple?
Is tuple mutable? Can I modify its items?
Is there any other useful operation we can do on a tuple?
Thanks in advance!
1] tuple.productArity
2] No.
3] Some interesting operations you can perform on tuples: (a short REPL session)
scala> val x = (3, "hello")
x: (Int, java.lang.String) = (3,hello)
scala> x.swap
res0: (java.lang.String, Int) = (hello,3)
scala> x.toString
res1: java.lang.String = (3,hello)
scala> val y = (3, "hello")
y: (Int, java.lang.String) = (3,hello)
scala> x == y
res2: Boolean = true
scala> x.productPrefix
res3: java.lang.String = Tuple2
scala> val xi = x.productIterator
xi: Iterator[Any] = non-empty iterator
scala> while(xi.hasNext) println(xi.next)
3
hello
See scaladocs of Tuple2, Tuple3 etc for more.
One thing that you can also do with a tuple is to extract the content using the match expression:
def tupleview( tup: Any ){
tup match {
case (a: String, b: String) =>
println("A pair of strings: "+a + " "+ b)
case (a: Int, b: Int, c: Int) =>
println("A triplet of ints: "+a + " "+ b + " " +c)
case _ => println("Unknown")
}
}
tupleview( ("Hello", "Freewind"))
tupleview( (1,2,3))
Gives:
A pair of strings: Hello Freewind
A triplet of ints: 1 2 3
Tuples are immutable, but, like all cases classes, they have a copy method that can be used to create a new Tuple with a few changed elements:
scala> (1, false, "two")
res0: (Int, Boolean, java.lang.String) = (1,false,two)
scala> res0.copy(_2 = true)
res1: (Int, Boolean, java.lang.String) = (1,true,two)
scala> res1.copy(_1 = 1f)
res2: (Float, Boolean, java.lang.String) = (1.0,true,two)
Concerning question 3:
A useful thing you can do with Tuples is to store parameter lists for functions:
def f(i:Int, s:String, c:Char) = s * i + c
List((3, "cha", '!'), (2, "bora", '.')).foreach(t => println((f _).tupled(t)))
//--> chachacha!
//--> borabora.
[Edit] As Randall remarks, you'd better use something like this in "real life":
def f(i:Int, s:String, c:Char) = s * i + c
val g = (f _).tupled
List((3, "cha", '!'), (2, "bora", '.')).foreach(t => println(g(t)))
In order to extract the values from tuples in the middle of a "collection transformation chain" you can write:
val words = List((3, "cha"),(2, "bora")).map{ case(i,s) => s * i }
Note the curly braces around the case, parentheses won't work.
Another nice trick ad question 3) (as 1 and 2 are already answered by others)
val tuple = ("Mike", 40, "New York")
tuple match {
case (name, age, city) =>{
println("Name: " + name)
println("Age: " + age)
println("City: " + city)
}
}
Edit: in fact it's rather a feature of pattern matching and case classes, a tuple is just a simple example of a case class...
You know the size of a tuple, it's part of it's type. For example if you define a function def f(tup: (Int, Int)), you know the length of tup is 2 because values of type (Int, Int) (aka Tuple2[Int, Int]) always have a length of 2.
No.
Not really. Tuples are useful for storing a fixed amount of items of possibly different types and passing them around, putting them into data structures etc. There's really not much you can do with them, other than creating tuples, and getting stuff out of tuples.
1 and 2 have already been answered.
A very useful thing that you can use tuples for is to return more than one value from a method or function. Simple example:
// Get the min and max of two integers
def minmax(a: Int, b: Int): (Int, Int) = if (a < b) (a, b) else (b, a)
// Call it and assign the result to two variables like this:
val (x, y) = minmax(10, 3) // x = 3, y = 10
Using shapeless, you easily get a lot of useful methods, that are usually available only on collections:
import shapeless.syntax.std.tuple._
val t = ("a", 2, true, 0.0)
val first = t(0)
val second = t(1)
// etc
val head = t.head
val tail = t.tail
val init = t.init
val last = t.last
val v = (2.0, 3L)
val concat = t ++ v
val append = t :+ 2L
val prepend = 1.0 +: t
val take2 = t take 2
val drop3 = t drop 3
val reverse = t.reverse
val zip = t zip (2.0, 2, "a", false)
val (unzip, other) = zip.unzip
val list = t.toList
val array = t.toArray
val set = t.to[Set]
Everything is typed as one would expect (that is first has type String, concat has type (String, Int, Boolean, Double, Double, Long), etc.)
The last method above (.to[Collection]) should be available in the next release (as of 2014/07/19).
You can also "update" a tuple
val a = t.updatedAt(1, 3) // gives ("a", 3, true, 0.0)
but that will return a new tuple instead of mutating the original one.