What is the diference between func1 and func2 body. Both compiles but I couldn't figure out how scala can recognize multiple statements in func2 as brace is putted before a => but the statements are after that.
val func1: Int => Int = a => {
val b = a * 2
b + 2
}
val func2: Int => Int = { a =>
val b = a * 2
b + 2
}
Even this one compiles
val func3: Int => Int = { a => val b = a * 2; b + 2 }
But this one doesn't
val func4: Int => Int = a => val b = a * 2; b + 2
It's because when you don't use {} you can only have one expression and val b = a * 2; b + 2 are two expressions. The parser considers it something like:
val func4: Int => Int = a => val b = a * 2;
b + 2
Which doesn't compile because there isn't a b in scope and the type of the expression val b = a * 2; is Unit, which doesn't satisfies func4s type.
Other than that, func 1 through 3 are equivalent.
Related
In order to use infix notation, I have the following example of scala code.
implicit class myclass(n:Int ){
private def mapCombineReduce(map : Int => Double, combine: (Double,Double) => Double, zero: Double )(a:Int, b:Double): Double =
if( a > b) zero else combine ( map(a), mapCombineReduce(map,combine,zero)(a+1,b) )
var default_value_of_z : Int = 0
def sum( z : Int = default_value_of_z) = mapReduce( x=>x , (x,y) => x+y+z, 0)(1,n)
def ! = mapCombineReduce( x=> x, (x,y) => x*y, 1)(1,n)
}
4 !
4 sum 1 //sum the elements from 1 to 7 and each time combine the result, add 1 to the result.
4 sum
Is there any way in scala 2.12 to run 4 sum without have a double declaration of the sum method inside myclass ?
No, because default arguments are only used if argument list is provided
def f(x: Int = 1) = x
f // interpreted as trying to do eta-expansion
In fact starting Scala 3 it will indeed eta-expand
scala> def f(x: Int = 1) = x
def f(x: Int): Int
scala> f
val res1: Int => Int = Lambda$7473/1229666909#61a1990e
so in your case you will have to write 4.sum() with argument list present.
Write a program in Scala that reads an String from the keyboard and counts the number of characters, ignoring if its UpperCase or LowerCase
ex: Avocado
R: A = 2; v = 1; o = 2; c = 1; d = 2;
So, i tried to do it with two fors iterating over the string, and then a conditional to transform the character in the position (x) to Upper and compare with the character in the position (y) which is the same position... basically i'm transforming the same character so i can increment in the counter ex: Ava -> A = 2; v = 1;
But with this logic when i print the result it comes with:
ex: Avocado
R: A = 2; v = 1; o = 2; c = 1; a = 2; d = 1; o = 2;
its repeting the same character Upper or Lower in the result...
so my teacher asked us to resolve this using the split method and yield of Scala but i dunno how to use the split without forEach() that he doesnt allow us to use.
sorry for the bad english
object ex8 {
def main(args: Array[String]): Unit = {
println("Write a string")
var string = readLine()
var cont = 0
for (x <- 0 to string.length - 1) {
for (y <- 0 to string.length - 1) {
if (string.charAt(x).toUpper == string.charAt(y).toUpper)
cont += 1
}
print(string.charAt(x) + " = " + cont + "; ")
cont = 0
}
}
}
But with this logic when i print the result it comes with:
ex: Avocado
R: A = 2; V = 1; o = 2; c = 1; a = 2; d = 1; o = 2;
Scala 2.13 has added a very handy method to cover this sort of thing.
inputStr.groupMapReduce(_.toUpper)(_ => 1)(_+_)
.foreach{case (k,v) => println(s"$k = $v")}
//A = 2
//V = 1
//C = 1
//O = 2
//D = 1
It might be easier to group the individual elements of the String (i.e. a collection of Chars, made case-insensitive with toLower) to aggregate their corresponding size using groupBy/mapValues:
"Avocado".groupBy(_.toLower).mapValues(_.size)
// res1: scala.collection.immutable.Map[Char,Int] =
// Map(a -> 2, v -> 1, c -> 1, o -> 2, d -> 1)
Scala 2.11
Tried with classic word count approach of map => group => reduce
val exampleStr = "Avocado R"
exampleStr.
toLowerCase.
trim.
replaceAll(" +","").
toCharArray.map(x => (x,1)).groupBy(_._1).
map(x => (x._1,x._2.length))
Answer :
exampleStr: String = Avocado R
res3: scala.collection.immutable.Map[Char,Int] =
Map(a -> 2, v -> 1, c -> 1, r -> 1, o -> 2, d -> 1)
I hava a class with custom Operators
case class Num(var value:Int) {
def add(x:Num) = Num(value + x.value)
def mul(x:Num) = Num(value * x.value)
}
So I can call them like this
val a = Num(2)
val b = Num(3)
val c = Num(4)
val m1 = a add b mul c
But how can I execute mul before add? I saw a solution like +| instead of add, but I want include letters in my Operator and +add and *mul not working. Also I want to include a pow function, so this needs an higher precidence than mul
You can use Parenthesis after add.
val m1 = a add (b mul c) = 14
val m1 = a add b mul c = 20
Update
you do not have any restrictions in naming your methods. For example, you can define methods +, -, * and etc. for a class.
case class Num(var value:Int) {
def + (x:Num) = Num(value + x.value)
def *(x:Num) = Num(value * x.value)
}
object Num extends App {
val a = Num(2)
val b = Num(3)
val c = Num(4)
val m1 = a + b * c
println(m1)
}
Output
Num(14)
I want to generate a list of Tuple2 objects. Each tuple (a,b) in the list should meeting the conditions:a and b both are perfect squares,(b/30)<a<b
and a>N and b>N ( N can even be a BigInt)
I am trying to write a scala function to generate the List of Tuples meeting the above requirements?
This is my attempt..it works fine for Ints and Longs..But for BigInt there is sqrt problem I am facing..Here is my approach in coding as below:
scala> def genTups(N:Long) ={
| val x = for(s<- 1L to Math.sqrt(N).toLong) yield s*s;
| val y = x.combinations(2).map{ case Vector(a,b) => (a,b)}.toList
| y.filter(t=> (t._1*30/t._2)>=1)
| }
genTups: (N: Long)List[(Long, Long)]
scala> genTups(30)
res32: List[(Long, Long)] = List((1,4), (1,9), (1,16), (1,25), (4,9), (4,16), (4,25), (9,16), (9,25), (16,25))
Improved this using BigInt square-root algorithm as below:
def genTups(N1:BigInt,N2:BigInt) ={
def sqt(n:BigInt):BigInt = {
var a = BigInt(1)
var b = (n>>5)+BigInt(8)
while((b-a) >= 0) {
var mid:BigInt = (a+b)>>1
if(mid*mid-n> 0) b = mid-1
else a = mid+1
}; a-1 }
val x = for(s<- sqt(N1) to sqt(N2)) yield s*s;
val y = x.combinations(2).map{ case Vector(a,b) => (a,b)}.toList
y.filter(t=> (t._1*30/t._2)>=1)
}
I appreciate any help to improve in my algorithm .
You can avoid sqrt in you algorithm by changing the way you calculate x to this:
val x = (BigInt(1) to N).map(x => x*x).takeWhile(_ <= N)
The final function is then:
def genTups(N: BigInt) = {
val x = (BigInt(1) to N).map(x => x*x).takeWhile(_ <= N)
val y = x.combinations(2).map { case Vector(a, b) if (a < b) => (a, b) }.toList
y.filter(t => (t._1 * 30 / t._2) >= 1)
}
You can also re-write this as a single chain of operations like this:
def genTups(N: BigInt) =
(BigInt(1) to N)
.map(x => x * x)
.takeWhile(_ <= N)
.combinations(2)
.map { case Vector(a, b) if a < b => (a, b) }
.filter(t => (t._1 * 30 / t._2) >= 1)
.toList
In a quest for performance, I came up with this recursive version that appears to be significantly faster
def genTups(N1: BigInt, N2: BigInt) = {
def sqt(n: BigInt): BigInt = {
var a = BigInt(1)
var b = (n >> 5) + BigInt(8)
while ((b - a) >= 0) {
var mid: BigInt = (a + b) >> 1
if (mid * mid - n > 0) {
b = mid - 1
} else {
a = mid + 1
}
}
a - 1
}
#tailrec
def loop(a: BigInt, rem: List[BigInt], res: List[(BigInt, BigInt)]): List[(BigInt, BigInt)] =
rem match {
case Nil => res
case head :: tail =>
val a30 = a * 30
val thisRes = rem.takeWhile(_ <= a30).map(b => (a, b))
loop(head, tail, thisRes.reverse ::: res)
}
val squares = (sqt(N1) to sqt(N2)).map(s => s * s).toList
loop(squares.head, squares.tail, Nil).reverse
}
Each recursion of the loop adds all the matching pairs for a given value of a. The result is built in reverse because adding to the front of a long list is much faster than adding to the tail.
Firstly create a function to check if number if perfect square or not.
def squareRootOfPerfectSquare(a: Int): Option[Int] = {
val sqrt = math.sqrt(a)
if (sqrt % 1 == 0)
Some(sqrt.toInt)
else
None
}
Then, create another func that will calculate this list of tuples according to the conditions mentioned above.
def generateTuples(n1:Int,n2:Int)={
for{
b <- 1 to n2;
a <- 1 to n1 if(b>a && squareRootOfPerfectSquare(b).isDefined && squareRootOfPerfectSquare(a).isDefined)
} yield ( (a,b) )
}
Then on calling the function with parameters generateTuples(5,10)
you will get an output as
res0: scala.collection.immutable.IndexedSeq[(Int, Int)] = Vector((1,4), (1,9), (4,9))
Hope that helps !!!
In this function "f" :
def f(x: => Int) : Int = x * x * x //> f: (x: => Int)Int
var y = 0 //> y : Int = 0
f {
y += 1
println("invoked")
y
} //> invoked
//| invoked
//| invoked
//| res0: Int = 6
"f" is invoked same amount of times as "x" parameter is multiplied.
But why is function invoked multiple times ?
Should "f" not expand to 1 * 1 * 1 not 1 * 2 * 3 ?
Your x is not a function, it is a by-name parameter, and its type is a parameterless method type.
Parameterless method type means the same as def x, something that is evaluated every time you reference it. By reference, we mean x and not x.apply() or x().
The expression you're passing to your function f is evaluated every time x is referenced in f. That expression is the whole thing in braces, a block expression. A block is a sequence of statements followed by the result expression at the end.
Here's another explanation: https://stackoverflow.com/a/13337382/1296806
But let's not call it a function, even if it behaves like one under the covers.
Here is the language used in the spec:
http://www.scala-lang.org/files/archive/spec/2.11/04-basic-declarations-and-definitions.html#by-name-parameters
It's not a value type because you can't write val i: => Int.
It was a big deal when they changed the implementation so you could pass a by-name arg to another method without evaluating it first. There was never a question that you can pass function values around like that. For example:
scala> def k(y: => Int) = 8
k: (y: => Int)Int
scala> def f(x: => Int) = k(x) // this used to evaluate x
f: (x: => Int)Int
scala> f { println("hi") ; 42 }
res8: Int = 8
An exception was made to "preserve the by-name behavior" of the incoming x.
This mattered to people because of eta expansion:
scala> def k(y: => Int)(z: Int) = y + y + z
k: (y: => Int)(z: Int)Int
scala> def f(x: => Int) = k(x)(_) // normally, evaluate what you can now
f: (x: => Int)Int => Int
scala> val g = f { println("hi") ; 42 }
g: Int => Int = <function1>
scala> g(6)
hi
hi
res11: Int = 90
The question is how many greetings do you expect?
More quirks:
scala> def f(x: => Int) = (1 to 5) foreach (_ => x)
f: (x: => Int)Unit
scala> def g(x: () => Int) = (1 to 5) foreach (_ => x())
g: (x: () => Int)Unit
scala> var y = 0
y: Int = 0
scala> y = 0 ; f { y += 1 ; println("hi") ; y }
hi
hi
hi
hi
hi
y: Int = 5
scala> y = 0 ; g { y += 1 ; println("hi") ; () => y }
hi
y: Int = 1
scala> y = 0 ; g { () => y += 1 ; println("hi") ; y }
hi
hi
hi
hi
hi
y: Int = 5
Functions don't cause this problem:
scala> object X { def f(i: Int) = i ; def f(i: => Int) = i+1 }
defined object X
scala> X.f(0)
res12: Int = 0
scala> trait Y { def f(i: Int) = i }
defined trait Y
scala> object X extends Y { def f(i: => Int) = i+1 }
defined object X
scala> X.f(0)
<console>:11: error: ambiguous reference to overloaded definition,
both method f in object X of type (i: => Int)Int
and method f in trait Y of type (i: Int)Int
match argument types (Int)
X.f(0)
^
Compare method types:
http://www.scala-lang.org/files/archive/spec/2.11/03-types.html#method-types
This is not a pedantic distinction; irrespective of the current implementation, it can be confusing to think of a by-name parameter as "really" a function.
Another way of saying what has already been said is that inside f you invoke the function x three times. The first time it increments the y var and returns 1. The second time it again increments y returning 2 and the third time it again increments y and returns 3.
If you want it invoked only once then you may want to do something like this:
def f(x: => Int) : Int = x * x * x
var y = 0
lazy val xx = {
y += 1
println("invoked")
y
}
f {xx}
This will print 'invoked' only once and result in a returned value of 1.
x: T means need a T value.
x: => T means need a T value, but it is call by name.
x: () => T This means need a function given nothing to T
However, this question is not related to the difference between function and method.
The reason is call by name is invoked every time you try to use it.
change to call by value def f(x: Int) : Int, it will only invoke once.
Because you increment y by 1 every time the argument is used inside f
The result which your function f() returns is changing, because there is a global variable that is incremented with every subsequent call to that function.
the x in f(x: => Int) is interpreted as "some function that returns Int". So it has to be called 3 times to evaluate the x*x*x expression. With every call, you increment the global variable and return the result, which is how you arrive at three subsequent natural numbers (because the global variable is initialized to 0). Hence 1*2*3.