Often times I have a desire to create variables scoped to an if statement. Some computations only relate to a particular 'if' statement - to pollute the outer scope with temporary variables smells bad.
What I would like to do:
val data = (whatever)
if (val x = data*2+5.4345/2.45; val y = data/128.4; x*y < 10)
x * y
else
x * 2
println(x) //ERROR!
One alternative is rather messy:
val data = (whatever)
if (data*2+5.4345/2.45*data/128.4 < 10)
data*2+5.4345/2.45*data/128.4
else
data*2+5.4345/2.45 * 2
The obvious alternative I'm trying to avoid:
val data = (whatever)
val x = data*2+5.4345/2.45
val y = data/128.4
if (x*y < 10)
x*y
else
x * 2
println(x) //OK
Is something like this possible in Scala? Is there a decent workaround? If not, what other languages support an idea like this?
Since if in Scala is an expression, i.e. it returns a value, normally you'd be setting some value to the result of your if expression. So your third alternative is just fine: put it in a code block, i.e.
val data = (whatever)
val myValue = {
val x = data*2+5.4345/2.45
val y = data/128.4
if (x*y < 10)
x*y
else
x * 2
}
None of the vals declared within the block are available outside it.
You can use a pattern match:
val data = 123
val (result, x) = (data*2+5.4345/2.45, data/128.4) match {
case (x, y) if x * y < 10 => (x * y, x)
case (x, _) => (x * 2, x)
}
println(x)
result contains the result of x * y or x * 2, depending on which computation ran, and x contains the value of data*2+5.4345/2.45 as desired.
You can create a scope for it...
{
val x = data*2+5.4345/2.45
val y = data/128.4;
if ( x*y < 10)
x * y
else
x * 2
}
Or, to make it clearer,
locally {
val x = data*2+5.4345/2.45
val y = data/128.4;
if ( x*y < 10)
x * y
else
x * 2
}
Related
var locations: List[Location] = List[Location]()
for (x <- 0 to 10; y <- 0 to 10) {
println("x: " + x + " y: " + y)
locations ::: List(Location(x, y))
println(locations)
}
The code above is supposed to concatenate some lists. But the result is an empty list. Why?
Your mistake is on the line locations ::: List(Location(x, y)). This is concatenating the lists, but the doing nothing with the result. If you replace it with locations = locations ::: List(Location(x, y)) you would have your desired result.
However there are more idiomatic ways to solve this problem in Scala. In Scala, writing immutable code is the preferred style (i.e. use val rather than var where possible).
Here's a couple of ways to do it:
Using yield:
val location = for (x <- 0 to 10; y <- 0 to 10) yield Location(x, y)
Using tabulate:
val location = List.tabulate(11, 11) { case (x, y) => Location(x, y) }
Even shorter:
val location = List.tabulate(11, 11)(Location)
Edit: just noticed you had 0 to 10 which is inclusive-inclusive. 0 until 10 is inclusive-exclusive. I've changed the args to tabulate to 11.
I try to implement a version of the Monte Carlo algorithm in Scala but i have a little problem.
In my first loop, i have a mismatch with Unit and Int, but I didn't know how to slove this.
Thank for your help !
import scala.math._
import scala.util.Random
import scala.collection.mutable.ListBuffer
object Main extends App{
def MonteCarlo(list: ListBuffer[Int]): List[Int] = {
for (i <- list) {
var c = 0.00
val X = new Random
val Y = new Random
for (j <- 0 until i) {
val x = X.nextDouble // in [0,1]
val y = Y.nextDouble // in [0,1]
if (x * x + y * y < 1) {
c = c + 1
}
}
c = c * 4
var p = c / i
var error = abs(Pi-p)
print("Approximative value of pi : $p \tError: $error")
}
}
var liste = ListBuffer (200, 2000, 4000)
MonteCarlo(liste)
}
A guy working usually with Python.
for loop does not return anything, so that's why your method returns Unit but expects List[Int] as return type is List[Int].
Second, you have not used scala interpolation correctly. It won't print the value of error. You forgot to use 's' before the string.
Third thing, if want to return list, you first need a list where you will accumulate all values of every iteration.
So i am assuming that you are trying to return error for all iterations. So i have created an errorList, which will store all values of error. If you want to return something else you can modify your code accordingly.
def MonteCarlo(list: ListBuffer[Int]) = {
val errorList = new ListBuffer[Double]()
for (i <- list) {
var c = 0.00
val X = new Random
val Y = new Random
for (j <- 0 until i) {
val x = X.nextDouble // in [0,1]
val y = Y.nextDouble // in [0,1]
if (x * x + y * y < 1) {
c = c + 1
}
}
c = c * 4
var p = c / i
var error = abs(Pi-p)
errorList += error
println(s"Approximative value of pi : $p \tError: $error")
}
errorList
}
scala> MonteCarlo(liste)
Approximative value of pi : 3.26 Error: 0.11840734641020667
Approximative value of pi : 3.12 Error: 0.02159265358979301
Approximative value of pi : 3.142 Error: 4.073464102067881E-4
res9: scala.collection.mutable.ListBuffer[Double] = ListBuffer(0.11840734641020667, 0.02159265358979301, 4.073464102067881E-4)
I have a function below that uses variable X and variable A.
How can I return both of these variables to be able to use these values further down the program.
val a = 1000;
val x = 5;
fun test (x,a) =
if (a<1) then(
x)
else(
test(x+1,a-1)
)
You just return a pair:
fun test (x, a) = if a < 1 then (x, a) else test (x+1, a-1)
You receive it by pattern matching:
val (y, z) = test (10, 11)
very simple question: I want to do something like this:
var arr1: Array[Double] = ...
var arr2: Array[Double] = ...
var arr3: Array[(Double,Double)] = arr1.zip(arr2)
arr3.foreach(x => {if (x._1 > treshold) {x._2 = x._2 * factor}})
I tried a lot differnt syntax versions, but I failed with all of them. How could I solve this? It can not be very difficult ...
Thanks!
Multiple approaches to solve this, consider for instance the use of collect which delivers an immutable collection arr4, as follows,
val arr4 = arr3.collect {
case (x, y) if x > threshold => (x ,y * factor)
case v => v
}
With a for comprehension like this,
for ((x, y) <- arr3)
yield (x, if (x > threshold) y * factor else y)
I think you want to do something like
scala> val arr1 = Array(1.1, 1.2)
arr1: Array[Double] = Array(1.1, 1.2)
scala> val arr2 = Array(1.1, 1.2)
arr2: Array[Double] = Array(1.1, 1.2)
scala> val arr3 = arr1.zip(arr2)
arr3: Array[(Double, Double)] = Array((1.1,1.1), (1.2,1.2))
scala> arr3.filter(_._1> 1.1).map(_._2*2)
res0: Array[Double] = Array(2.4)
I think there are two problems:
You're using foreach, which returns Unit, where you want to use map, which returns an Array[B].
You're trying to update an immutable value, when you want to return a new, updated value. This is the difference between _._2 = _._2 * factor and _._2 * factor.
To filter the values not meeting the threshold:
arr1.zip(arr2).filter(_._1 > threshold).map(_._2 * factor)
To keep all values, but only multiply the ones meeting the threshold:
arr1.zip(arr2).map {
case (x, y) if x > threshold => y * factor
case (_, y) => y
}
You can do it with this,
arr3.map(x => if (x._1 > threshold) (x._1, x._2 * factor) else x)
How about this?
arr3.map { case(x1, x2) => // extract first and second value
if (x1 > treshold) (x1, x2 * factor) // if first value is greater than threshold, 'change' x2
else (x1, x2) // otherwise leave it as it is
}.toMap
Scala is generally functional, which means you do not change values, but create new values, for example you do not write x._2 = …, since tuple is immutable (you can't change it), but create a new tuple.
This will do what you need.
arr3.map(x => if(x._1 > treshold) (x._1, x._2 * factor) else x)
The key here is that you can return tuple from the map lambda expression by putting two variable into (..).
Edit: You want to change every element of an array without creating a new array. Then you need to do the next.
arr3.indices.foreach(x => if(arr3(x)._1 > treshold) (arr3(x)._1, arr3(x)._2 * factor) else x)
Here is some imperative code:
var sum = 0
val spacing = 6
var x = spacing
for(i <- 1 to 10) {
sum += x * x
x += spacing
}
Here are two of my attempts to "functionalize" the above code:
// Attempt 1
(1 to 10).foldLeft((0, 6)) {
case((sum, x), _) => (sum + x * x, x + spacing)
}
// Attempt 2
Stream.iterate ((0, 6)) { case (sum, x) => (sum + x * x, x + spacing) }.take(11).last
I think there might be a cleaner and better functional way to do this. What would be that?
PS: Please note that the above is just an example code intended to illustrate the problem; it is not from the real application code.
Replacing 10 by N, you have spacing * spacing * N * (N + 1) * (2 * N + 1) / 6
This is by noting that you're summing (spacing * i)^2 for the range 1..N. This sum factorizes as spacing^2 * (1^2 + 2^2 + ... + N^2), and the latter sum is well-known to be N * (N + 1) * (2 * N + 1) / 6 (see Square Pyramidal Number)
I actually like idea of lazy sequences in this case. You can split your algorithm in 2 logical steps.
At first you want to work on all natural numbers (ok.. not all, but up to max int), so you define them like this:
val naturals = 0 to Int.MaxValue
Then you need to define knowledge about how numbers, that you want to sum, can be calculated:
val myDoubles = (naturals by 6 tail).view map (x => x * x)
And putting this all together:
val naturals = 0 to Int.MaxValue
val myDoubles = (naturals by 6 tail).view map (x => x * x)
val mySum = myDoubles take 10 sum
I think it's the way mathematician will approach this problem. And because all collections are lazily evaluated - you will not get out of memory.
Edit
If you want to develop idea of mathematical notation further, you can actually define this implicit conversion:
implicit def math[T, R](f: T => R) = new {
def ∀(range: Traversable[T]) = range.view map f
}
and then define myDoubles like this:
val myDoubles = ((x: Int) => x * x) ∀ (naturals by 6 tail)
My personal favourite would have to be:
val x = (6 to 60 by 6) map {x => x*x} sum
Or given spacing as an input variable:
val x = (spacing to 10*spacing by spacing) map {x => x*x} sum
or
val x = (1 to 10) map (spacing*) map {x => x*x} sum
There are two different directions to go. If you want to express yourself, assuming that you can't use the built-in range function (because you actually want something more complicated):
Iterator.iterate(spacing)(x => x+spacing).take(10).map(x => x*x).foldLeft(0)(_ + _)
This is a very general pattern: specify what you start with and how to get the next given the previous; then take the number of items you need; then transform them somehow; then combine them into a single answer. There are shortcuts for almost all of these in simple cases (e.g. the last fold is sum) but this is a way to do it generally.
But I also wonder--what is wrong with the mutable imperative approach for maximal speed? It's really quite clear, and Scala lets you mix the two styles on purpose:
var x = spacing
val last = spacing*10
val sum = 0
while (x <= last) {
sum += x*x
x += spacing
}
(Note that the for is slower than while since the Scala compiler transforms for loops to a construct of maximum generality, not maximum speed.)
Here's a straightforward translation of the loop you wrote to a tail-recursive function, in an SML-like syntax.
val spacing = 6
fun loop (sum: int, x: int, i: int): int =
if i > 0 then loop (sum+x*x, x+spacing, i-1)
else sum
val sum = loop (0, spacing, 10)
Is this what you were looking for? (What do you mean by a "cleaner" and "better" way?)
What about this?
def toSquare(i: Int) = i * i
val spacing = 6
val spaceMultiples = (1 to 10) map (spacing *)
val squares = spaceMultiples map toSquare
println(squares.sum)
You have to split your code in small parts. This can improve readability a lot.
Here is a one-liner:
(0 to 10).reduceLeft((u,v)=>u + spacing*spacing*v*v)
Note that you need to start with 0 in order to get the correct result (else the first value 6 would be added only, but not squared).
Another option is to generate the squares first:
(1 to 2*10 by 2).scanLeft(0)(_+_).sum*spacing*spacing