How to derive this "converter" in Scala in compile time? - scala

We have an API that receives client queries in JSON format and executes them against as: Seq[A] of case class A(x: Int, y: Int). For example, query {x : 0} works as as.filter(_.x == 0), { "and" : {x = 0, y = 1} } works as as.filter(a => a.x == 0 && a.y == 1), etc.
The problem is that case class A is changing and it breaks the queries.
For example, now I need to execute query { "and" : {x = 0, y = 1} } against Seq[A1] of case class A1(x1: Int, y: String). So I have to convert that query to { "and" : {x1 = 0, y = "1"} }.
I don't want to write such query converters manually every time I change case class A to case class A1. I want to write some "mapping" between A and A1 (not sure how to do it) and then "derive" the query converter in compile time. How would you suggest do that ? Should I use macros or a library for type-level programming ?

Related

I want to create new type of simple operator class kotlin

What I want to do is save operator with number in list
and evaluate it after pushing it to list
need 2 types of operator
Add and Multiply
Plus(x) (x is number)
Multiply(x) (x is number)
lis = [ Plus(3), Multiply(3), Plus(4), Multiply(2) ]
When Plus Start
lis.fold(0) { total, operator -> blar..blar }
-> (((0 + 3) * 3) + 4) * 2
When Multiply Start
lis.fold(1) { total, operator -> blar..blar }
how to declare abstract operator
I think Plus and Multiply Operator has one infix function
something like below
infix fun <T>evaluate(value: T): T = this+value
or
infix fun <T>evaluate(value: T): T = this*value
In order to allow the distinction whether to use a 0 or a 1 depending on the first element, you could define your operation with an identity element:
open class Operation(
val op: (Double) -> Double,
val identity: Double,
)
class Plus(operand: Double) : Operation({ it + operand }, 0.0)
class Multiply(operand: Double) : Operation({ it * operand }, 1.0)
Then, the application of all elements in a list can be defined as:
fun List<Operation>.apply(): Double = if (isEmpty()) 1.0 else
fold(first().identity) { total, operation -> operation.op(total) }
The application of all elements as in your example can then be invoked using apply:
val lis = listOf(Plus(3.0), Multiply(3.0), Plus(4.0), Multiply(2.0))
lis.apply()
If you do not like that an operator is defined and applied via its property op (e.g. the application of Plus(3.0) to a value 5.0 is done by Plus(3.0).op(5.0)), we can alternatively define Operation and the concrete subclasses in the following way:
abstract class Operation(val identity: Double) {
abstract operator fun invoke(x: Double): Double
}
class Plus(private val operand: Double) : Operation(0.0) {
override fun invoke(x: Double): Double = x + operand
}
class Multiply(private val operand: Double) : Operation(1.0) {
override fun invoke(x: Double): Double = x * operand
}
Then, the apply function would also change:
fun List<Operation>.apply(): Double = if (isEmpty()) 1.0 else
fold(first().identity) { total, operation -> operation(total) }
which might look a bit more intuitive.
The invoke operator allows us to apply the Operation like a function to a value, e.g. you can also write Plus(3.0)(5.0) to apply _ + 3.0 to the value 5.0.

Is it possible to decompose Scala match statements using partial functions?

I have a match statement like this:
i match {
case x if x == 0 ⇒
romanNumeral
case x if x >= 1000 ⇒
this.roman(i - 1000, s"${romanNumeral}M")
case x if x >= 900 ⇒
this.roman(i - 900, s"${romanNumeral}CM")
// etc.
Depending on how this is formatted, it could be about 30 lines of redundant code.
So I'm wondering whether it is possible to make this code more dry.
So I created a partial function:
private def toRoman(upperGuard: Int, token: String, romanNumeral: String): String = {
case value: Int if value >= upperGuard ⇒
this.roman(upperGuard - 1, s"$romanNumeral$token")
}
which I then tried to include into the above matcher like:
i match {
case x if x == 0 ⇒
romanNumeral
toRoman(1000, "M", romanNumeral)
toRoman(900, "CM", romanNumeral)
// etc.
... But this won't work because Scala's compiler doesn't recognize these functions are the case statements it is looking for.
Is there any way to make this work?
You could create your cases as partial functions:
private def toRoman(upperGuard: Int, token: String, romanNumeral: String): PartialFunction[Int, String] = {
case value if value >= upperGuard =>
this.roman(upperGuard - 1, s"$romanNumeral$token")
}
val firstCase: PartialFunction[Int, String] = {
case x if x == 0 => romanNumeral
}
And then compose it like this:
val toRomanPartialFunction = firstCase
.orElse(toRoman(1000, "M", romanNumeral))
.orElse(toRoman(900, "CM", romanNumeral))
After that you could use it like regular function:
toRomanPartialFunction.apply(443)
toRomanPartialFunction(443)
So you basically want to make the code less repetitive, first things first:
Scala´s compiler doesn't recognise these functions as case statements because they are functions;
Second thing, that function is not actually a partial function;
Finally, those matches are ok to be like that, if you find a way of reducing lines of code by grouping in some kind of function like you did great but I don't find them redundant, unless you can redo it with a function like I said.
Ps: That function don't do the some thing as the case
You don't have a pattern matching expression, you are just writing an if else statement. If you want to return a function, you would write: x => this.something(x,"foo-bar") You have no need for case x. If you are only having one expression on the LHS case x you have no need for pattern matching, and can just use if else. Pattern matching without a class hierarchy doesn't make sense.

Continuation-passing style in Scala

I have superficially read a couple of blog articles/Wikipedia about continuation-passing style. My high-level goal is to find a systematic technique to make any recursive function (or, if there are restrictions, being aware of them) tail-recursive. However, I have trouble articulating my thoughts and I'm not sure if what my attempts of it make any sense.
For the purpose of the example, I'll propose a simple problem. The goal is, given a sorted list of unique characters, to output all possible words made out of these characters in alphabetical order. For example, sol("op".toList, 3) should return ooo,oop,opo,opp,poo,pop,ppo,ppp.
My recursive solution is the following:
def sol(chars: List[Char], n: Int) = {
def recSol(n: Int): List[List[Char]] = (chars, n) match {
case (_ , 0) => List(Nil)
case (Nil, _) => Nil
case (_ , _) =>
val tail = recSol(n - 1)
chars.map(ch => tail.map(ch :: _)).fold(Nil)(_ ::: _)
}
recSol(n).map(_.mkString).mkString(",")
}
I did try to rewrite this by adding a function as a parameter but I did not manage to make something I was convinced to be tail-recursive. I prefer not including my attempt(s) in the question as I'm ashamed of their naiveness, so please excuse me for this.
Therefore the question is basically: how would the function above be written in CPS ?
Try that:
import scala.annotation.tailrec
def sol(chars: List[Char], n: Int) = {
#tailrec
def recSol(n: Int)(cont: (List[List[Char]]) => List[List[Char]]): List[List[Char]] = (chars, n) match {
case (_ , 0) => cont(List(Nil))
case (Nil, _) => cont(Nil)
case (_ , _) =>
recSol(n-1){ tail =>
cont(chars.map(ch => tail.map(ch :: _)).fold(Nil)(_ ::: _))
}
}
recSol(n)(identity).map(_.mkString).mkString(",")
}
The first order of business in performing the CPS transform is deciding on a representation for continuations. We can think of continuations as a suspended computation with a "hole". When the hole is filled in with a value, the remainder of the computation can be computed. So functions are a natural choice for representing continuations, at least for toy examples:
type Cont[Hole,Result] = Hole => Result
Here Hole represents the type of the hole that needs to be filled in, and Result represents the type of value the computation ultimately computes.
Now that we have a way to represent continuations, we can worry about the CPS transform itself. Basically, this involves the following steps:
The transformation is applied recursively to an expression, stopping at "trivial" expressions / function calls. In this context, "trivial" includes functions defined by Scala (since they are not CPS-transformed, and thus do not have a continuation parameter).
We need to add a parameter of type Cont[Return,Result] to each function, where Return is the return type of the untransformed function and Result is the type of the ultimate result of the computation as a whole. This new parameter represents the current continuation. The return type for the transformed function is also changed to Result.
Every function call needs to be transformed to accommodate the new continuation parameter. Everything after the call needs to be put into a continuation function, which is then added to the parameter list.
For example, a function:
def f(x : Int) : Int = x + 1
becomes:
def fCps[Result](x : Int)(k : Cont[Int,Result]) : Result = k(x + 1)
and
def g(x : Int) : Int = 2 * f(x)
becomes:
def gCps[Result](x : Int)(k : Cont[Int,Result]) : Result = {
fCps(x)(y => k(2 * y))
}
Now gCps(5) returns (via currying) a function that represents a partial computation. We can extract the value from this partial computation and use it by supplying a continuation function. For example, we can use the identity function to extract the value unchanged:
gCps(5)(x => x)
// 12
Or, we can print it by using println instead:
gCps(5)(println)
// prints 12
Applying this to your code, we obtain:
def solCps[Result](chars : List[Char], n : Int)(k : Cont[String, Result]) : Result = {
#scala.annotation.tailrec
def recSol[Result](n : Int)(k : Cont[List[List[Char]], Result]) : Result = (chars, n) match {
case (_ , 0) => k(List(Nil))
case (Nil, _) => k(Nil)
case (_ , _) =>
recSol(n - 1)(tail =>
k(chars.map(ch => tail.map(ch :: _)).fold(Nil)(_ ::: _)))
}
recSol(n)(result =>
k(result.map(_.mkString).mkString(",")))
}
As you can see, although recSol is now tail-recursive, it comes with the cost of building a more complex continuation at each iteration. So all we've really done is trade space on the JVM's control stack for space on the heap -- the CPS transform does not magically reduce the space complexity of an algorithm.
Also, recSol is only tail-recursive because the recursive call to recSol happens to be the first (non-trivial) expression recSol performs. In general, though, recursive calls would be take place inside a continuation. In the case where there is one recursive call, we can work around that by transforming only calls to the recursive function to CPS. Even so, in general, we would still just be trading stack space for heap space.

Generate all IP addresses given a string in scala

I was trying my hand at writing an IP generator given a string of numbers. The generator would take as an input a string of number such as "17234" and will return all possible list of ips as follows:
1.7.2.34
1.7.23.4
1.72.3.4
17.2.3.4
I attempted to write a snippet to do the generation as follows:
def genip(ip:String):Unit = {
def legal(ip:String):Boolean = (ip.size == 1) || (ip.size == 2) || (ip.size == 3)
def genips(ip:String,portion:Int,accum:String):Unit = portion match {
case 1 if legal(ip) => println(accum+ip)
case _ if portion > 1 => {
genips(ip.drop(1),portion-1,if(accum.size == 0) ip.take(1)+"." else accum+ip.take(1)+".")
genips(ip.drop(2),portion-1,if(accum.size == 0) ip.take(2)+"." else accum+ip.take(2)+".")
genips(ip.drop(3),portion-1,if(accum.size == 0) ip.take(3)+"." else accum+ip.take(3)+".")
}
case _ => return
}
genips(ip,4,"")
}
The idea is to partition the string into four octets and then further partition the octet into strings of size "1","2" and "3" and then recursively descend into the remaining string.
I am not sure if I am on the right track but it would be great if somebody could suggest a more functional way of accomplishing the same.
Thanks
Here is an alternative version of the attached code:
def generateIPs(digits : String) : Seq[String] = generateIPs(digits, 4)
private def generateIPs(digits : String, partsLeft : Int) : Seq[String] = {
if ( digits.size < partsLeft || digits.size > partsLeft * 3) {
Nil
} else if(partsLeft == 1) {
Seq(digits)
} else {
(1 to 3).map(n => generateIPs(digits.drop(n), partsLeft - 1)
.map(digits.take(n) + "." + _)
).flatten
}
}
println("Results:\n" + generateIPs("17234").mkString("\n"))
Major changes:
Methods now return the collection of strings (rather than Unit), so they are proper functions (rather than working of side effects) and can be easily tested;
Avoiding repeating the same code 3 times depending on the size of the bunch of numbers we take;
Not passing accumulated interim result as a method parameter - in this case it doesn't have sense since you'll have at most 4 recursive calls and it's easier to read without it, though as you're loosing the tail recursion in many case it might be reasonable to leave it.
Note: The last map statement is a good candidate to be replaced by for comprehension, which many developers find easier to read and reason about, though I will leave it as an exercise :)
You code is the right idea; I'm not sure making it functional really helps anything, but I'll show both functional and side-effecting ways to do what you want. First, we'd like a good routine to chunk off some of the numbers, making sure an okay number are left for the rest of the chunking, and making sure they're in range for IPs:
def validSize(i: Int, len: Int, more: Int) = i + more <= len && i + 3*more >= len
def chunk(s: String, more: Int) = {
val parts = for (i <- 1 to 3 if validSize(i, s.length, more)) yield s.splitAt(i)
parts.filter(_._1.toInt < 256)
}
Now we need to use chunk recursively four times to generate the possibilities. Here's a solution that is mutable internally and iterative:
def genIPs(digits: String) = {
var parts = List(("", digits))
for (i <- 1 to 4) {
parts = parts.flatMap{ case (pre, post) =>
chunk(post, 4-i).map{ case (x,y) => (pre+x+".", y) }
}
}
parts.map(_._1.dropRight(1))
}
Here's one that recurses using Iterator:
def genIPs(digits: String) = Iterator.iterate(List((3,"",digits))){ _.flatMap{
case(j, pre, post) => chunk(post, j).map{ case(x,y) => (j-1, pre+x+".", y) }
}}.dropWhile(_.head._1 >= 0).next.map(_._2.dropRight(1))
The logic is the same either way. Here it is working:
scala> genIPs("1238516")
res2: List[String] = List(1.23.85.16, 1.238.5.16, 1.238.51.6,
12.3.85.16, 12.38.5.16, 12.38.51.6,
123.8.5.16, 123.8.51.6, 123.85.1.6)

Polish notation evaluate function

I am new to Scala and I am having hard-time with defining, or more likely translating my code from Ruby to evaluate calculations described as Polish Notations,
f.e. (+ 3 2) or (- 4 (+ 3 2))
I successfully parse the string to form of ArrayBuffer(+, 3, 2) or ArrayBuffer(-, 4, ArrayBuffer(+, 3 2)).
The problem actually starts when I try to define a recursive eval function ,which simply takes ArrayBuffer as argument and "return" an Int(result of evaluated application).
IN THE BASE CASE:
I want to simply check if 2nd element is an instanceOf[Int] and 3rd element is instanceOf[Int] then evaluate them together (depending on sign operator - 1st element) and return Int.
However If any of the elements is another ArrayBuffer, I simply want to reassign that element to returned value of recursively called eval function. like:
Storage(2) = eval(Storage(2)). (** thats why i am using mutable ArrayBuffer **)
The error ,which I get is:
scala.collection.mutable.ArrayBuffer cannot be cast to java.lang.Integer
I am of course not looking for any copy-and-paste answers but for some advices and observations.
Constructive Criticism fully welcomed.
****** This is the testing code I am using only for the addition ******
def eval(Input: ArrayBuffer[Any]):Int = {
if(ArrayBuffer(2).isInstaceOf[ArrayBuffer[Any]]) {
ArrayBuffer(2) = eval(ArrayBuffer(2))
}
if(ArrayBuffer(3).isInstaceOf[ArrayBuffer[Any]]) {
ArrayBuffer(3) = eval(ArrayBuffer(3))
}
if(ArrayBuffer(2).isInstaceOf[Int] && ArrayBuffer(3).isInstanceOf[Int]) {
ArrayBuffer(2).asInstanceOf[Int] + ArrayBuffer(3).asInstanceOf[Int]
}
}
A few problems with your code:
ArrayBuffer(2) means "construct an ArrayBuffer with one element: 2". Nowhere in your code are you referencing your parameter Input. You would need to replace instances of ArrayBuffer(2) with Input(2) for this to work.
ArrayBuffer (and all collections in Scala) are 0-indexed, so if you want to access the second thing in the collection, you would do input(1).
If you leave the the final if there, then the compiler will complain since your function won't always return an Int; if the input contained something unexpected, then that last if would evaluate to false, and you have no else to fall to.
Here's a direct rewrite of your code: fixing the issues:
def eval(input: ArrayBuffer[Any]):Int = {
if(input(1).isInstanceOf[ArrayBuffer[Any]])
input(1) = eval(input(1).asInstanceOf[ArrayBuffer[Any]])
if(input(2).isInstanceOf[ArrayBuffer[Any]])
input(2) = eval(input(2).asInstanceOf[ArrayBuffer[Any]])
input(1).asInstanceOf[Int] + input(2).asInstanceOf[Int]
}
(note also that variable names, like input, should be lowercased.)
That said, the procedure of replacing entries in your input with their evaluations is probably not the best route because it destroys the input in the process of evaluating. You should instead write a function that takes the ArrayBuffer and simply recurses through it without modifying the original.
You'll want you eval function to check for specific cases. Here's a simple implementation as a demonstration:
def eval(e: Seq[Any]): Int =
e match {
case Seq("+", a: Int, b: Int) => a + b
case Seq("+", a: Int, b: Seq[Any]) => a + eval(b)
case Seq("+", a: Seq[Any], b: Int) => eval(a) + b
case Seq("+", a: Seq[Any], b: Seq[Any]) => eval(a) + eval(b)
}
So you can see that for the simple case of (+ arg1 arg2), there are 4 cases. In each case, if the argument is an Int, we use it directly in the addition. If the argument itself is a sequence (like ArrayBuffer), then we recursively evaluate before adding. Notice also that Scala's case syntax lets to do pattern matches with types, so you can skip the isInstanceOf and asInstanceOf stuff.
Now there definitely style improvements you'd want to make down the line (like using Either instead of Any and not hard coding the "+"), but this should get you on the right track.
And here's how you would use it:
eval(Seq("+", 3, 2))
res0: Int = 5
scala> eval(Seq("+", 4, Seq("+", 3, 2)))
res1: Int = 9
Now, if you want to really take advantage of Scala features, you could use an Eval extractor:
object Eval {
def unapply(e: Any): Option[Int] = {
e match {
case i: Int => Some(i)
case Seq("+", Eval(a), Eval(b)) => Some(a + b)
}
}
}
And you'd use it like this:
scala> val Eval(result) = 2
result: Int = 2
scala> val Eval(result) = ArrayBuffer("+", 2, 3)
result: Int = 5
scala> val Eval(result) = ArrayBuffer("+", 2, ArrayBuffer("+", 2, 3))
result: Int = 7
Or you could wrap it in an eval function:
def eval(e: Any): Int = {
val Eval(result) = e
result
}
Here is my take on right to left stack-based evaluation:
def eval(expr: String): Either[Throwable, Int] = {
import java.lang.NumberFormatException
import scala.util.control.Exception._
def int(s: String) = catching(classOf[NumberFormatException]).opt(s.toInt)
val symbols = expr.replaceAll("""[^\d\+\-\*/ ]""", "").split(" ").toSeq
allCatch.either {
val results = symbols.foldRight(List.empty[Int]) {
(symbol, operands) => int(symbol) match {
case Some(op) => op :: operands
case None => val x :: y :: ops = operands
val result = symbol match {
case "+" => x + y
case "-" => x - y
case "*" => x * y
case "/" => x / y
}
result :: ops
}
}
results.head
}
}