Fetch elements from Option[Any] of List - list

scala> val a = jsonMap.get("L2_ID")
a: Option[Any] = Some(List(24493, 22774, 23609, 20517, 22829, 23646, 22779, 23578, 22765, 23657))
I want to fetch the first element of list i.e 24493. So, tried below code:
scala> var b = a.map(_.toString)
b: Option[String] = Some(List(24493, 22774, 23609, 20517, 22829, 23646, 22779, 23578, 22765, 23657))
scala>
scala> var c = b.map(_.split(",")).toList.flatten
c: List[String] = List(List(24493, " 22774", " 23609", " 20517", " 22829", " 23646", " 22779", " 23578", " 22765", " 23657)")
scala> c(0)
res34: String = List(24493
This is not returning as expected.

I suggest you use pattern matching.
To be defensive, i also added a Try to protect against the case of your json not being a List of numbers.
Code below returns an Option[Int] and you can call .getOrElse(0) on it - or some other default value, if you like.
import scala.util.Try
val first = a match {
case Some(h :: _) => Try(h.toString.toInt).toOption
case _ => None
}

So, you have an Option, and List inside of it. Then scala> var b = a.map(_.toString) converts the contents of the Option (a List) into a String. That's not what you want.
Look at the types of the results of your transformations, they are there to provide pretty good hints for you. b: Option[String], for example, tells you that you have lost the list ...
a.map(_.map(_.toString))
has the type Option[List[String]] on the other hand: you have converted every element of the list to a string.
If you are just looking for the first element, there is no need to convert all of them though. Something like this will do:
a
.flatMap(_.headOption) // Option[Int], containing first element or None if list was empty or id a was None
.map(_.toString) // convert Int inside of Option (if any) to String
.getOrElse("") // get the contents of the Option, or empty string if it was None

If you are certain that it's a Some, and that the list is non-empty, then you can unwrap the option and get the List[Int] using .get. Then you can access the first element of the list using .head:
val x: Option[List[Int]] = ???
x.get.head
If you are not in the REPL, and if you aren't sure whether it's a Some or None, and whether the List has any elements, then use
x.flatMap(_.headOption).getOrElse(yourDefaultValueEg0)
"Stringly-typed" programming is certainly not necessary in a language with such a powerful type system, so converting everything to string and splitting by commas was a seriously flawed approach.

Related

How can I get the elements of list using ListBuffer

I am facing a very weird problem while getting elements of a list
Below is the piece of code where I am passing arguments as "bc" and "mn"
val list1 = List("abc", "def", "mnp")
val list2 = List(args(0), args(1))
val header1=list1.filter(x => list2.exists(y => x.contains(y)))
println(header1)
Output-List("abc","mnp")
I am trying to do it in a different way (by passing the same arguments)but getting an empty List
val list1 = List("abc", "def", "mnp")
//val list2 = List(args(0), args(1))
val ipList1= new ListBuffer[Any]
for(i <- 0 to 1){
ipList1 +=args(i)
}
val list2=ipList1.toList
println(list2)
val header1=list1.filter(x => list2.exists(y => x.contains(y)))
println(header1)
Output-List(bc, mn)
List()-->This is the empty List I am getting
Can Someone please tell where I am doing it wrong and How to make it right?
The problem is that x.contains(y) does not mean what you think it means. String has a contains method that checks whether another String is a substring of this String. But in your code y doesn't have type String, but type Any. So the contains method of String isn't called. It's the contains method of WrappedString which treats the String x as though it's a Seq[Char]. That method doesn't check whether any substring is equal to y but whether any character is equal to y.
The solution, obviously, is to use a ListBuffer[String].
The problem is that you are using a ListBuffer[Any] thus the elements lost their type information from String to Any and apparently that changes the semantics of the code.
You may either do this:
val ipList1 = new ListBuffer[String]
for (i <- 0 to 1) {
ipList1 += args(i).toString
}
val list2 = ipList1.toList
Or even better just:
val list2 = args.slice(0, 2).toList

Immutable Lists in Scala

I am just trying to figure out how immutable things like a List are working, and how I can add things to it?
I am very sorry for asking such dumb questions, but why is here my list always empty when printing it out?
var end = false
val list = List()
while (!end) {
val input = scala.io.StdIn.readLine("input:")
if (input == "stop" ) end = true
else input :: list
}
println(list)
}
Sorry for my inconvenience and this rather stupid question!
I am just trying to figure out how immutable things like a List are working, and how I can add things to it?
You can't. That's what immutable means, after all. If Latin is not your cup of tea, the English translation of immutable is unchangeable. It should be clear now, why you can't change something that is unchangeable.
I am very sorry for asking such dumb questions, but why is here my list always empty when printing it out?
You create an empty list, and you never change it (because it cannot be changed anyway). So, of course it is empty.
What can you can do, however, is create a new list which is almost exactly like the old list, except with a new item prepended to the front. That's what you are doing here:
input :: list
However, you don't assign this new list anywhere, you don't return it, you completely ignore it.
If you want to actually use your list in any way, you need to remember it somehow. The most obvious solution would be to assign it to a variable:
var end = false
var list: List[String] = List() // note: `var` instead of `val`
while (!end) {
val input = scala.io.StdIn.readLine("input:")
if (input == "stop" ) end = true
else list = input :: list // note: assign to `list`
}
println(list)
However, that's not very idiomatic. After all, we have now taken an immutable list and assigned it to a mutable variable … IOW, we have just moved the mutability around.
Instead, we could use a recursive solution:
def buildListFromInput(list: List[String] = List()): List[String] = {
val input = scala.io.StdIn.readLine("input:")
if (input == "stop") list else buildListFromInput(input :: list)
}
println(buildListFromInput())
This solution is not only recursive, the recursive call is also in tail position (IOW, the method is tail-recursive), which means that it will be just as efficient as a while loop (in fact, it will be compiled into a while loop, or more precisely, into a GOTO). The Scala Language Specification guarantees that all implementations of Scala must eliminate direct tail-recursion.
The reason
println(list)
is only printing out an empty list is because the bit
input :: list
isn't actually mutating the list itself. It is simply, in this case, very temporarily, creating a list containing the input at the front.
Try
println(input :: list)
or
val newList = input :: list
println(newList)
and you'll see what I mean.
Try rewriting the code in more functional way. Every operation on Immutable data structures return new instance with change. So :: operator creates new List with input on front. You might want to try rewrite this code as tail recursive function as follows.
#tailrec
def scanInput(continue: Boolean,acc: List[String]): List[String] = {
val input = scala.io.StdIn.readLine("input:")
if(!continue) acc
else scanInput(input != "stop", input :: acc)
}
Above code has no mutating state and it suits more Scala functional style.
In scala List is immutable.
Then how can I add items to the list?
When you add an item to list a new List instance is crated with a item as its head and its tail now contains the previous list.
If you have list of "1,2,3" called intList internally it is represented as
List(3, List(2, List(1, Nil) ) )
If you add an element 4 to this intList
List(4, intList )
Lets call this newList
Note intList still contains List(3, List(2, List(1, Nil) ) ).
If you want the intList to refer the newList You will have to do
intList = intList.add(4)
How can I fix my code
Change list from val to var. Then you can assign resulting List to list variable
list = input :: list
Source: Online course on Scala called Functional Programming Principles in Scala
Thank you for all your help, I appreciate your help so much from all of you!
I should have taken a closer look at recursion since it seems to be really important like in Scala!
But trough your help I am getting an better idea of how it works!
I just tried to figure out how your solutions are working and created my own:
val list = List()
def scanInput(acc: List[String]): List[String] = {
val input = scala.io.StdIn.readLine("input:")
input match {
case "stop" => acc
case _ => scanInput(input :: acc)
}
}
println(scanInput(list))

Pattern matching list prefixes in Scala

I have questions regarding pattern matching of list prefixes (i.e. the first couple of elements of a list).
This compiles, but it does not work as expected:
val l = List(1,2,3)
val test = { m: List[Int] =>
m match {
case l :: tail => println("tail: "+tail.mkString(","))
case _ => println("no match")
}
}
test(List(1,2,3,4,5))
Output is tail: 2,3,4,5. I'd expect it to say either tail: 4,5, or to fail to match, or to fail at compile time. What makes this work as it does?
My second question is: How can I match a list prefix using a list? I know that this works as I expect:
case 1 :: 2 :: 3 :: tail => println("tail: "+tail.mkString(","))
I have, however, my prefixes as lists, and cannot hard-code them. Is pattern matching even the right thing here?
I know I could do something like
if (m startsWith l) {
val tail = m drop l.size
}
But a long list of these seems rather inelegant, especially in Scala.
Regarding the output of your first code snippet, the l inside of the match is actually a new value that shadows the outer scoped l and captures 1 during execution.
The problem you are encountering is that :: is the unapply for List to break it into exactly a single head value and the tail, deconstructing the linked list.
While there is a ::: operation to go along with ::: to concatenate two lists, it does not have a corresponding unapply which would let you use it in a pattern match in the way you desire.
I don't think this is possible. Closest syntax i could propose based on this workaround:
import collection.SeqLike
implicit class PrefixMatcher[T](prefix: Seq[T]) {
object then {
def unapply[S <: SeqLike[T,S]](seq: S): Option[S] =
if (seq startsWith prefix) Some(seq drop prefix.length) else None
}
}
Then you could use it as
val test: List[Int] => Unit = {
case l.then(tail) => println("tail: " + tail.mkString(","))
case _ => println("no match")
}
Addresing the first part of your question: As #Arne mentions, case l is not being matched against your list, but captures a new list. For the former you'd need to enclose it on backticks, but even then I don't see how you can achieve what you want, the closes I can think of is:
case `l` :+ x => println(s"tail is $s") //but this just works when the tail is just one element long.
For the final part of your question, maybe pattern matching is not the right thing to do here, how about:
val prefixes = List (List (1,2,3), List(4,5,6)...)
def findTail(li:List[Int]) = prefixes.collectFirst{ case p if li.startsWith(p) => li.drop(p.size) } //Option[List[Int]]
This will find the tail for the first prefix that matches the testList or None if there was no match.You can easily generalize it to work with more than just Ints.

Scala: can not print list elements

I did "mkString" but still can not print list of strings. With input line:
9002194187,2644,54,100,3,4,2,5
I get the following output:
Line: 9002194187,2644,54,100,3,4,2,5
StrArr: 9002194187,2644,54,100,3,4,2,5
Lst: [Ljava.lang.String;#223d2e6c
Lst again: List([Ljava.lang.String;#223d2e6c)
Lst1: [Ljava.lang.String;#223d2e6c
Result: foo
From the code below:
def mkRecord(line: String) : String = {
val klass = "foo"
val strArr = line.split(",") // convert string to array of strings
println("Line: "+line)
println("StrArr: "+strArr.mkString(","))
val lst = List(strArr)
println("Lst: "+lst.mkString(" - "))
println("Lst again: "+lst)
val lst1 = lst.tail ++ List(klass) // attribute list except for the first one, plus new klass attribute
println("Lst1: "+lst.mkString(" , "))
val result = lst1.mkString(",") // attribute string
println("Result: "+ result)
return result
}
Please, help. I am at complete loss (
The constructor for List (actually, the apply method on the List companion object) takes parameters in the form of scala's "varargs" equivalent:
def apply[A](xs: A*): List[A] // some irrelevant details have been elided
In java, this would be written something like:
public static List<A> apply(A... args)
In scala this can be called using any Seq (or subclass), but using a special notation. The line you used:
val lst = List(strArr)
creates a List[Array[String]], with a single entry - the array strArr. To tell the compiler to turn the array into a varargs when passing it to the List apply method, add : _* on the end (the space is optional):
val lst = List(strArr: _*)
This change in your code will result in:
scala> mkRecord(chkStr)
Line: 9002194187,2644,54,100,3,4,2,5
StrArr: 9002194187,2644,54,100,3,4,2,5
Lst: 9002194187 - 2644 - 54 - 100 - 3 - 4 - 2 - 5
Lst again: List(9002194187, 2644, 54, 100, 3, 4, 2, 5)
Lst1: 9002194187 , 2644 , 54 , 100 , 3 , 4 , 2 , 5
Result: 2644,54,100,3,4,2,5,foo
res1: String = 2644,54,100,3,4,2,5,foo
You can turn any array into a list with its toList operator, avoiding this problem (the nature of which Shadowlands has explained). You can do string -> array -> list in one line:
line.split(',').toList
Using a collection's toList method is often going to be faster than extracting all the elements into a sequence and then converting that sequence into a list, not least because you'll be using a method optimised for the source collection. However, that's an optimisation which you can worry about after the priorities of success and clarity.

Scala: alternative List syntax (with square brackets, if possible)

is there an alternative 'List' syntax in Scala?
Is it possible to define one aditional class/type/operator* called '[' and ']'?
I know 'square brackets' are used to indicate Type, but they are perfect to the repetitive task of declaring lists.
A ';' or '?' would be good also, as a last resource.
Thanks
obs.:
after much search the only alternative I found was to use 'cons':
val list = 1 :: 2 :: 3 :: Nil
but it doesn't reduce any key typing at all.
I am still learning those things in Scala
EDIT:
Just to clarify: Performance is not a priority in my case. And yes, shift is not welcome. :P
Motivation behind the scenes: I like Haskell style, but cannot use it directly with Java.
EDIT 2:
Final solution based on both Rex Kerr solutions
implementing object Types:
package a
object Types {
type \[z] = List[z]
implicit def make_lists[A](z: A) = new ListMaker(z)
class ListMaker[A](a0: A) {
private[this] val buffer = List.newBuilder[A]
buffer += a0
def \(z: A) = {
buffer += z;
this
}
def \\ = buffer.result
}
}
using object Types:
package a
import a.Types._
object Whatever {
def listInListOut (l: \[Int]) = {
1\2\\
}
}
[ and ] are reserved symbols in Scala which are used for type annotations. You can't use them for lists. ; is reserved for end of line. You could use ? in many cases, but it would be awkward.
I recommend that you learn to use the :: notation (and get used to typing the : symbol fast twice in succession) because it really makes the list operations visually clear, plus it is a great syntactic reminder that lists are weird because you put things on the head of the list.
However, if you cannot tolerate this, your best option is probably to define a one-letter list symbol. For example,
List(1,2,3,4)
is a list of the numbers from 1 to 4. What if you could just type L instead of List? It turns out that you can, since this is not a fancy constructor or static method, but a singleton companion object to the class List. So you just
val L = List
L(1,2,3,4)
and you are just one character worse off than your suggestion of brackets.
Define
def l[A](a:A*) = List(a:_*)
Then you can do
l(1,2,3)
which is only one character more than [1,2,3]
I can't help pointing out another way to go here for lists where all the elements are the same type, if you really hate the shift key and don't care if other people can understand your code:
class ListMaker[A](a0: A) {
private[this] val buffer = List.newBuilder[A]
buffer += a0
def \(a: A) = { buffer += a; this }
def \\ = buffer.result
}
implicit def make_lists[A](a: A) = new ListMaker(a)
Now you can list to your heart's content, without ever touching the shift key!
scala> val a = 1\2\3\4\5\\
a: List[Int] = List(1, 2, 3, 4, 5)
scala> val b = 'a'\'b'\\
b: List[Char] = List(a, b)
scala> val c = false\true\false\false\false\false\true\\
c: List[Boolean] = List(false, true, false, false, false, false, true)
This uses exactly as many characters as brackets would. (It doesn't nest well, however.)
Welcome to Scala version 2.10.0.r24777-b20110419020105 (Java HotSpot(TM) Client VM, Java 1.6.0
Type in expressions to have them evaluated.
Type :help for more information.
scala> class LM[A](x: A) {
| def \(y: A) = List(x,y)
| }
defined class LM
scala> implicit def a2l[A](x: A): LM[A] = new LM(x)
a2l: [A](x: A)LM[A]
scala> class LX[A](xs: List[A]) {
| def \(y: A) = xs:::List(y)
| }
defined class LX
scala> implicit def l2lx[A](xs: List[A]): LX[A] = new LX(xs)
l2lx: [A](xs: List[A])LX[A]
scala> 1\2
res0: List[Int] = List(1, 2)
scala> 1\2\3
res1: List[Int] = List(1, 2, 3)
scala>
Not exactly an alternative syntax, but it is by far the most portable solution:
In Intellij IDEA it is possible to create "Live Templates";
press Ctrl+Alt+s; search for "template"; go to "Live Templates" section;
just add one new item named "l" inside Scala entry, add a random description and the following code:
List($END$)
Press Enter, go to the editor, press L followed by Tab.
It is the end of your typing pains.
Do the same for Arrays.