Scala for loop multiple counters - scala

I'm new to Scala, and I'm trying to convert this for loop from Java:
for(int x=1, y=2; x<=5; x++, y+=2)
System.out.println(x+y);
I'm trying to zip the values in Scala since I can't find a way to have multiple counters which are non-nested:
val a = Seq(1 to 5)
val b = Seq(2 to 10 by 2)
for((x,y) <- a.zip(b))
println(x+y)
But the above code is giving this error:
type mismatch; found: scala.collection.immutable.Range required: String
Does anyone know how to fix this? I would prefer to do with for loop only, not while loop.

Try this, no need to wrap the Range in a Seq:
val a = 1 to 5
val b = 2 to 10 by 2
for(
(x,y) <- a.zip(b)
)
println(x+y)

You might try . . .
((1 to 5) zip (2 to 10 by 2)).foreach(x => println(x._1+x._2))
Because Scala for comprehensions are sufficiently different from for() loops in other languages, it's often a good idea for beginners to avoid them until they've gained a sufficient knowledge of map, flatMap, and foreach.

In your example you want x to range from 1 to 5 and y is always 2*x. Using for loops is easy for those coming from Java:
for(x <- 1 to 5; y = x*2) {
println(s"x = $x, y = $y, x+y = ${x+y}")
}
Here is solution to a more generic problem - iterating over elements in a collection using multiple counters (=indices or pointers), like if you want to compare each 2 pairs:
val c = List("a", "b", "c", "d") //or any collection
val end = c.length - 1
for(i <- 0 to end-1; j <- i+1 to end)
//compare or operate with each pair
println(c(i)+c(j))
... prints:
ab
ac
ad
bc
bd
cd

Related

find out if a number is a good number in scala

Hi I am new to scala functional programming methodology. I want to input a number to my function and check if it is a good number or not.
A number is a good number if its every digit is larger than the sum of digits which are on the right side of that digit. 
For example:
9620  is good as (2 > 0, 6 > 2+0, 9 > 6+2+0)
steps I am using to solve this is
1. converting a number to string and reversing it
2. storing all digits of the reversed number as elements of a list
3. applying for loop from i equals 1 to length of number - 1
4. calculating sum of first i digits as num2
5. extracting ith digit from the list as digit1 which is one digit ahead of the first i numbers for which we calculated sum because list starts from zero.
6. comparing output of 4th and 5th step. if num1 is greater than num2 then we will break the for loop and come out of the loop to print it is not a good number.
please find my code below
val num1 = 9521.toString.reverse
val list1 = num1.map(_.todigit).toList
for (i <- 1 to num1.length - 1) {
val num2 = num1.take(i).map(_.toDigits) sum
val digit1 = list1(i)
if (num2 > digit1) {
print("number is not a good number")
break
}
}
I know this is not the most optimized way to solve this problem. Also I am looking for a way to code this using tail recursion where I pass two numbers and get all the good numbers falling in between those two numbers.
Can this be done in more optimized way?
Thanks in advance!
No String conversions required.
val n = 9620
val isGood = Stream.iterate(n)(_/10)
.takeWhile(_>0)
.map(_%10)
.foldLeft((true,-1)){ case ((bool,sum),digit) =>
(bool && digit > sum, sum+digit)
}._1
Here is a purely numeric version using a recursive function.
def isGood(n: Int): Boolean = {
#tailrec
def loop(n: Int, sum: Int): Boolean =
(n == 0) || (n%10 > sum && loop(n/10, sum + n%10))
loop(n/10, n%10)
}
This should compile into an efficient loop.
Using this function:(This will be the efficient way as the function forall will not traverse the entire list of digits. it stops when it finds the false condition immediately ( ie., when v(i)>v.drop(i+1).sum becomes false) while traversing from left to right of the vector v. )
def isGood(n: Int)= {
val v1 = n.toString.map(_.asDigit)
val v = if(v1.last!=0) v1 else v1.dropRight(1)
(0 to v.size-1).forall(i=>v(i)>v.drop(i+1).sum)
}
If we want to find good numbers in an interval of integers ranging from n1 to n2 we can use this function:
def goodNums(n1:Int,n2:Int) = (n1 to n2).filter(isGood(_))
In Scala REPL:
scala> isGood(9620)
res51: Boolean = true
scala> isGood(9600)
res52: Boolean = false
scala> isGood(9641)
res53: Boolean = false
scala> isGood(9521)
res54: Boolean = true
scala> goodNums(412,534)
res66: scala.collection.immutable.IndexedSeq[Int] = Vector(420, 421, 430, 510, 520, 521, 530, 531)
scala> goodNums(3412,5334)
res67: scala.collection.immutable.IndexedSeq[Int] = Vector(4210, 5210, 5310)
This is a more functional way. pairs is a list of tuples between a digit and the sum of the following digits. It is easy to create these tuples with drop, take and slice (a combination of drop and take) methods.
Finally I can represent my condition in an expressive way with forall method.
val n = 9620
val str = n.toString
val pairs = for { x <- 1 until str.length } yield (str.slice(x - 1, x).toInt, str.drop(x).map(_.asDigit).sum)
pairs.forall { case (a, b) => a > b }
If you want to be functional and expressive avoid to use break. If you need to check a condition for each element is a good idea to move your problem to collections, so you can use forAll.
This is not the case, but if you want performance (if you don't want to create an entire pairs collection because the condition for the first element is false) you can change your for collection from a Range to Stream.
(1 until str.length).toStream
Functional style tends to prefer monadic type things, such as maps and reduces. To make this look functional and clear, I'd do something like:
def isGood(value: Int) =
value.toString.reverse.map(digit=>Some(digit.asDigit)).
reduceLeft[Option[Int]]
{
case(sum, Some(digit)) => sum.collectFirst{case sum if sum < digit => sum+digit}
}.isDefined
Instead of using tail recursion to calculate this for ranges, just generate the range and then filter over it:
def goodInRange(low: Int, high: Int) = (low to high).filter(isGood(_))

Dynamic For Comprehensions

I'm relatively new to Scala so I'm not super confident with the language and I need your help to solve a problem. I know that the for-comprehension is just a syntactic sugar to simplify complex map/flatMap hierarchies.
Now, consider to have 3 different Range intervals, which should be combined in order to create all the possible combinations (respecting the intervals) of values.
Example:
Using the for-comprehension the problem can be solved as:
val intervalX = 1 to 5
val intervalY = 6 to 13
val intervalZ = 20 to 50
for {
x <- intervalX;
y <- intervalY;
z <- intervalZ
} yield (x,y,z)
Which is converted by the Scala compiler as:
intervalX.flatMap{x =>
intervalY.flatMap{y =>
intervalZ.map{z => (x,y,z)}
}
}
However, the problem is harder if you are given in input a variable number d of intervals. Is it possible to perform the same operation, obtaining all the possible d-tuples? I think that it could be solved using the foldLeft operation, but I am not able to write it correctly at the moment. Can you help me?
Thanks
If you can live without tuples as a result then a version using foldLeft and returning lists representing combinations could be:
val intervalX = 1 to 5
val intervalY = 6 to 13
val intervalZ = 20 to 50
val ranges = intervalZ :: intervalY :: intervalX :: Nil
val combos = ranges.foldLeft(Iterable[Seq[Int]](Nil)) { case (c, e) =>
for {
i <- e
j <- c
} yield i +: j
}
combos foreach { println(_) }

Scala comprehension from input

I am new to Scala and I am having troubles constructing a Map from inputs.
Here is my problem :
I am getting an input for elevators information. It consists of n lines, each one has the elevatorFloor number and the elevatorPosition on the floor.
Example:
0 5
1 3
4 5
So here I have 3 elevators, first one is on floor 0 at position 5, second one at floor 1 position 3 etc..
Is there a way in Scala to put it in a Map without using var ?
What I get so far is a Vector of all the elevators' information :
val elevators = {
for{i <- 0 until n
j <- readLine split " "
} yield j.toInt
}
I would like to be able split the lines in two variables "elevatorFloor" and "elevatorPos" and group them in a data structure (my guess is Map would be the appropriate choice) I would like to get something looking like:
elevators: SomeDataStructure[Int,Int] = ( 0->5, 1 -> 3, 4 -> 5)
I would like to clarify that I know I could write Javaish code, initialise a Map and then add the values to it, but I am trying to keep as close to functionnal programming as possible.
Thanks for the help or comments
You can do:
val res: Map[Int, Int] =
Source.fromFile("myfile.txt")
.getLines
.map { line =>
Array(floor, position) = line.split(' ')
(floor.toInt -> position.toInt)
}.toMap

How does "for (i <- 1 to x; j <- 1 to y)" actually increment the variables in Scala?

How is below loop being incremented ?
for(i <- 1 to 3; j <- 1 to 3) print((10 * i + j) + " ")
Is there an implicit counter using 'to' ?
for is actually shorthand for applying a bunch of collections methods. In particular, if you are not using yield, each statement in the for selector is translated to foreach. So
for (i <- 1 to 3; j <- 1 to 4) f(i,j)
turns into
(1 to 3).foreach{ i => (1 to 4).foreach{ j => f(i,j) } }
foreach is a method on all collections--Range included, which is what 1 to 3 turns into--which loops through each item in the collection, calling a provided function each time. A Range's items are the numbers listed (endpoints included, in this case)--in fact, Range doesn't actually store the numbers in a separate list, so it's main purpose is precisely to hold ranges of numbers for exactly this sort of iteration.
In scala, the for construct is like the "foreach" construct in Java. The following sets i to be each successive item in the given Iterable.
scala> for(i <- Seq(1, 2, 3)) println(i)
1
2
3
The to operator, as in 1 to 3 constructs a Range from 1 to 3:
scala> 1 to 3
res3: scala.collection.immutable.Range.Inclusive = Range(1, 2, 3)
There is an implicit conversion from Int to RichInt.
RichInt defines the function to() which returns a Range.
Range is a collection, and has foreach() hence it can be used in a for comprehension (which is just syntactic sugar for foreach()).

Incrementing the for loop (loop variable) in scala by power of 5

I had asked this question on Javaranch, but couldn't get a response there. So posting it here as well:
I have this particular requirement where the increment in the loop variable is to be done by multiplying it with 5 after each iteration. In Java we could implement it this way:
for(int i=1;i<100;i=i*5){}
In scala I was trying the following code-
var j=1
for(i<-1.to(100).by(scala.math.pow(5,j).toInt))
{
println(i+" "+j)
j=j+1
}
But its printing the following output:
1 1
6 2
11 3
16 4
21 5
26 6
31 7
36 8
....
....
Its incrementing by 5 always. So how do I got about actually multiplying the increment by 5 instead of adding it.
Let's first explain the problem. This code:
var j=1
for(i<-1.to(100).by(scala.math.pow(5,j).toInt))
{
println(i+" "+j)
j=j+1
}
is equivalent to this:
var j = 1
val range: Range = Predef.intWrapper(1).to(100)
val increment: Int = scala.math.pow(5, j).toInt
val byRange: Range = range.by(increment)
byRange.foreach {
println(i+" "+j)
j=j+1
}
So, by the time you get to mutate j, increment and byRange have already been computed. And Range is an immutable object -- you can't change it. Even if you produced new ranges while you did the foreach, the object doing the foreach would still be the same.
Now, to the solution. Simply put, Range is not adequate for your needs. You want a geometric progression, not an arithmetic one. To me (and pretty much everyone else answering, it seems), the natural solution would be to use a Stream or Iterator created with iterate, which computes the next value based on the previous one.
for(i <- Iterator.iterate(1)(_ * 5) takeWhile (_ < 100)) {
println(i)
}
EDIT: About Stream vs Iterator
Stream and Iterator are very different data structures, that share the property of being non-strict. This property is what enables iterate to even exist, since this method is creating an infinite collection1, from which takeWhile will create a new2 collection which is finite. Let's see here:
val s1 = Stream.iterate(1)(_ * 5) // s1 is infinite
val s2 = s1.takeWhile(_ < 100) // s2 is finite
val i1 = Iterator.iterate(1)(_ * 5) // i1 is infinite
val i2 = i1.takeWhile(_ < 100) // i2 is finite
These infinite collections are possible because the collection is not pre-computed. On a List, all elements inside the list are actually stored somewhere by the time the list has been created. On the above examples, however, only the first element of each collection is known in advance. All others will only be computed if and when required.
As I mentioned, though, these are very different collections in other respects. Stream is an immutable data structure. For instance, you can print the contents of s2 as many times as you wish, and it will show the same output every time. On the other hand, Iterator is a mutable data structure. Once you used a value, that value will be forever gone. Print the contents of i2 twice, and it will be empty the second time around:
scala> s2 foreach println
1
5
25
scala> s2 foreach println
1
5
25
scala> i2 foreach println
1
5
25
scala> i2 foreach println
scala>
Stream, on the other hand, is a lazy collection. Once a value has been computed, it will stay computed, instead of being discarded or recomputed every time. See below one example of that behavior in action:
scala> val s2 = s1.takeWhile(_ < 100) // s2 is finite
s2: scala.collection.immutable.Stream[Int] = Stream(1, ?)
scala> println(s2)
Stream(1, ?)
scala> s2 foreach println
1
5
25
scala> println(s2)
Stream(1, 5, 25)
So Stream can actually fill up the memory if one is not careful, whereas Iterator occupies constant space. On the other hand, one can be surprised by Iterator, because of its side effects.
(1) As a matter of fact, Iterator is not a collection at all, even though it shares a lot of the methods provided by collections. On the other hand, from the problem description you gave, you are not really interested in having a collection of numbers, just in iterating through them.
(2) Actually, though takeWhile will create a new Iterator on Scala 2.8.0, this new iterator will still be linked to the old one, and changes in one have side effects on the other. This is subject to discussion, and they might end up being truly independent in the future.
In a more functional style:
scala> Stream.iterate(1)(i => i * 5).takeWhile(i => i < 100).toList
res0: List[Int] = List(1, 5, 25)
And with more syntactic sugar:
scala> Stream.iterate(1)(_ * 5).takeWhile(_ < 100).toList
res1: List[Int] = List(1, 5, 25)
Maybe a simple while-loop would do?
var i=1;
while (i < 100)
{
println(i);
i*=5;
}
or if you want to also print the number of iterations
var i=1;
var j=1;
while (i < 100)
{
println(j + " : " + i);
i*=5;
j+=1;
}
it seems you guys likes functional so how about a recursive solution?
#tailrec def quints(n:Int): Unit = {
println(n);
if (n*5<100) quints(n*5);
}
Update: Thanks for spotting the error... it should of course be power, not multiply:
Annoyingly, there doesn't seem to be an integer pow function in the standard library!
Try this:
def pow5(i:Int) = math.pow(5,i).toInt
Iterator from 1 map pow5 takeWhile (100>=) toList
Or if you want to use it in-place:
Iterator from 1 map pow5 takeWhile (100>=) foreach {
j => println("number:" + j)
}
and with the indices:
val iter = Iterator from 1 map pow5 takeWhile (100>=)
iter.zipWithIndex foreach { case (j, i) => println(i + " = " + j) }
(0 to 2).map (math.pow (5, _).toInt).zipWithIndex
res25: scala.collection.immutable.IndexedSeq[(Int, Int)] = Vector((1,0), (5,1), (25,2))
produces a Vector, with i,j in reversed order.