Scala trying to count instances of a digit in a number - scala

Question

This is my first day using scala. I am trying to make a string of the number of times each digit is represented in a string. For instance, the number 4310227 would return "1121100100" because 0 appears once, 1 appears once, 2 appears twice and so on...
def pow(n:Int) : String = {
val cubed = (n * n * n).toString
val digits = 0 to 9
val str = ""
for (a <- digits) {
println(a)
val b = cubed.count(_==a.toString)
println(b)
}
return cubed
}
and it doesn't seem to work. would like some scalay reasons why and to know whether I should even be going about it in this manner. Thanks!

Answer 1

When you iterate over strings, which is what you are doing when you call String#count(), you are working with Chars, not Strings. You don't want to compare these two with ==, since they aren't the same type of object.
One way to solve this problem is to call Char#toString() before performing the comparison, e.g., amend your code to read cubed.count(_.toString==a.toString).

Answer 2

As Rado and cheeken said, you're comparing a Char with a String, which will never be be equal. An alternative to cheekin's answer of converting each character to a string is to create a range from chars, ie '0' to '9':
val digits = '0' to '9'
...
val b = cubed.count(_ == a)
Note that if you want the Int that a Char represents, you can call char.asDigit.
Aleksey's, Ren's and Randall's answers are something you will want to strive towards as they separate out the pure solution to the problem. However, given that it's your first day with Scala, depending on what background you have, you might need a bit more context before understanding them.

Answer 3

Fairly simple:
scala> ("122333abc456xyz" filter (_.isDigit)).foldLeft(Map.empty[Char, Int]) ((histo, c) => histo + (c -> (histo.getOrElse(c, 0) + 1)))
res1: scala.collection.immutable.Map[Char,Int] = Map(4 -> 1, 5 -> 1, 6 -> 1, 1 -> 1, 2 -> 2, 3 -> 3)

Answer 4

This is perhaps not the fastest approach because intermediate datatype like String and Char are used but one of the most simplest:
def countDigits(n: Int): Map[Int, Int] =
n.toString.groupBy(x => x) map { case (n, c) => (n.asDigit, c.size) }
Example:
scala> def countDigits(n: Int): Map[Int, Int] = n.toString.groupBy(x => x) map { case (n, c) => (n.asDigit, c.size) }
countDigits: (n: Int)Map[Int,Int]
scala> countDigits(12345135)
res0: Map[Int,Int] = Map(5 -> 2, 1 -> 2, 2 -> 1, 3 -> 2, 4 -> 1)

Answer 5

Where myNumAsString is a String, eg "15625"
myNumAsString.groupBy(x => x).map(x => (x._1, x._2.length))
Result = Map(2 -> 1, 5 -> 2, 1 -> 1, 6 -> 1)
ie. A map containing the digit with its corresponding count.
What this is doing is taking your list, grouping the values by value (So for the initial string of "15625", it produces a map of 1 -> 1, 2 -> 2, 6 -> 6, and 5 -> 55.). The second bit just creates a map of the value to the count of how many times it occurs.

Answer 6

The counts for these hundred digits happen to fit into a hex digit.
scala> val is = for (_ <- (1 to 100).toList) yield r.nextInt(10)
is: List[Int] = List(8, 3, 9, 8, 0, 2, 0, 7, 8, 1, 6, 9, 9, 0, 3, 6, 8, 6, 3, 1, 8, 7, 0, 4, 4, 8, 4, 6, 9, 7, 4, 6, 6, 0, 3, 0, 4, 1, 5, 8, 9, 1, 2, 0, 8, 8, 2, 3, 8, 6, 4, 7, 1, 0, 2, 2, 6, 9, 3, 8, 6, 7, 9, 5, 0, 7, 6, 8, 7, 5, 8, 2, 2, 2, 4, 1, 2, 2, 6, 8, 1, 7, 0, 7, 6, 9, 5, 5, 5, 3, 5, 8, 2, 5, 1, 9, 5, 7, 2, 3)
scala> (new Array[Int](10) /: is) { case (a, i) => a(i) += 1 ; a } map ("%x" format _) mkString
warning: there were 1 feature warning(s); re-run with -feature for details
res7: String = a8c879caf9
scala> (new Array[Int](10) /: is) { case (a, i) => a(i) += 1 ; a } sum
warning: there were 1 feature warning(s); re-run with -feature for details
res8: Int = 100
I was going to point out that no one used a char range, but now I see Kristian did.
def pow(n:Int) : String = {
val cubed = (n * n * n).toString
val cnts = for (a <- '0' to '9') yield cubed.count(_ == a)
(cnts map (c => ('0' + c).toChar)).mkString
}