I have a List[Int] from 1 to 10 and want to make a List[List[Int]] containing two List[Int]: one list containing even numbers and the other containing odd numbers. The result should be like this:
List(List(2,4,6,8,10),List(1,3,5,7,9))
I tried these things:
1.to(10).toList.span((x:Int) => x % 2 == 0)
and
val lst = 1.to(10).toList; lst span (_%2==0)
However, neither of these worked.
Can someone help me on this matter?
The method you need to use is partition, not span:
scala> (1 to 10).partition(_ % 2 == 0)
res0: (IndexedSeq[Int], IndexedSeq[Int]) = (Vector(2, 4, 6, 8, 10),Vector(1, 3, 5, 7, 9))
Since you want a List[List[Int]], you could do this:
val lst = (1 to 10).toList
val (evens, odds) = lst.partition(_ % 2 == 0)
val newList = List(evens,odds) // List(List(2, 4, 6, 8, 10), List(1, 3, 5, 7, 9))
The span method can only be used to split a sequence at a single point:
scala> (1 to 10).span(_ < 5)
res1: (Range, Range) = (Range(1, 2, 3, 4),Range(5, 6, 7, 8, 9, 10))
When you tried lst.span(_ % 2 == 0), the program found that the first item, 1, did not pass the test (_ % 2 == 0), so all the elements were put in the second list, leaving none in the first.
Related
Trivial example
List.range(1, 10).filter(_ % 2 == 0)
This will yield List(2,4,6,8).
What's the best way to also get what was filtered out: List(1,3,5,7,9)
If you want to know both what passes some filter and what doesn't, instead of filter use partition:
val (even, odd) = List.range(1, 10).partition(_ % 2 == 0)
// even: List[Int] = List(2, 4, 6, 8)
// odd: List[Int] = List(1, 3, 5, 7, 9)
For example I have following Scala list, I want get a sublist until there is a requirement can be met.
val list = Seq(1,2,3,4,5,5,4,1,2,5)
The requirement is the number is 5, so I want the result as:
Seq(1,2,3,4)
Currently I use Scala collection's indexWhere and splitAt to return:
list.splitAt(list.indexWhere(x => x == 5))
(Seq(1,2,3,4), Seq(5,5,4,1,2,5))
I am not sure there are more better ways to achieve the same with better Scala collection's method I didn't realise?
You can use takeWhile:
scala> val list = Seq(1,2,3,4,5,5,4,1,2,5)
list: Seq[Int] = List(1, 2, 3, 4, 5, 5, 4, 1, 2, 5)
scala> list.takeWhile(_ != 5)
res30: Seq[Int] = List(1, 2, 3, 4)
Use span like this,
val (l,r) = list.span(_ != 5)
which delivers
l: List(1, 2, 3, 4)
r: List(5, 5, 4, 1, 2, 5)
Alternatively, you can write
val l = list.span(_ != 5)._1
to access only the first element of the resulting tuple.
This bisects the list at the first element that does not hold the condition.
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!
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).
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.
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)
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)
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.
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
}
And I have a comparison function "compr" already in the code to compare two values.
I want something like this:
Sorting.stableSort(arr[i,j] , compr)
where arr[i,j] is a range of element in array.
Take the slice as a view, sort and copy it back (or take a slice as a working buffer).
scala> val vs = Array(3,2,8,5,4,9,1,10,6,7)
vs: Array[Int] = Array(3, 2, 8, 5, 4, 9, 1, 10, 6, 7)
scala> vs.view(2,5).toSeq.sorted.copyToArray(vs,2)
scala> vs
res31: Array[Int] = Array(3, 2, 4, 5, 8, 9, 1, 10, 6, 7)
Outside the REPL, the extra .toSeq isn't needed:
vs.view(2,5).sorted.copyToArray(vs,2)
Updated:
scala 2.13.8> val vs = Array(3, 2, 8, 5, 4, 9, 1, 10, 6, 7)
val vs: Array[Int] = Array(3, 2, 8, 5, 4, 9, 1, 10, 6, 7)
scala 2.13.8> vs.view.slice(2,5).sorted.copyToArray(vs,2)
val res0: Int = 3
scala 2.13.8> vs
val res1: Array[Int] = Array(3, 2, 4, 5, 8, 9, 1, 10, 6, 7)
Split array into three parts, sort middle part and then concat them, not the most efficient way, but this is FP who cares about performance =)
val sorted =
for {
first <- l.take(FROM)
sortingPart <- l.slice(FROM, UNTIL)
lastPart <- l.takeRight(UNTIL)
} yield (first ++ Sorter.sort(sortingPart) ++ lastPart)
Something like that:
def stableSort[T](x: Seq[T], i: Int, j: Int, comp: (T,T) => Boolean ):Seq[T] = {
x.take(i) ++ x.slice(i,j).sortWith(comp) ++ x.drop(i+j-1)
}
def comp: (Int,Int) => Boolean = { case (x1,x2) => x1 < x2 }
val x = Array(1,9,5,6,3)
stableSort(x,1,4, comp)
// > res0: Seq[Int] = ArrayBuffer(1, 5, 6, 9, 3)
If your class implements Ordering it would be less cumbersome.
This should be as good as you can get without reimplementing the sort. Creates just one extra array with the size of the slice to be sorted.
def stableSort[K:reflect.ClassTag](xs:Array[K], from:Int, to:Int, comp:(K,K) => Boolean) : Unit = {
val tmp = xs.slice(from,to)
scala.util.Sorting.stableSort(tmp, comp)
tmp.copyToArray(xs, from)
}
As I noted in the title, how to compare the element of index N with element of index N+1, if elements compared are exactly the same, yield element only once.
I know I can use toSet, to get a set of unique elements, but this does not help me because, my list can contain duplicated elements but duplicated element can't be the next element in my list.
val ll = List(1, 2, 3, 6, 3, 7, 5, 5, 6, 3)
// Desired output: List(1, 2, 3, 6, 3, 7, 5, 6, 3)
I got a "near working solution" using zipWithIndex.collect, but when I compare inside it, index runs OutOfBounds. I can make this to work if I can use two conditions inside, first check maximum index to be index = (list.size-1) then I can compare list(index) != list(index+1) then yield list(index)
What I have tried without success (because of OutOfBounds), is:
times.zipWithIndex.collect
{
case (element, index)
// index+1 will be incremented out of my list
if (times(index) != times(index+1)) => times(index)
}
This can work if I can use one more condition to limit index, but does not work with two conditions:
times.zipWithIndex.collect
{
case (element, index)
if (index < times.size)
if (times(index) != times(index+1)) => times(index)
}
I appreciate any kind of alternative.
how about
ll.foldLeft(List[Int]())((acc, x) => acc match {case Nil => List(x) case y => if (y.last == x) y else y :+ x})
Here's my alternative using the sliding function:
val ll = List(1, 2, 3, 6, 3, 7, 5, 5, 6, 3)
ll.sliding(2)
.filter( t => t.length > 1 && t(0) != t(1) )
.map( t => t(0) )
.toList :+ ll.last
You can use zip the list with itself, dropping the first element so that you compare elements at index N with N + 1. You only need to append the last element (you may want to use a ListBuffer as appending the last element requires to copy the list).
val r = times.zip(times.drop(1)).withFilter(t => t._1 != t._2).map(_._1) :+ times.last
scala> val times = List(1, 2, 3, 6, 3, 7, 5, 5, 6, 3)
times: List[Int] = List(1, 2, 3, 6, 3, 7, 5, 5, 6, 3)
scala> val r = times.zip(times.drop(1)).withFilter(t => t._1 != t._2).map(_._1) :+ times.last
r: List[Int] = List(1, 2, 3, 6, 3, 7, 5, 6, 3)