I have a MutableList and I want to remove an element from it but I cannot find the appropriate method. There is a method to remove element from ListBuffer like this:
val x = ListBuffer(1, 2, 3, 4, 5, 6, 7, 8, 9)
x -= 5
I am unable to find an equivalent method on MutableList.
MutableList lacks -= and --= because it does not extend the Shrinkable trait. Various motivations for this can be found here.
MutableList does have diff, filter, and other methods which can help you in case you are in a situation where reassigning a variable (or instantiating a new variable) might be an option, and performance concerns aren't paramount:
var mylist = MutableList(1, 2, 3)
mylist = mylist diff Seq(1)
val myNewList = mylist.filter(_ != 2)
val indexFiltered = mylist.zipWithIndex.collect { case (el, ind) if ind != 1 => el }
You can often use ListBuffer instead of MutableList, which will unlock the desired -= and --= methods:
val mylist = ListBuffer(1, 2, 3)
mylist -= 1 //mylist is now ListBuffer(2, 3)
mylist --= Seq(2, 3) //mylist is now empty
It's not the answer, just to warn you about problems (at least in 2.11.x):
//street magic
scala> val a = mutable.MutableList(1,2,3)
a: scala.collection.mutable.MutableList[Int] = MutableList(1, 2, 3)
scala> a += 4
res7: a.type = MutableList(1, 2, 3, 4)
scala> a
res8: scala.collection.mutable.MutableList[Int] = MutableList(1, 2, 3, 4)
scala> a ++= List(8,9,10)
res9: a.type = MutableList(1, 2, 3, 4, 8, 9, 10)
scala> val b = a.tail
b: scala.collection.mutable.MutableList[Int] = MutableList(2, 3, 4, 8, 9, 10)
scala> b.length
res10: Int = 6
scala> a.length
res11: Int = 7
scala> a ++= List(8,9,10)
res12: a.type = MutableList(1, 2, 3, 4, 8, 9, 10, 8, 9, 10)
scala> b += 7
res13: b.type = MutableList(2, 3, 4, 8, 9, 10, 7)
scala> a
res14: scala.collection.mutable.MutableList[Int] = MutableList(1, 2, 3, 4, 8, 9, 10, 7)
scala> b
res15: scala.collection.mutable.MutableList[Int] = MutableList(2, 3, 4, 8, 9, 10, 7)
scala> a ++= List(8,9,10)
res16: a.type = MutableList(1, 2, 3, 4, 8, 9, 10, 7)
This example is taken from some gist - I've posted it on facebook with #devid_blein #street_magic tags, but can't find original link on the internet.
Related
Can you please explain what init method performs with respect to below list
i can see the result of new list says that last sequence is being omitted from the existing list.
val numbers = List(1, 2, 3, 4, 5)
val result = numbers.init
println(result)
.init and .last are the compliments to the .head and .tail methods.
val nums = List(1,2,3,4)
nums.head //res0: Int = 1
nums.tail //res1: List[Int] = List(2, 3, 4)
nums.init //res2: List[Int] = List(1, 2, 3)
nums.last //res3: Int = 4
def init: List[A] which selects all elements except the last.
l: List[Int] = List(1, 2, 3, 4, 5, 6, 7, 8)
scala> l.last
res58: Int = 8
scala> l.init
res59: List[Int] = List(1, 2, 3, 4, 5, 6, 7)
scala> val testList = List(1,2,3,4,5)
testList: List[Int] = List(1, 2, 3, 4, 5)
scala> testList.init
res0: List[Int] = List(1, 2, 3, 4)
scala> testList.last
res1: Int = 5
scala> testList.head
res2: Int = 1
scala> testList.tail
res3: List[Int] = List(2, 3, 4, 5)
Let's say I have Scala list like this:
val mylist = List(4,2,5,6,4,4,2,6,5,6,6,2,5,4,4)
How can I transform it into list of count and list of element? For example, I want to convert mylist into:
val count = List(3,5,3,4)
val elements = List(2,4,5,6)
Which means, in mylist, I have three occurrences of 2, five occurrences of 4, etc.
In procedural, this is easy as I can just make two empty lists (for count and elements) and fill them while doing iteration. However, I have no idea on how to achieve this in Scala.
Arguably a shortest version:
val elements = mylist.distinct
val count = elements map (e => mylist.count(_ == e))
Use .groupBy(identity) to create a Map regrouping elements with their occurences:
scala> val mylist = List(4,2,5,6,4,4,2,6,5,6,6,2,5,4,4)
mylist: List[Int] = List(4, 2, 5, 6, 4, 4, 2, 6, 5, 6, 6, 2, 5, 4, 4)
scala> mylist.groupBy(identity)
res0: scala.collection.immutable.Map[Int,List[Int]] = Map(2 -> List(2, 2, 2), 5 -> List(5, 5, 5), 4 -> List(4, 4, 4, 4, 4), 6 -> List(6, 6, 6, 6))
Then you can use .mapValues(_.length) to change the 'value' part of the map to the size of the list:
scala> mylist.groupBy(identity).mapValues(_.length)
res1: scala.collection.immutable.Map[Int,Int] = Map(2 -> 3, 5 -> 3, 4 -> 5, 6 -> 4)
If you want to get 2 lists out of this you can use .unzip, which returns a tuple, the first part being the keys (ie the elements), the second being the values (ie the number of instances of the element in the original list):
scala> val (elements, counts) = mylist.groupBy(identity).mapValues(_.length).unzip
elements: scala.collection.immutable.Iterable[Int] = List(2, 5, 4, 6)
counts: scala.collection.immutable.Iterable[Int] = List(3, 3, 5, 4)
One way would be to use groupBy and then check the size of each "group":
val withSizes = mylist.groupBy(identity).toList.map { case (v, l) => (v, l.size) }
val count = withSizes.map(_._2)
val elements = withSizes.map(_._1)
You can try like this as well alternative way of doing the same.
Step - 1
scala> val mylist = List(4,2,5,6,4,4,2,6,5,6,6,2,5,4,4)
mylist: List[Int] = List(4, 2, 5, 6, 4, 4, 2, 6, 5, 6, 6, 2, 5, 4, 4)
// Use groupBy { x => x } returns a "Map[Int, List[Int]]"
step - 2
scala> mylist.groupBy(x => (x))
res0: scala.collection.immutable.Map[Int,List[Int]] = Map(2 -> List(2, 2, 2), 5 -> List(5, 5, 5), 4 -> List(4, 4, 4, 4, 4), 6 -> List(6, 6, 6, 6))
step - 3
scala> mylist.groupBy(x => (x)).map{case(num,times) =>(num,times.size)}.toList
res1: List[(Int, Int)] = List((2,3), (5,3), (4,5), (6,4))
step -4 - sort by num
scala> mylist.groupBy(x => (x)).map{case(num,times) =>(num,times.size)}.toList.sortBy(_._1)
res2: List[(Int, Int)] = List((2,3), (4,5), (5,3), (6,4))
step -5 - unzip to beak into to list it return tuple
scala> mylist.groupBy(x => (x)).map{case(num,times) =>(num,times.size)}.toList.sortBy(_._1).unzip
res3: (List[Int], List[Int]) = (List(2, 4, 5, 6),List(3, 5, 3, 4))
scala> List(1,2,3,4,5,6,7).takeWhile(i=>i<5)
res1: List[Int] = List(1, 2, 3, 4)
What if I also need to include 5 in the result?
Assuming that the function that you will be using is more complicated than the taking first 5 elements then,
You can do
scala> List(1,2,3,4,5,6,7)
res5: List[Int] = List(1, 2, 3, 4, 5, 6, 7)
scala> res5.takeWhile(_<5) ++ res5.dropWhile(_<5).take(1)
res7: List[Int] = List(1, 2, 3, 4, 5)
Also
scala> res5.span(_<5)
res8: (List[Int], List[Int]) = (List(1, 2, 3, 4),List(5, 6, 7))
scala> res8._1 ++ res8._2.take(1)
res10: List[Int] = List(1, 2, 3, 4, 5)
Also
scala> res5.take(res5.segmentLength(_<5, 0) + 1)
res17: List[Int] = List(1, 2, 3, 4, 5)
scala> res5.take(res5.indexWhere(_>5))
res18: List[Int] = List(1, 2, 3, 4, 5)
Edit
If this is not a parallelized computation and you won't be using the parallel collections:
var last = myList.head
val rem = myList.takeWhile{ x=>
last = x
x < 5
}
last :: rem
anonymous function forming a closure around the solution you want.
Previous Answer
I'd settle for the far less complicated:
.takeWhile(_ <= 5)
wherein you're just using the less than or equal operator.
List(1,2,3,4,5,6,7,8).takeWhile(_ <= 5)
This is best optimal solution for it
Is this a bug in indexWhere method, or is there a meaningful explanation for the first four lines in the example bellow?
scala> List(1, 1, 4, 4, 4).indexWhere(_ > 3, -4)
res0: Int = -2
scala> List(1, 1, 4, 4, 4).indexWhere(_ > 3, -3)
res1: Int = -1
scala> List(1, 1, 4, 4, 4).indexWhere(_ > 3, -2)
res2: Int = 0
scala> List(1, 1, 4, 4, 4).indexWhere(_ > 3, -1)
res3: Int = 1
scala> List(1, 1, 4, 4, 4).indexWhere(_ > 3, 0)
res4: Int = 2
scala> List(1, 1, 4, 4, 4).indexWhere(_ > 3, 1)
res5: Int = 2
scala> List(1, 1, 4, 4, 4).indexWhere(_ > 3, 2)
res6: Int = 2
scala> List(1, 1, 4, 4, 4).indexWhere(_ > 3, 3)
res7: Int = 3
scala> List(1, 1, 4, 4, 4).indexWhere(_ > 3, 4)
res8: Int = 4
scala> List(1, 1, 4, 4, 4).indexWhere(_ > 3, 5)
res9: Int = -1
I think ideally you'd see an IllegalArgumentException if you passed a negative number as the start index (what's that supposed to mean anyway?), but maybe it was felt that the overhead of a bounds check wasn't worth it. I don't see any specified behaviour in the docs for this, so not sure I'd call it a bug; more a case of "rubbish in, rubbish out".
Source code looks like this:
override /*SeqLike*/
def indexWhere(p: A => Boolean, from: Int): Int = {
var i = from
var these = this drop from
while (these.nonEmpty) {
if (p(these.head))
return i
i += 1
these = these.tail
}
-1
}
So there is no input check, neither zeroing of i variable. I would say it is a bug. If you can, write it on https://groups.google.com/forum/#!forum/scala-language where devs can take a look at it.
As a future solution, you might want to implement your own version that does check. You might make it return an Option as well instead of -1, which is also unsafe:
implicit class WithIndexWhereSafe[T](seq: Seq[T]) {
def indexWhereSafe(p: T => Boolean, from: Int) = {
assert(from >= 0, "from must be >= 0")
val i = seq.indexWhere(p, from)
if (i != -1)
Some(i)
else
None
}
}
List(1, 1, 4, 4, 4).indexWhereSafe(_ > 3, 2) // Some(2)
List(1, 1, 4, 4, 4).indexWhereSafe(_ > 3, 5) // None
List(1, 1, 4, 4, 4).indexWhereSafe(_ > 3, -4) // AssertionError
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)
}