If I wan't to see if each element in a list corresponds correctly to an element of the same index in another list, could I use forall to do this? For example something like
val p=List(2,4,6)
val q=List(1,2,3)
p.forall(x=>x==q(x)/2)
I understand that the x isn't an index of of q, and thats the problem I'm having, is there any way to make this work?
The most idiomatic way to handle this situation would be to zip the two lists:
scala> p.zip(q).forall { case (x, y) => x == y * 2 }
res0: Boolean = true
You could also use zipped, which can be slightly more efficient in some situations, as well as letting you be a bit more concise (or maybe just obfuscated):
scala> (p, q).zipped.forall(_ == _ * 2)
res1: Boolean = true
Note that both of these solutions will silently ignore extra elements if the lists don't have the same length, which may or may not be what you want.
Your best bet is probably to use zip
p.zip(q).forall{case (fst, snd) => fst == snd * 2}
Sequences from scala collection library have corresponds method which does exactly what you need:
p.corresponds(q)(_ == _ * 2)
It will return false if p and q are of different length.
Related
Using two Lists, element wise multiplication of these lists and sum of resultant list can be calculated in following way.
(List1 , List2).zipped.foldLeft(0.0) { case (a, (b, c)) => a + b * c }
How can I preform this operation for two iterators in Scala in an optimal and fast way?
(iterator1 zip iterator2).foldLeft(0.0) { case (a, (b, c)) => a + b * c }
is okay I suppose. If you want to squeeze the last bit of performance out of it, use arrays and a while loop.
You can use this piece of code that should work with any collection and any numeric type.
It tries to be efficient by doing everything in one traversal. However, as #Martijn said, if you need it to be the most efficient solution then just use plain Arrays of a primitive type like Int or Double and a while.
def dotProduct[N : Numeric](l1: IterableOnce[N], l2: IterableOnce[N]): N =
l1.iterator.zip(l2).map {
case (x, y) => Numeric[N].times(x, y)
}.sum
(note: this code is intended for 2.13+, for 2.12- you may use Iterable instead of IterableOnce)
I'm new to Scala and trying to figure out the best way to filter & map a collection. Here's a toy example to explain my problem.
Approach 1: This is pretty bad since I'm iterating through the list twice and calculating the same value in each iteration.
val N = 5
val nums = 0 until 10
val sqNumsLargerThanN = nums filter { x: Int => (x * x) > N } map { x: Int => (x * x).toString }
Approach 2: This is slightly better but I still need to calculate (x * x) twice.
val N = 5
val nums = 0 until 10
val sqNumsLargerThanN = nums collect { case x: Int if (x * x) > N => (x * x).toString }
So, is it possible to calculate this without iterating through the collection twice and avoid repeating the same calculations?
Could use a foldRight
nums.foldRight(List.empty[Int]) {
case (i, is) =>
val s = i * i
if (s > N) s :: is else is
}
A foldLeft would also achieve a similar goal, but the resulting list would be in reverse order (due to the associativity of foldLeft.
Alternatively if you'd like to play with Scalaz
import scalaz.std.list._
import scalaz.syntax.foldable._
nums.foldMap { i =>
val s = i * i
if (s > N) List(s) else List()
}
The typical approach is to use an iterator (if possible) or view (if iterator won't work). This doesn't exactly avoid two traversals, but it does avoid creation of a full-sized intermediate collection. You then map first and filter afterwards and then map again if needed:
xs.iterator.map(x => x*x).filter(_ > N).map(_.toString)
The advantage of this approach is that it's really easy to read and, since there are no intermediate collections, it's reasonably efficient.
If you are asking because this is a performance bottleneck, then the answer is usually to write a tail-recursive function or use the old-style while loop method. For instance, in your case
def sumSqBigN(xs: Array[Int], N: Int): Array[String] = {
val ysb = Array.newBuilder[String]
def inner(start: Int): Array[String] = {
if (start >= xs.length) ysb.result
else {
val sq = xs(start) * xs(start)
if (sq > N) ysb += sq.toString
inner(start + 1)
}
}
inner(0)
}
You can also pass a parameter forward in inner instead of using an external builder (especially useful for sums).
I have yet to confirm that this is truly a single pass, but:
val sqNumsLargerThanN = nums flatMap { x =>
val square = x * x
if (square > N) Some(x) else None
}
You can use collect which applies a partial function to every value of the collection that it's defined at. Your example could be rewritten as follows:
val sqNumsLargerThanN = nums collect {
case (x: Int) if (x * x) > N => (x * x).toString
}
A very simple approach that only does the multiplication operation once. It's also lazy, so it will be executing code only when needed.
nums.view.map(x=>x*x).withFilter(x => x> N).map(_.toString)
Take a look here for differences between filter and withFilter.
Consider this for comprehension,
for (x <- 0 until 10; v = x*x if v > N) yield v.toString
which unfolds to a flatMap over the range and a (lazy) withFilter onto the once only calculated square, and yields a collection with filtered results. To note one iteration and one calculation of square is required (in addition to creating the range).
You can use flatMap.
val sqNumsLargerThanN = nums flatMap { x =>
val square = x * x
if (square > N) Some(square.toString) else None
}
Or with Scalaz,
import scalaz.Scalaz._
val sqNumsLargerThanN = nums flatMap { x =>
val square = x * x
(square > N).option(square.toString)
}
The solves the asked question of how to do this with one iteration. This can be useful when streaming data, like with an Iterator.
However...if you are instead wanting the absolute fastest implementation, this is not it. In fact, I suspect you would use a mutable ArrayList and a while loop. But only after profiling would you know for sure. In any case, that's for another question.
Using a for comprehension would work:
val sqNumsLargerThanN = for {x <- nums if x*x > N } yield (x*x).toString
Also, I'm not sure but I think the scala compiler is smart about a filter before a map and will only do 1 pass if possible.
I am also beginner did it as follows
for(y<-(num.map(x=>x*x)) if y>5 ) { println(y)}
Ok so I thought this would be a snap, trying to practice Scala's collection operators and my example is a list of points.
The class can calculate and return the distance to another point (as double).
However, fold left doesn't seem to be the right solution - considering elements e1, e2, e3.. I need a moving window to calculate, I need the last element looked at to carry forward in the function - not just the sum
Sum {
e1.dist(e2)
e2.dist(e3)
etc
}
Reading the API I noticed a function called "sliding", perhaps that's the correct solution in conjunction with another operator. I know how to do this with loops of course, but trying to learn the scala way.
Thanks
import scala.math._
case class Point(x:Int, y:Int) {
def dist(p:Point) = sqrt( (p.x-x)^2+(p.y-y)^2 )
}
object Point {
//Unsure how to define this?
def dist(l:Seq[Point]) =l.foldLeft(0.0)((sum:Double,p:Point)=>)
}
I'm not quite sure what you want to do, but assuming you want the sum of the distances:
l.zip(l.tail).map { case (x,y) => x.dist(y) }.sum
Or with sliding:
l.sliding(2).map {
case List(fst,snd) => fst.dist(snd)
case _ => 0
}.sum
If you want to do it as a fold, you can, but you need the accumulator to keep both the total and the previous element:
l.foldLeft(l.head, 0.0){
case ((prev, sum), p) => (p, sum + p.dist(prev))
}._2
You finish with a tuple consiting of the last element and sum, so use ._2 to get the sum part.
btw, ^ on Int is bitwise logical XOR, not power. Use math.pow.
The smartest way is probably using zipped, which is a kind of iterator so you don't traverse the list more than once as you would using zip:
(l, l.tail).zipped.map( _ dist _ ).sum
Simply, I have two lists and I need to extract the new elements added to one of them.
I have the following
val x = List(1,2,3)
val y = List(1,2,4)
val existing :List[Int]= x.map(xInstance => {
if (!y.exists(yInstance =>
yInstance == xInstance))
xInstance
})
Result :existing: List[AnyVal] = List((), (), 3)
I need to remove all other elements except the numbers with the minimum cost.
Pick a suitable data structure, and life becomes a lot easier.
scala> x.toSet -- y
res1: scala.collection.immutable.Set[Int] = Set(3)
Also beware that:
if (condition) expr1
Is shorthand for:
if (condition) expr1 else ()
Using the result of this, which will usually have the static type Any or AnyVal is almost always an error. It's only appropriate for side-effects:
if (condition) buffer += 1
if (condition) sys.error("boom!")
retronym's solution is okay IF you don't have repeated elements that and you don't care about the order. However you don't indicate that this is so.
Hence it's probably going to be most efficient to convert y to a set (not x). We'll only need to traverse the list once and will have fast O(log(n)) access to the set.
All you need is
x filterNot y.toSet
// res1: List[Int] = List(3)
edit:
also, there's a built-in method that is even easier:
x diff y
(I had a look at the implementation; it looks pretty efficient, using a HashMap to count ocurrences.)
The easy way is to use filter instead so there's nothing to remove;
val existing :List[Int] =
x.filter(xInstance => !y.exists(yInstance => yInstance == xInstance))
val existing = x.filter(d => !y.exists(_ == d))
Returns
existing: List[Int] = List(3)
I'm coding up my first Scala script to get a feel for the language, and I'm a bit stuck as to the best way to achieve something.
My situation is the following, I have a method which I need to call N times, this method returns an Int on each run (might be different, there's a random component to the execution), and I want to keep the best run (the smallest value returned on these runs).
Now, coming from a Java/Python background, I would simply initialize the variable with null/None, and compare in the if, something like:
best = None
for...
result = executionOfThings()
if(best is None or result < best):
best = result
And that's that (pardon for the semi-python pseudo-code).
Now, on Scala, I'm struggling a bit. I've read about the usage of Option and pattern matching to achieve the same effect, and I guess I could code up something like (this was the best I could come up with):
best match {
case None => best = Some(res)
case Some(x) if x > res => best = Some(res)
case _ =>
}
I believe this works, but I'm not sure if it's the most idiomatic way of writing it. It's clear enough, but a bit verbose for such a simple "use-case".
Anyone that could shine a functional light on me?
Thanks.
For this particular problem, not in general, I would suggest initializing with Int.MaxValue as long as you're guaranteed that N >= 1. Then you just
if (result < best) best = result
You could also, with best as an option,
best = best.filter(_ >= result).orElse( Some(result) )
if the optionality is important (e.g. it is possible that N == 0, and you don't take a distinct path through the code in that case). This is a more general way to deal with optional values that may get replaced: use filter to keep the non-replaced cases, and orElse to fill in the replacement if needed.
Just use the min function:
(for (... executionOfThings()).min
Example:
((1 to 5).map (x => 4 * x * x - (x * x * x))).min
edit: adjusted to #user-unknown's suggestion
I would suggest you to rethink you whole computation to be more functional. You mutate state which should be avoided. I could think of a recursive version of your code:
def calcBest[A](xs: List[A])(f: A => Int): Int = {
def calcBest(xs: List[A], best: Int = Int.MaxValue): Int = xs match {
// will match an empty list
case Nil => best
// x will hold the head of the list and rest the rest ;-)
case x :: rest => calcBest(rest, math.min(f(x), best))
}
calcBest(xs)
}
callable with calcBest(List(7,5,3,8,2))(_*2) // => res0: Int = 4
With this you have no mutable state at all.
Another way would be to use foldLeft on the list:
list.foldLeft(Int.MaxValue) { case (best,x) => math.min(calculation(x),best) }
foldLeft takes a B and a PartialFunction of Tuple2[B,A] => B and returns B
Both ways are equivalent. The first one is probably faster, the second is more readable. Both traverse a list call a function on each value and return the smallest. Which from your snippet is what you want, right?
I thought I would offer another idiomatic solution. You can use Iterator.continually to create an infinite-length iterator that's lazily evaluated, take(N) to limit the iterator to N elements, and use min to find the winner.
Iterator.continually { executionOfThings() }.take(N).min