I'm looking for an elegant way to combine every element of a Seq with the rest for a large collection.
Example: Seq(1,2,3).someMethod should produce something like
Iterator(
(1,Seq(2,3)),
(2,Seq(1,3)),
(3,Seq(1,2))
)
Order of elements doesn't matter. It doesn't have to be a tuple, a Seq(Seq(1),Seq(2,3)) is also acceptable (although kinda ugly).
Note the emphasis on large collection (which is why my example shows an Iterator).
Also note that this is not combinations.
Ideas?
Edit:
In my use case, the numbers are expected to be unique. If a solution can eliminate the dupes, that's fine, but not at additional cost. Otherwise, dupes are acceptable.
Edit 2: In the end, I went with a nested for-loop, and skipped the case when i == j. No new collections were created. I upvoted the solutions that were correct and simple ("simplicity is the ultimate sophistication" - Leonardo da Vinci), but even the best ones are quadratic just by the nature of the problem, and some create intermediate collections by usage of ++ that I wanted to avoid because the collection I'm dealing with has close to 50000 elements, 2.5 billion when quadratic.
The following code has constant runtime (it does everything lazily), but accessing every element of the resulting collections has constant overhead (when accessing each element, an index shift must be computed every time):
def faceMap(i: Int)(j: Int) = if (j < i) j else j + 1
def facets[A](simplex: Vector[A]): Seq[(A, Seq[A])] = {
val n = simplex.size
(0 until n).view.map { i => (
simplex(i),
(0 until n - 1).view.map(j => simplex(faceMap(i)(j)))
)}
}
Example:
println("Example: facets of a 3-dimensional simplex")
for ((i, v) <- facets((0 to 3).toVector)) {
println(i + " -> " + v.mkString("[", ",", "]"))
}
Output:
Example: facets of a 3-dimensional simplex
0 -> [1,2,3]
1 -> [0,2,3]
2 -> [0,1,3]
3 -> [0,1,2]
This code expresses everything in terms of simplices, because "omitting one index" corresponds exactly to the face maps for a combinatorially described simplex. To further illustrate the idea, here is what the faceMap does:
println("Example: how `faceMap(3)` shifts indices")
for (i <- 0 to 5) {
println(i + " -> " + faceMap(3)(i))
}
gives:
Example: how `faceMap(3)` shifts indices
0 -> 0
1 -> 1
2 -> 2
3 -> 4
4 -> 5
5 -> 6
The facets method uses the faceMaps to create a lazy view of the original collection that omits one element by shifting the indices by one starting from the index of the omitted element.
If I understand what you want correctly, in terms of handling duplicate values (i.e., duplicate values are to be preserved), here's something that should work. Given the following input:
import scala.util.Random
val nums = Vector.fill(20)(Random.nextInt)
This should get you what you need:
for (i <- Iterator.from(0).take(nums.size)) yield {
nums(i) -> (nums.take(i) ++ nums.drop(i + 1))
}
On the other hand, if you want to remove dups, I'd convert to Sets:
val numsSet = nums.toSet
for (num <- nums) yield {
num -> (numsSet - num)
}
seq.iterator.map { case x => x -> seq.filter(_ != x) }
This is quadratic, but I don't think there is very much you can do about that, because in the end of the day, creating a collection is linear, and you are going to need N of them.
import scala.annotation.tailrec
def prems(s : Seq[Int]):Map[Int,Seq[Int]]={
#tailrec
def p(prev: Seq[Int],s :Seq[Int],res:Map[Int,Seq[Int]]):Map[Int,Seq[Int]] = s match {
case x::Nil => res+(x->prev)
case x::xs=> p(x +: prev,xs, res+(x ->(prev++xs)))
}
p(Seq.empty[Int],s,Map.empty[Int,Seq[Int]])
}
prems(Seq(1,2,3,4))
res0: Map[Int,Seq[Int]] = Map(1 -> List(2, 3, 4), 2 -> List(1, 3, 4), 3 -> List(2, 1, 4),4 -> List(3, 2, 1))
I think you are looking for permutations. You can map the resulting lists into the structure you are looking for:
Seq(1,2,3).permutations.map(p => (p.head, p.tail)).toList
res49: List[(Int, Seq[Int])] = List((1,List(2, 3)), (1,List(3, 2)), (2,List(1, 3)), (2,List(3, 1)), (3,List(1, 2)), (3,List(2, 1)))
Note that the final toList call is only there to trigger the evaluation of the expressions; otherwise, the result is an iterator as you asked for.
In order to get rid of the duplicate heads, toMap seems like the most straight-forward approach:
Seq(1,2,3).permutations.map(p => (p.head, p.tail)).toMap
res50: scala.collection.immutable.Map[Int,Seq[Int]] = Map(1 -> List(3, 2), 2 -> List(3, 1), 3 -> List(2, 1))
Related
Suppose I have an imperative algorithm that keeps two indices left and right and moves them from left to right and from right to left
var left = 0
var right = array.length - 1
while (left < right) { .... } // move left and right inside the loop
Now I would like to write this algorithm without mutable indices.
How can I do that ? Do you have any examples of such algorithms ? I would prefer a non-recursive approach.
You can map pairs of elements between your list and its reverse, then go from left to right through that list of pairs and keep taking as long as your condition is satisfied:
val list = List(1, 2, 3, 4, 5)
val zipped = list zip list.reverse
val filtered = zipped takeWhile { case (a, b) => (a < b) }
Value of filtered is List((1, 5), (2, 4)).
Now you can do whatever you need with those elements:
val result = filtered map {
case (a, b) =>
// do something with each left-right pair, e.g. sum them
a + b
}
println(result) // List(6, 6)
If you need some kind of context dependant operation (that is, each
iteration depends on the result of the previous one) then you have to
use a more powerful abstraction (monad), but let's not go there if
this is enough for you. Even better would be to simply use recursion, as pointed out by others, but you said that's not an option.
EDIT:
Version without extra pass for reversing, only constant-time access for elem(length - index):
val list = List(1, 2, 3, 4, 5)
val zipped = list.view.zipWithIndex
val filtered = zipped takeWhile { case (a, index) => (a < list(list.length - 1 - index)) }
println(filtered.toList) // List((1, 0), (2, 1))
val result = filtered map {
case (elem, index) => // do something with each left-right pair, e.g. sum them
val (a, b) = (elem, list(list.length - 1 - index))
a + b
}
println(result.toList) // List(6, 6)
Use reverseIterator:
scala> val arr = Array(1,2,3,4,5)
arr: Array[Int] = Array(1, 2, 3, 4, 5)
scala> arr.iterator.zip(arr.reverseIterator).foreach(println)
(1,5)
(2,4)
(3,3)
(4,2)
(5,1)
This function is efficient on IndexedSeq collections, which Array is implicitly convertible to.
It really depends on what needs to be done at each iteration, but here's something to think about.
array.foldRight(0){case (elem, index) =>
if (index < array.length/2) {
/* array(index) and elem are opposite elements in the array */
/* do whatever (note: requires side effects) */
index+1
} else index // do nothing
} // ignore result
Upside: Traverse the array only once and no mutable variables.
Downside: Requires side effects (but that was implied in your example). Also, it'd be better if it traversed only half the array, but that would require early breakout and Scala doesn't offer an easy/elegant solution for that.
myarray = [1,2,3,4,5,6]
rmyarray = myarray[::-1]
Final_Result = []
for i in range(len(myarray)//2):
Final_Result.append(myarray[i])
Final_Result.append(rmyarray[i])
print(Final_Result)
# This is the simple approach I think 😉.
Beginner here.
Sorry but I did'nt found an answer so I ask the question here.
I want to know how to do this by using the Scala API :
(blabla))( -> List(('(',2),(')',2))
Currently I have this :
"(blabla))(".toCharArray.toList.filter(p => (p == '(' || p == ')')).sortBy(x => x)
Output :
List((, (, ), ))
Now how can I map each character to the tuples I describe ?
Example for a general case :
"t:e:s:t" -> List(('t',2),('e',1),('s',1),(':',3))
Thanks
val source = "ok:ok:k::"
val chars = source.toList
val shorter = chars.distinct.map( c => (c, chars.count(_ == c)))
//> shorter : List[(Char, Int)] = List((o,2), (k,3), (:,4))
Classic groupBy . mapValues use case:
scala> val str = "ok:ok:k::"
str: String = ok:ok:k::
scala> str.groupBy(identity).mapValues(_.size) // identity <=> (x => x)
res0: scala.collection.immutable.Map[Char,Int] = Map(k -> 3, : -> 4, o -> 2)
I like sschaef's solution very much, but I was wondering if anyone could weigh in on how efficient that solution is compared to this one:
scala> val str = "ok:ok:k::"
str: String = ok:ok:k::
scala> str.foldLeft(Map[Char,Int]().withDefaultValue(0))((current, c) => current.updated(c, current(c) + 1))
res29: scala.collection.immutable.Map[Char,Int] = Map(o -> 2, k -> 3, : -> 4)
I think my solution is slower. If we have n total occurrences and m unique values:
My solution: we have the fold left over all occurrences or n. For each of these occurrences we look up once to find the current count and then again to create the updated the map. I'm assuming that the creating of the updated map is constant time.
Total complexity: n * 2m or O(n*m)
sschaef's solution: we have the groupBy which I'm assuming just adds entries onto a list without checking the map (so for all values this would be a constant time look up plus appending to the list) so n. Then for the mapValues it probably iterates over the unique values and grabs the size for each key's list. I'm assuming that getting the size of each entry's list is constant time.
Total complexity: O(n + m)
Does this seem correct or am I mistaken in my assumptions?
Suppose you have
val docs = List(List("one", "two"), List("two", "three"))
where e.g. List("one", "two") represents a document containing terms "one" and "two", and you want to build a map with the document frequency for every term, i.e. in this case
Map("one" -> 1, "two" -> 2, "three" -> 1)
How would you do that in Scala? (And in an efficient way, assuming a much larger dataset.)
My first Java-like thought is to use a mutable map:
val freqs = mutable.Map.empty[String,Int]
for (doc <- docs)
for (term <- doc)
freqs(term) = freqs.getOrElse(term, 0) + 1
which works well enough but I'm wondering how you could do that in a more "functional" way, without resorting to a mutable map?
Try this:
scala> docs.flatten.groupBy(identity).mapValues(_.size)
res0: Map[String,Int] = Map(one -> 1, two -> 2, three -> 1)
If you are going to be accessing the counts many times, then you should avoid mapValues since it is "lazy" and, thus, would recompute the size on every access. This version gives you the same result but won't require the recomputations:
docs.flatten.groupBy(identity).map(x => (x._1, x._2.size))
The identity function just means x => x.
docs.flatten.foldLeft(new Map.WithDefault(Map[String,Int](),Function.const(0))){
(m,x) => m + (x -> (1 + m(x)))}
What a train wreck!
[Edit]
Ah, that's better!
docs.flatten.foldLeft(Map[String,Int]() withDefaultValue 0){
(m,x) => m + (x -> (1 + m(x)))}
Starting Scala 2.13, after flattening the list of lists, we can use groupMapReduce which is a one-pass alternative to groupBy/mapValues:
// val docs = List(List("one", "two"), List("two", "three"))
docs.flatten.groupMapReduce(identity)(_ => 1)(_ + _)
// Map[String,Int] = Map("one" -> 1, "three" -> 1, "two" -> 2)
This:
flattens the List of Lists as a List
groups list elements (identity) (group part of groupMapReduce)
maps each grouped value occurrence to 1 (_ => 1) (map part of groupMapReduce)
reduces values within a group of values (_ + _) by summing them (reduce part of groupMapReduce).
Apologies: I'm well noob
I have an items class
class item(ind:Int,freq:Int,gap:Int){}
I have an ordered list of ints
val listVar = a.toList
where a is an array
I want a list of items called metrics where
ind is the (unique) integer
freq is the number of times that ind appears in list
gap is the minimum gap between ind and the number in the list before it
so far I have:
def metrics = for {
n <- 0 until 255
listVar filter (x == n) count > 0
}
yield new item(n, (listVar filter == n).count,0)
It's crap and I know it - any clues?
Well, some of it is easy:
val freqMap = listVar groupBy identity mapValues (_.size)
This gives you ind and freq. To get gap I'd use a fold:
val gapMap = listVar.sliding(2).foldLeft(Map[Int, Int]()) {
case (map, List(prev, ind)) =>
map + (ind -> (map.getOrElse(ind, Int.MaxValue) min ind - prev))
}
Now you just need to unify them:
freqMap.keys.map( k => new item(k, freqMap(k), gapMap.getOrElse(k, 0)) )
Ideally you want to traverse the list only once and in the course for each different Int, you want to increment a counter (the frequency) as well as keep track of the minimum gap.
You can use a case class to store the frequency and the minimum gap, the value stored will be immutable. Note that minGap may not be defined.
case class Metric(frequency: Int, minGap: Option[Int])
In the general case you can use a Map[Int, Metric] to lookup the Metric immutable object. Looking for the minimum gap is the harder part. To look for gap, you can use the sliding(2) method. It will traverse the list with a sliding window of size two allowing to compare each Int to its previous value so that you can compute the gap.
Finally you need to accumulate and update the information as you traverse the list. This can be done by folding each element of the list into your temporary result until you traverse the whole list and get the complete result.
Putting things together:
listVar.sliding(2).foldLeft(
Map[Int, Metric]().withDefaultValue(Metric(0, None))
) {
case (map, List(a, b)) =>
val metric = map(b)
val newGap = metric.minGap match {
case None => math.abs(b - a)
case Some(gap) => math.min(gap, math.abs(b - a))
}
val newMetric = Metric(metric.frequency + 1, Some(newGap))
map + (b -> newMetric)
case (map, List(a)) =>
map + (a -> Metric(1, None))
case (map, _) =>
map
}
Result for listVar: List[Int] = List(2, 2, 4, 4, 0, 2, 2, 2, 4, 4)
scala.collection.immutable.Map[Int,Metric] = Map(2 -> Metric(4,Some(0)),
4 -> Metric(4,Some(0)), 0 -> Metric(1,Some(4)))
You can then turn the result into your desired item class using map.toSeq.map((i, m) => new Item(i, m.frequency, m.minGap.getOrElse(-1))).
You can also create directly your Item object in the process, but I thought the code would be harder to read.
I'm learning Scala as my first functional-ish language. As one of the problems, I was trying to find a functional way of generating the sequence S up to n places. S is defined so that S(1) = 1, and S(x) = the number of times x appears in the sequence. (I can't remember what this is called, but I've seen it in programming books before.)
In practice, the sequence looks like this:
S = 1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7 ...
I can generate this sequence pretty easily in Scala using an imperative style like this:
def genSequence(numItems: Int) = {
require(numItems > 0, "numItems must be >= 1")
var list: List[Int] = List(1)
var seq_no = 2
var no = 2
var no_nos = 0
var num_made = 1
while(num_made < numItems) {
if(no_nos < seq_no) {
list = list :+ no
no_nos += 1
num_made += 1
} else if(no % 2 == 0) {
no += 1
no_nos = 0
} else {
no += 1
seq_no += 1
no_nos = 0
}
}
list
}
But I don't really have any idea how to write this without using vars and the while loop.
Thanks!
Pavel's answer has come closest so far, but it's also inefficient. Two flatMaps and a zipWithIndex are overkill here :)
My understanding of the required output:
The results contain all the positive integers (starting from 1) at least once
each number n appears in the output (n/2) + 1 times
As Pavel has rightly noted, the solution is to start with a Stream then use flatMap:
Stream from 1
This generates a Stream, a potentially never-ending sequence that only produces values on demand. In this case, it's generating 1, 2, 3, 4... all the way up to Infinity (in theory) or Integer.MAX_VALUE (in practice)
Streams can be mapped over, as with any other collection. For example: (Stream from 1) map { 2 * _ } generates a Stream of even numbers.
You can also use flatMap on Streams, allowing you to map each input element to zero or more output elements; this is key to solving your problem:
val strm = (Stream from 1) flatMap { n => Stream.fill(n/2 + 1)(n) }
So... How does this work? For the element 3, the lambda { n => Stream.fill(n/2 + 1)(n) } will produce the output stream 3,3. For the first 5 integers you'll get:
1 -> 1
2 -> 2, 2
3 -> 3, 3
4 -> 4, 4, 4
5 -> 5, 5, 5
etc.
and because we're using flatMap, these will be concatenated, yielding:
1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, ...
Streams are memoised, so once a given value has been calculated it'll be saved for future reference. However, all the preceeding values have to be calculated at least once. If you want the full sequence then this won't cause any problems, but it does mean that generating S(10796) from a cold start is going to be slow! (a problem shared with your imperative algorithm). If you need to do this, then none of the solutions so far is likely to be appropriate for you.
The following code produces exactly the same sequence as yours:
val seq = Stream.from(1)
.flatMap(Stream.fill(2)(_))
.zipWithIndex
.flatMap(p => Stream.fill(p._1)(p._2))
.tail
However, if you want to produce the Golomb sequence (that complies with the definition, but differs from your sample code result), you may use the following:
val seq = 1 #:: a(2)
def a(n: Int): Stream[Int] = (1 + seq(n - seq(seq(n - 2) - 1) - 1)) #:: a(n + 1)
You may check my article for more examples of how to deal with number sequences in functional style.
Here is a translation of your code to a more functional style:
def genSequence(numItems: Int): List[Int] = {
genSequenceR(numItems, 2, 2, 0, 1, List[Int](1))
}
def genSequenceR(numItems: Int, seq_no: Int, no:Int, no_nos: Int, numMade: Int, list: List[Int]): List[Int] = {
if(numMade < numItems){
if(no_nos < seq_no){
genSequenceR(numItems, seq_no, no, no_nos + 1, numMade + 1, list :+ no)
}else if(no % 2 == 0){
genSequenceR(numItems, seq_no, no + 1, 0, numMade, list)
}else{
genSequenceR(numItems, seq_no + 1, no + 1, 0, numMade, list)
}
}else{
list
}
}
The genSequenceR is the recursive function that accumulates values in the list and calls the function with new values based on the conditions. Like the while loop, it terminates, when numMade is less than numItems and returns the list to genSequence.
This is a fairly rudimentary functional translation of your code. It can be improved and there are better approaches typically used. I'd recommend trying to improve it with pattern matching and then work towards the other solutions that use Stream here.
Here's an attempt from a Scala tyro. Keep in mind I don't really understand Scala, I don't really understand the question, and I don't really understand your algorithm.
def genX_Ys[A](howMany : Int, ofWhat : A) : List[A] = howMany match {
case 1 => List(ofWhat)
case _ => ofWhat :: genX_Ys(howMany - 1, ofWhat)
}
def makeAtLeast(startingWith : List[Int], nextUp : Int, howMany : Int, minimumLength : Int) : List[Int] = {
if (startingWith.size >= minimumLength)
startingWith
else
makeAtLeast(startingWith ++ genX_Ys( howMany, nextUp),
nextUp +1, howMany + (if (nextUp % 2 == 1) 1 else 0), minimumLength)
}
def genSequence(numItems: Int) = makeAtLeast(List(1), 2, 2, numItems).slice(0, numItems)
This seems to work, but re-read the caveats above. In particular, I am sure there is a library function that performs genX_Ys, but I couldn't find it.
EDIT Could be
def genX_Ys[A](howMany : Int, ofWhat : A) : Seq[A] =
(1 to howMany) map { x => ofWhat }
Here is a very direct "translation" of the definition of the Golomb seqence:
val it = Iterator.iterate((1,1,Map(1->1,2->2))){ case (n,i,m) =>
val c = m(n)
if (c == 1) (n+1, i+1, m + (i -> n) - n)
else (n, i+1, m + (i -> n) + (n -> (c-1)))
}.map(_._1)
println(it.take(10).toList)
The tripel (n,i,m) contains the actual number n, the index i and a Map m, which contains how often an n must be repeated. When the counter in the Map for our n reaches 1, we increase n (and can drop n from the map, as it is not longer needed), else we just decrease n's counter in the map and keep n. In every case we add the new pair i -> n into the map, which will be used as counter later (when a subsequent n reaches the value of the current i).
[Edit]
Thinking about it, I realized that I don't need indexes and not even a lookup (because the "counters" are already in the "right" order), which means that I can replace the Map with a Queue:
import collection.immutable.Queue
val it = 1 #:: Iterator.iterate((2, 2, Queue[Int]())){
case (n,1,q) => (n+1, q.head, q.tail + (n+1))
case (n,c,q) => (n,c-1,q + n)
}.map(_._1).toStream
The Iterator works correctly when starting by 2, so I had to add a 1 at the beginning. The second tuple argument is now the counter for the current n (taken from the Queue). The current counter could be kept in the Queue as well, so we have only a pair, but then it's less clear what's going on due to the complicated Queue handling:
val it = 1 #:: Iterator.iterate((2, Queue[Int](2))){
case (n,q) if q.head == 1 => (n+1, q.tail + (n+1))
case (n,q) => (n, ((q.head-1) +: q.tail) + n)
}.map(_._1).toStream