I am new to Scala and new OOP too. How can I update a particular element in a list while creating a new list.
val numbers= List(1,2,3,4,5)
val result = numbers.map(_*2)
I need to update third element only -> multiply by 2. How can I do that by using map?
You can use zipWithIndex to map the list into a list of tuples, where each element is accompanied by its index. Then, using map with pattern matching - you single out the third element (index = 2):
val numbers = List(1,2,3,4,5)
val result = numbers.zipWithIndex.map {
case (v, i) if i == 2 => v * 2
case (v, _) => v
}
// result: List[Int] = List(1, 2, 6, 4, 5)
Alternatively - you can use patch, which replaces a sub-sequence with a provided one:
numbers.patch(from = 2, patch = Seq(numbers(2) * 2), replaced = 1)
I think the clearest way of achieving this is by using updated(index: Int, elem: Int). For your example, it could be applied as follows:
val result = numbers.updated(2, numbers(2) * 2)
list.zipWithIndex creates a list of pairs with original element on the left, and index in the list on the right (indices are 0-based, so "third element" is at index 2).
val result = number.zipWithIndex.map {
case (n, 2) => n*2
case n => n
}
This creates an intermediate list holding the pairs, and then maps through it to do your transformation. A bit more efficient approach is to use iterator. Iterators a 'lazy', so, rather than creating an intermediate container, it will generate the pairs one-by-one, and send them straight to the .map:
val result = number.iterator.zipWithIndex.map {
case (n, 2) => n*2
case n => n
}.toList
1st and the foremost scala is FOP and not OOP. You can update any element of a list through the keyword "updated", see the following example for details:
Signature :- updated(index,value)
val numbers= List(1,2,3,4,5)
print(numbers.updated(2,10))
Now here the 1st argument is the index and the 2nd argument is the value. The result of this code will modify the list to:
List(1, 2, 10, 4, 5).
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 😉.
I'm trying to calculate the vector product between two vector using the map and reduce functions.
Let's see what happens in the REPL of Scala:
First of all I define 2 vectors with same length
scala> val v1 = Array(1,4,5,2)
v1: Array[Int] = Array(1, 4, 5, 2)
scala> val v2 = Array (3,5,1,5)
v2: Array[Int] = Array(3, 5, 1, 5)
Now I create a new array vecZip using the zip function
scala> val vecZip = v1 zip v2
vecZip: Array[(Int, Int)] = Array((1,3), (4,5), (5,1), (2,5))
Now I'd like to apply the reduce method
(to do the product of each tuple) for each element of this array.
I thought this:
val vecToSum = vecZip.map(x=>(List(x).reduce(_*_)))
I want to get a list (vecToSum) where apply the reduce method to calculate the total result. However I get this error:
scala> val vecToSum = vecZip.map(x=>(List(x).reduce(_*_)))
<console>:10: error: value * is not a member of (Int, Int)
val vecToSum = vecZip.map(x=>(List(x).reduce(_*_)))
^
You just need to call map and multiply the tuples values with each other, like this:
val vecToSum = vecZip.map(x => x._1 * x._2)
vecToSum is a List of tuples, so x is a Tuple of (Int, Int). Therefore if you call List(x).reduce(...), you're creating a List with the only value being the tuple, so that's not really what you want.
What your code is actually trying to do is it creates a list of a single tuple element, and then tries to reduce it. It would never work this way, as there is nothing to reduce - there is already single element in a list - a tuple.
Instead you need to map your vecZip array elements (tuples) via multiplying their elements:
vecZip.map { case (x, y) => x * y }
You don't need to reduce here. Reducing an Array[(Int, Int)] would mean performing some associative binary operation on all tuples inside the array. Note that it could be performing the operation on the first couple of tuples, then on the result of that and the third tuple, then on the result of that and the fourth tuple etc. but also, due to associativity, it could perform the operation on first and second tuple, simultaneously on third and fourth tuple, and then on their results etc., which is nice for parallelization (and frameworks such as Spark rely on it heavily)).
For example you could sum all first elements and all second elements of each tuple:
val reduced = vecZip.reduce((pair1, pair2) => (pair1._1 + pair2._1, pair1._2 + pair2._2))
What you want however is to simply map each tuple into the product of its elements:
val vecToSum = vecZip.map { case (x, y) => x * y }
Note that I used the partial function (see that case over there) in order to perform pattern matching on the tuple; without the partial function it would look like this:
val vecToSum = vecZip.map(tuple => tuple._1 * tuple._2)
I was wondering what is the most elegant way of getting the increasing prefix of a given sequence. My idea is as follows, but it is not purely functional or any elegant:
val sequence = Seq(1,2,3,1,2,3,4,5,6)
var currentElement = sequence.head - 1
val increasingPrefix = sequence.takeWhile(e =>
if (e > currentElement) {
currentElement = e
true
} else
false)
The result of the above is:
List(1,2,3)
You can take your solution, #Samlik, and effectively zip in the currentElement variable, but then map it out when you're done with it.
sequence.take(1) ++ sequence.zip(sequence.drop(1)).
takeWhile({case (a, b) => a < b}).map({case (a, b) => b})
Also works with infinite sequences:
val sequence = Seq(1, 2, 3).toStream ++ Stream.from(1)
sequence is now an infinite Stream, but we can peek at the first 10 items:
scala> sequence.take(10).toList
res: List[Int] = List(1, 2, 3, 1, 2, 3, 4, 5, 6, 7)
Now, using the above snippet:
val prefix = sequence.take(1) ++ sequence.zip(sequence.drop(1)).
takeWhile({case (a, b) => a < b}).map({case (a, b) => b})
Again, prefix is a Stream, but not infinite.
scala> prefix.toList
res: List[Int] = List(1, 2, 3)
N.b.: This does not handle the cases when sequence is empty, or when the prefix is also infinite.
If by elegant you mean concise and self-explanatory, it's probably something like the following:
sequence.inits.dropWhile(xs => xs != xs.sorted).next
inits gives us an iterator that returns the prefixes longest-first. We drop all the ones that aren't sorted and take the next one.
If you don't want to do all that sorting, you can write something like this:
sequence.scanLeft(Some(Int.MinValue): Option[Int]) {
case (Some(last), i) if i > last => Some(i)
case _ => None
}.tail.flatten
If the performance of this operation is really important, though (it probably isn't), you'll want to use something more imperative, since this solution still traverses the entire collection (twice).
And, another way to skin the cat:
val sequence = Seq(1,2,3,1,2,3,4,5,6)
sequence.head :: sequence
.sliding(2)
.takeWhile{case List(a,b) => a <= b}
.map(_(1)).toList
// List[Int] = List(1, 2, 3)
I will interpret elegance as the solution that most closely resembles the way we humans think about the problem although an extremely efficient algorithm could also be a form of elegance.
val sequence = List(1,2,3,2,3,45,5)
val increasingPrefix = takeWhile(sequence, _ < _)
I believe this code snippet captures the way most of us probably think about the solution to this problem.
This of course requires defining takeWhile:
/**
* Takes elements from a sequence by applying a predicate over two elements at a time.
* #param xs The list to take elements from
* #param f The predicate that operates over two elements at a time
* #return This function is guaranteed to return a sequence with at least one element as
* the first element is assumed to satisfy the predicate as there is no previous
* element to provide the predicate with.
*/
def takeWhile[A](xs: Traversable[A], f: (Int, Int) => Boolean): Traversable[A] = {
// function that operates over tuples and returns true when the predicate does not hold
val not = f.tupled.andThen(!_)
// Maybe one day our languages will be better than this... (dependant types anyone?)
val twos = sequence.sliding(2).map{case List(one, two) => (one, two)}
val indexOfBreak = twos.indexWhere(not)
// Twos has one less element than xs, we need to compensate for that
// An intuition is the fact that this function should always return the first element of
// a non-empty list
xs.take(i + 1)
}
Let's say we have this list of tuples:
val data = List(('a', List(1, 0)), ('b', List(1, 1)), ('c', List(0)))
The list has this signature:
List[(Char, List[Int])]
My task is to get the "List[Int]" element from a tuple inside "data" whose key is, for instance, letter "b". If I implement a method like "findIntList(data, 'b')", then I expect List(1, 1) as a result. I have tried the following approaches:
data.foreach { elem => if (elem._1 == char) return elem._2 }
data.find(x=> x._1 == ch)
for (elem <- data) yield elem match {case (x, y: List[Bit]) => if (x == char) y}
for (x <- data) yield if (x._1 == char) x._2
With all the approaches (except Approach 1, where I employ an explicit "return"), I get either a List[Option] or List[Any] and I don't know how to extract the "List[Int]" out of it.
One of many ways:
data.toMap.get('b').get
toMap converts a list of 2-tuples into a Map from the first element of the tuples to the second. get gives you the value for the given key and returns an Option, thus you need another get to actually get the list.
Or you can use:
data.find(_._1 == 'b').get._2
Note: Only use get on Option when you can guarantee that you'll have a Some and not a None. See http://www.scala-lang.org/api/current/index.html#scala.Option for how to use Option idiomatic.
Update: Explanation of the result types you see with your different approaches
Approach 2: find returns an Option[List[Int]] because it can not guarantee that a matching element gets found.
Approach 3: here you basically do a map, i.e. you apply a function to each element of your collection. For the element you are looking for the function returns your List[Int] for all other elements it contains the value () which is the Unit value, roughly equivalent to void in Java, but an actual type. Since the only common super type of ´List[Int]´ and ´Unit´ is ´Any´ you get a ´List[Any]´ as the result.
Approach 4 is basically the same as #3
Another way is
data.toMap.apply('b')
Or with one intermediate step this is even nicer:
val m = data.toMap
m('b')
where apply is used implicitly, i.e., the last line is equivalent to
m.apply('b')
There are multiple ways of doing it. One more way:
scala> def listInt(ls:List[(Char, List[Int])],ch:Char) = ls filter (a => a._1 == ch) match {
| case Nil => List[Int]()
| case x ::xs => x._2
| }
listInt: (ls: List[(Char, List[Int])], ch: Char)List[Int]
scala> listInt(data, 'b')
res66: List[Int] = List(1, 1)
You can try something like(when you are sure it exists) simply by adding type information.
val char = 'b'
data.collect{case (x,y:List[Int]) if x == char => y}.head
or use headOption if your not sure the character exists
data.collect{case (x,y:List[Int]) if x == char => y}.headOption
You can also solve this using pattern matching. Keep in mind you need to make it recursive though. The solution should look something like this;
def findTupleValue(tupleList: List[(Char, List[Int])], char: Char): List[Int] = tupleList match {
case (k, list) :: _ if char == k => list
case _ :: theRest => findTupleValue(theRest, char)
}
What this will do is walk your tuple list recursively. Check whether the head element matches your condition (the key you are looking for) and then returns it. Or continues with the remainder of the list.
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.