How can I validate that a sequence of numbers falls into an allowed range of numbers in a more scala/functional (not java like) way?
val valuesRight = Seq(1, 2, 3, 4, 5, 6, 7)
val valuesWrong = Seq(1, 2, 5, 6, 7, 8, 9)
val allowedValues = Range(1, 8)
def containsNotAllowedValues(allowed: Range, input: Seq[Int]): Boolean = {
var containsNotAllowedValue = false
// pseudo code of how to do it in java, But how in scala / functional way?
while next element is available
validate if it is contained in allowed
if not allowed set containsNotAllowedValue to true and break the loop early
}
containsNotAllowedValue
}
containsNotAllowedValues(allowedValues, valuesRight) // expected false as no wrong element contained
containsNotAllowedValues(allowedValues, valuesWrong) // expected true as at least single wrong element is contained
You can use the forall function on Seq. It checks whether the given predicate is true for all elements in the Seq. If that function returns false with the predicate allowed contains _, then your Seq contains an illegal value.
def containsNotAllowedValues(allowed: Range, input: Seq[Int]): Boolean =
!input.forall { allowed contains _ }
You could do something as simple as this:
def containsNotAllowedValues(allowed: Range, input: Seq[Int]): Boolean = {
!(allowed.head <= input.min && allowed.last >= input.max)
}
val valuesRight = Seq(1, 2, 3, 4, 5, 6, 7)
val valuesWrong = Seq(1, 2, 5, 6, 7, 8, 9)
val allowedValues = Range(1, 8)
containsNotAllowedValues(allowedValues,valuesRight) //false
containsNotAllowedValues(allowedValues,valuesWrong) //true
This will fail if the Range has only 1 value like Range(1,1)
If you want you can add
allowed.size>1
to catch those cases
I have a Scala List that contains some repeated numbers. I want to count the number of times a specific number will repeat itself. For example:
val list = List(1,2,3,3,4,2,8,4,3,3,5)
val repeats = list.takeWhile(_ == List(3,3)).size
And the val repeats would equal 2.
Obviously the above is pseudo-code and takeWhile will not find two repeated 3s since _ represents an integer. I tried mixing both takeWhile and take(2) but with little success. I also referred code from How to find count of repeatable elements in scala list but it appears the author is looking to achieve something different.
Thanks for your help.
This will work in this case:
val repeats = list.sliding(2).count(_.forall(_ == 3))
The sliding(2) method gives you an iterator of lists of elements and successors and then we just count where these two are equal to 3.
Question is if it creates the correct result to List(3, 3, 3)? Do you want that to be 2 or just 1 repeat.
val repeats = list.sliding(2).toList.count(_==List(3,3))
and more generally the following code returns tuples of element and repeats value for all elements:
scala> list.distinct.map(x=>(x,list.sliding(2).toList.count(_.forall(_==x))))
res27: List[(Int, Int)] = List((1,0), (2,0), (3,2), (4,0), (8,0), (5,0))
which means that the element '3' repeats 2 times consecutively at 2 places and all others 0 times.
and also if we want element repeats 3 times consecutively we just need to modify the code as follows:
list.distinct.map(x=>(x,list.sliding(3).toList.count(_.forall(_==x))))
in SCALA REPL:
scala> val list = List(1,2,3,3,3,4,2,8,4,3,3,3,5)
list: List[Int] = List(1, 2, 3, 3, 3, 4, 2, 8, 4, 3, 3, 3, 5)
scala> list.distinct.map(x=>(x,list.sliding(3).toList.count(_==List(x,x,x))))
res29: List[(Int, Int)] = List((1,0), (2,0), (3,2), (4,0), (8,0), (5,0))
Even sliding value can be varied by defining a function as:
def repeatsByTimes(list:List[Int],n:Int) =
list.distinct.map(x=>(x,list.sliding(n).toList.count(_.forall(_==x))))
Now in REPL:
scala> val list = List(1,2,3,3,4,2,8,4,3,3,5)
list: List[Int] = List(1, 2, 3, 3, 4, 2, 8, 4, 3, 3, 5)
scala> repeatsByTimes(list,2)
res33: List[(Int, Int)] = List((1,0), (2,0), (3,2), (4,0), (8,0), (5,0))
scala> val list = List(1,2,3,3,3,4,2,8,4,3,3,3,2,4,3,3,3,5)
list: List[Int] = List(1, 2, 3, 3, 3, 4, 2, 8, 4, 3, 3, 3, 2, 4, 3, 3, 3, 5)
scala> repeatsByTimes(list,3)
res34: List[(Int, Int)] = List((1,0), (2,0), (3,3), (4,0), (8,0), (5,0))
scala>
We can go still further like given a list of integers and given a maximum number
of consecutive repetitions that any of the element can occur in the list, we may need a list of 3-tuples representing (the element, number of repetitions of this element, at how many places this repetition occurred). this is more exhaustive information than the above. Can be achieved by writing a function like this:
def repeats(list:List[Int],maxRep:Int) =
{ var v:List[(Int,Int,Int)] = List();
for(i<- 1 to maxRep)
v = v ++ list.distinct.map(x=>
(x,i,list.sliding(i).toList.count(_.forall(_==x))))
v.sortBy(_._1) }
in SCALA REPL:
scala> val list = List(1,2,3,3,3,4,2,8,4,3,3,3,2,4,3,3,3,5)
list: List[Int] = List(1, 2, 3, 3, 3, 4, 2, 8, 4, 3, 3, 3, 2, 4, 3, 3, 3, 5)
scala> repeats(list,3)
res38: List[(Int, Int, Int)] = List((1,1,1), (1,2,0), (1,3,0), (2,1,3),
(2,2,0), (2,3,0), (3,1,9), (3,2,6), (3,3,3), (4,1,3), (4,2,0), (4,3,0),
(5,1,1), (5,2,0), (5,3,0), (8,1,1), (8,2,0), (8,3,0))
scala>
These results can be understood as follows:
1 times the element '1' occurred at 1 places.
2 times the element '1' occurred at 0 places.
............................................
............................................
.............................................
2 times the element '3' occurred at 6 places..
.............................................
3 times the element '3' occurred at 3 places...
............................................and so on.
Thanks to Luigi Plinge I was able to use methods in run-length encoding to group together items in a list that repeat. I used some snippets from this page here: http://aperiodic.net/phil/scala/s-99/
var n = 0
runLengthEncode(totalFrequencies).foreach{ o =>
if(o._1 > 1 && o._2==subjectNumber) n+=1
}
n
The method runLengthEncode is as follows:
private def pack[A](ls: List[A]): List[List[A]] = {
if (ls.isEmpty) List(List())
else {
val (packed, next) = ls span { _ == ls.head }
if (next == Nil) List(packed)
else packed :: pack(next)
}
}
private def runLengthEncode[A](ls: List[A]): List[(Int, A)] =
pack(ls) map { e => (e.length, e.head) }
I'm not entirely satisfied that I needed to use the mutable var n to count the number of occurrences but it did the trick. This will count the number of times a number repeats itself no matter how many times it is repeated.
If you knew your list was not very long you could do it with Strings.
val list = List(1,2,3,3,4,2,8,4,3,3,5)
val matchList = List(3,3)
(matchList.mkString(",")).r.findAllMatchIn(list.mkString(",")).length
From you pseudocode I got this working:
val pairs = list.sliding(2).toList //create pairs of consecutive elements
val result = pairs.groupBy(x => x).map{ case(x,y) => (x,y.size); //group pairs and retain the size, which is the number of occurrences.
result will be a Map[List[Int], Int] so you can the count number like:
result(List(3,3)) // will return 2
I couldn't understand if you also want to check lists of several sizes, then you would need to change the parameter to sliding to the desired size.
def pack[A](ls: List[A]): List[List[A]] = {
if (ls.isEmpty) List(List())
else {
val (packed, next) = ls span { _ == ls.head }
if (next == Nil) List(packed)
else packed :: pack(next)
}
}
def encode[A](ls: List[A]): List[(Int, A)] = pack(ls) map { e => (e.length, e.head) }
val numberOfNs = list.distinct.map{ n =>
(n -> list.count(_ == n))
}.toMap
val runLengthPerN = runLengthEncode(list).map{ t => t._2 -> t._1}.toMap
val nRepeatedMostInSuccession = runLengthPerN.toList.sortWith(_._2 <= _._2).head._1
Where runLength is defined as below from scala's 99 problems problem 9 and scala's 99 problems problem 10.
Since numberOfNs and runLengthPerN are Maps, you can get the population count of any number in the list with numberOfNs(number) and the length of the longest repitition in succession with runLengthPerN(number). To get the runLength, just compute as above with runLength(list).map{ t => t._2 -> t._1 }.
I'm trying to search a scala collection for an item in a list that matches some predicate. I don't necessarily need the return value, just testing if the list contains it.
In Java, I might do something like:
for ( Object item : collection ) {
if ( condition1(item) && condition2(item) ) {
return true;
}
}
return false;
In Groovy, I can do something like:
return collection.find { condition1(it) && condition2(it) } != null
What's the idiomatic way to do this in Scala? I could of course convert the Java loop style to Scala, but I feel like there's a more functional way to do this.
Use filter:
scala> val collection = List(1,2,3,4,5)
collection: List[Int] = List(1, 2, 3, 4, 5)
// take only that values that both are even and greater than 3
scala> collection.filter(x => (x % 2 == 0) && (x > 3))
res1: List[Int] = List(4)
// you can return this in order to check that there such values
scala> res1.isEmpty
res2: Boolean = false
// now query for elements that definitely not in collection
scala> collection.filter(x => (x % 2 == 0) && (x > 5))
res3: List[Int] = List()
scala> res3.isEmpty
res4: Boolean = true
But if all you need is to check use exists:
scala> collection.exists( x => x % 2 == 0 )
res6: Boolean = true
Testing if value matching predicate exists
If you're just interested in testing if a value exists, you can do it with.... exists
scala> val l=(1 to 4) toList
l: List[Int] = List(1, 2, 3, 4)
scala> l exists (_>5)
res1: Boolean = false
scala> l exists (_<2)
res2: Boolean = true
scala> l exists (a => a<2 || a>5)
res3: Boolean = true
Other methods (some based on comments):
Counting matching elements
Count elements that satisfy predicate (and check if count > 0)
scala> (l count (_ < 3)) > 0
res4: Boolean = true
Returning first matching element
Find the first element that satisfies predicate (as suggested by Tomer Gabel and Luigi Plinge this should be more efficient because it returns as soon as it finds one element that satisfies the predicate, rather than traversing the whole List anyway)
scala> l find (_ < 3)
res5: Option[Int] = Some(1)
// also see if we found some element by
// checking if the returned Option has a value in it
scala> l.find(_ < 3) isDefined
res6: Boolean = true
Testing if exact value exists
For the simple case where we're actually only checking if one specific element is in the list
scala> l contains 2
res7: Boolean = true
The scala way would be to use exists:
collection.exists(item => condition1(item) && condition2(item))
And since java 8 you can use anyMatch:
collection.stream().anyMatch(item -> condition1(item) && condition2(item));
which is much better than a plain for or foreach.
Filter and exists keywords to get the matching values from Lists
val values = List(1,2,3,4,5,6,7,8,9,10,....,1000) //List -1
val factors= List(5,7) // List -2
//To get the factors of List-2 from List-1
values .filter(a => factors.exists(b => a%b == 0)) //Apply different logic for our convenience
Given code helps to get the matching values from 2 different lists
How do you replace an element by index with an immutable List.
E.g.
val list = 1 :: 2 ::3 :: 4 :: List()
list.replace(2, 5)
If you want to replace index 2, then
list.updated(2,5) // Gives 1 :: 2 :: 5 :: 4 :: Nil
If you want to find every place where there's a 2 and put a 5 in instead,
list.map { case 2 => 5; case x => x } // 1 :: 5 :: 3 :: 4 :: Nil
In both cases, you're not really "replacing", you're returning a new list that has a different element(s) at that (those) position(s).
In addition to what has been said before, you can use patch function that replaces sub-sequences of a sequence:
scala> val list = List(1, 2, 3, 4)
list: List[Int] = List(1, 2, 3, 4)
scala> list.patch(2, Seq(5), 1) // replaces one element of the initial sequence
res0: List[Int] = List(1, 2, 5, 4)
scala> list.patch(2, Seq(5), 2) // replaces two elements of the initial sequence
res1: List[Int] = List(1, 2, 5)
scala> list.patch(2, Seq(5), 0) // adds a new element
res2: List[Int] = List(1, 2, 5, 3, 4)
You can use list.updated(2,5) (which is a method on Seq).
It's probably better to use a scala.collection.immutable.Vector for this purpose, becuase updates on Vector take (I think) constant time.
You can use map to generate a new list , like this :
# list
res20: List[Int] = List(1, 2, 3, 4, 4, 5, 4)
# list.map(e => if(e==4) 0 else e)
res21: List[Int] = List(1, 2, 3, 0, 0, 5, 0)
It can also be achieved using patch function as
scala> var l = List(11,20,24,31,35)
l: List[Int] = List(11, 20, 24, 31, 35)
scala> l.patch(2,List(27),1)
res35: List[Int] = List(11, 20, 27, 31, 35)
where 2 is the position where we are looking to add the value, List(27) is the value we are adding to the list and 1 is the number of elements to be replaced from the original list.
If you do a lot of such replacements, it is better to use a muttable class or Array.
following is a simple example of String replacement in scala List, you can do similar for other types of data
scala> val original: List[String] = List("a","b")
original: List[String] = List(a, b)
scala> val replace = original.map(x => if(x.equals("a")) "c" else x)
replace: List[String] = List(c, b)