Is the a better functional idiom alternative to the code below? ie Is there a neater way to get the value j without having to use a var?
var j = i + 1
while (j < idxs.length && idxs(j) == x) j += 1
val j = idxs.drop(i).indexWhere(_ != x) + i
Or, as suggested by #kosii in the comments, use the indexWhere overload that takes an index from where to start searching:
val j = idxs.indexWhere(_ != x, i)
Edit
Since j must equal the length of idxs in case all items following i are equal to x:
val index = idxs.indexWhere(_ != x, i)
val j = if(index < 0) idxs.length else index
// or
val j = if (idxs.drop(i).forall(_ == x)) idxs.length
else idxs.indexWhere(_ != x, i)
Maybe with streams, something like:
((i + 1) to idxs.length).toStream.takeWhile(j => idxs(j) == x).last
Related
I've been working on the scala recursion problem. I used to develop the program using loops and then use the concept of recursion to convert the existing loop problem in a recursive solution.
So I have written the following code to find the perfect number using loops.
def isPerfect(n: Int): Boolean = {
var sum = 1
// Find all divisors and add them
var i = 2
while ( {
i * i <= n
}) {
if (n % i == 0) if (i * i != n) sum = sum + i + n / i
else sum = sum + i
i += 1
}
// If sum of divisors is equal to
// n, then n is a perfect number
if (sum == n && n != 1) return true
false
}
Here is my attempt to convert it into a recursive solution. But I'm getting the incorrect result.
def isPerfect(n: Int): Boolean = {
var sum = 1
// Find all divisors and add them
var i = 2
def loop(i:Int, n:Int): Any ={
if(n%i == 0) if (i * i != n) return sum + i + n / i
else
return loop(i+1, sum+i)
}
val sum_ = loop(2, n)
// If sum of divisors is equal to
// n, then n is a perfect number
if (sum_ == n && n != 1) return true
false
}
Thank you in advance.
Here is a tail-recursive solution
def isPerfectNumber(n: Int): Boolean = {
#tailrec def loop(d: Int, acc: List[Int]): List[Int] = {
if (d == 1) 1 :: acc
else if (n % d == 0) loop(d - 1, d :: acc)
else loop(d - 1, acc)
}
loop(n-1, Nil).sum == n
}
As a side-note, functions that have side-effects such as state mutation scoped locally are still considered pure functions as long as the mutation is not visible externally, hence having while loops in such functions might be acceptable.
Below is an example of code which will generate Correlation Matrix but I need to add column header and row header in front and top of matrix.
For example in the above matrix amber coloured objects are the labels which i need to add to the blue coloured data generated by Correlation matrix whose code i have attached below.
In Scala breeze is there a way to add labels to matrix ? The problem is DenseMatrix is Double and labels are character so i am not able to add any char label to matrix object.
def getCorMatrix(c :String,d :String,n :Int) :breeze.linalg.DenseMatrix[Double] = {
CorMatrixlogger.info("Inside generating Correlation Matrix")
val query = MongoDBObject("RunDate" -> d) ++ ("Country" -> c)
CorMatrixlogger.info("Query Object created for {}", c)
val dbObject = for (d <- price.find(query)) yield(d)
val objectReader = (dbObject map {x => objectRead(x)}).toList
val fetchAssetData = objectReader map {x => x.Symbol} map { x=> assetStats(x,n) } filterNot {x => x.length < n-1}
CorMatrixlogger.info("Asset Data fetched")
val excessReturnMatrix = DenseMatrix((for(i <- fetchAssetData) yield i.excessReturn).map(_.toArray):_*)
CorMatrixlogger.info("Excess Return matrix generated")
val transposeExcessreturnMatrix = excessReturnMatrix.t
val vcvMatrix = breeze.numerics.rint(((excessReturnMatrix * transposeExcessreturnMatrix):/ (n-1).toDouble ) :* 1000000.0) :/ 1000000.0
CorMatrixlogger.info("VcV Matrix Generated")
val transposeStDevVector = DenseMatrix(for (i <- fetchAssetData ) yield i.sigma)
val stDevVector = transposeStDevVector.t
val stDevMatrix = breeze.numerics.rint(( stDevVector * transposeStDevVector) :* 1000000.0) :/ 1000000.0
CorMatrixlogger.info("Correlation Matrix Generated")
lowerTriangular(breeze.numerics.rint((vcvMatrix :/ stDevMatrix) :* 10000.0) :/ 10000.0)
}
Edit
Thanks David. Your solution really worked well for me.
val ma = DenseMatrix((1.0,2.0,3.0), (3.0,4.0,5.0),(6.0,7.0,8.0))
val im = DenseMatrix.tabulate(ma.rows,ma.cols)(ma(_,_).toString)
val head = DenseVector("a","b","c")
val thead = head.t
val withHeader:DenseMatrix[String] = DenseMatrix.tabulate(im.rows+1, im.cols+1) { (i, j) =>
if (i == 0 && j == 0) " "
else if (i == 0) head(j -1)
else if (j == 0 ) thead (i -1)
else im(i-1,j-1)
} //> withHeader : breeze.linalg.DenseMatrix[String] = a b c
//| a 1.0 2.0 3.0
//| b 3.0 4.0 5.0
//| c 6.0 7.0 8.0
There's nothing built in, sadly. You could do something like
val withHeader:DenseMatrix[Any] = DenseMatrix.tabulate(n+1, m+1){ (i, j) =>
if (i == 0 && j == 0) ""
else if (i == 0) colHeaders(j - 1)
else if (j == 0) rowHeaders(i - 1)
else orig(i - 1, j - 1)
}
You lose all typing information that way, of course, but if you just need to toString something, it's probably the quickest way in current Breeze.
I have just started learning Scala and sideways I am doing some algorithms also. Below is an implementation of merge sort in Scala. I know it isn't very "scala" in nature, and some might even reckon that I have tried to write java in scala. I am not totally familiar with scala, i just know some basic syntax and i keep googling if i need something more. So please give me some pointers on to what can i do in this code to make it more functional and in accord with scala conventions and best practices. Please dont just give correct/optimized code, i will like to do it myself. Any suggestions are welcomed !
def mergeSort(list: Array[Int]): Array[Int] = {
val len = list.length
if (len == 1) list
else {
var x, y = new Array[Int](len / 2)
val z = new Array[Int](len)
Array.copy(list, 0, x, 0, len / 2)
Array.copy(list, len / 2, y, 0, len / 2)
x = mergeSort(x)
y = mergeSort(y)
var i, j = 0
for (k <- 0 until len) {
if (j >= y.length || (i < x.length && x(i) < y(j))) {
z(k) = x(i)
i = i + 1
} else {
z(k) = y(j)
j = j + 1
}
}
z
}
}
[EDIT]
This code works fine and I have assumed for now that input array will always be of even length.
UPDATE
Removed vars x and y
def mergeSort(list: Array[Int]): Array[Int] = {
val len = list.length
if (len == 1) list
else {
val z = new Array[Int](len)
val x = mergeSort(list.dropRight(len/2))
val y = mergeSort(list.drop(len/2))
var i, j = 0
for (k <- 0 until len) {
if (j >= y.length || (i < x.length && x(i) < y(j))) {
z(k) = x(i)
i = i + 1
} else {
z(k) = y(j)
j = j + 1
}
}
z
}
}
Removing the var x,y = ... would be a good start to being functional. Prefer immutability to mutable datasets.
HINT: a method swap that takes two values and returns them ordered using a predicate
Also consider removing the for loop(or comprehension).
I am trying to swap every pair of values in my array using for and yield and so far I am very unsuccessful. What I have tried is as follows:
val a = Array(1,2,3,4,5) //What I want is Array(2,1,4,3,5)
for(i<-0 until (a.length-1,2),r<- Array(i+1,i)) yield r
The above given snippet returns the vector 2,1,4,3(and the 5 is omitted)
Can somebody point out what I am doing wrong here and how to get the correct reversal using for and yields?
Thanks
a.grouped(2).flatMap(_.reverse).toArray
or if you need for/yield (much less concise in this case, and in fact expands to the same code):
(for {b <- a.grouped(2); c <- b.reverse} yield c).toArray
It would be easier if you didin't use for/yield:
a.grouped(2)
.flatMap{
case Array(x,y) => Array(y,x)
case Array(x) => Array(x)
}.toArray // Array(2, 1, 4, 3, 5)
I don't know if the OP is reading Scala for the Impatient, but this was exercise 3.3 .
I like the map solution, but we're not on that chapter yet, so this is my ugly implementation using the required for/yield. You can probably move some yield logic into a guard/definition.
for( i <- 0 until(a.length,2); j <- (i+1).to(i,-1) if(j<a.length) ) yield a(j)
I'm a Java guy, so I've no confirmation of this assertion, but I'm curious what the overhead of the maps/grouping and iterators are. I suspect it all compiles down to the same Java byte code.
Another simple, for-yield solution:
def swapAdjacent(array: ArrayBuffer[Int]) = {
for (i <- 0 until array.length) yield (
if (i % 2 == 0)
if (i == array.length - 1) array(i) else array(i + 1)
else array(i - 1)
)
}
Here is my solution
def swapAdjacent(a: Array[Int]):Array[Int] =
(for(i <- 0 until a.length) yield
if (i%2==0 && (i+1)==a.length) a(i) //last element for odd length
else if (i%2==0) a(i+1)
else a(i-1)
).toArray
https://github.com/BasileDuPlessis/scala-for-the-impatient/blob/master/src/main/scala/com/basile/scala/ch03/Ex03.scala
If you are doing exercises 3.2 and 3.3 in Scala for the Impatient here are both my answers. They are the same with the logic moved around.
/** Excercise 3.2 */
for (i <- 0 until a.length if i % 2 == 1) {val t = a(i); a(i) = a(i-1); a(i-1) = t }
/** Excercise 3.3 */
for (i <- 0 until a.length) yield { if (i % 2 == 1) a(i-1) else if (i+1 <= a.length-1) a(i+1) else a(i) }
for (i <- 0 until arr.length-1 by 2) { val tmp = arr(i); arr(i) = arr(i+1); arr(i+1) = tmp }
I have started to learn Scala recently and all solutions from the book Scala for the Impatient (1st edition) are available at my github:
Chapter 2
https://gist.github.com/carloscaldas/51c01ccad9d86da8d96f1f40f7fecba7
Chapter 3
https://gist.github.com/carloscaldas/3361321306faf82e76c967559b5cea33
I have my solution, but without yield. Maybe someone will found it usefull.
def swap(x: Array[Int]): Array[Int] = {
for (i <- 0 until x.length-1 by 2){
var left = x(i)
x(i) = x(i+1)
x(i+1) = left
}
x
}
Assuming array is not empty, here you go:
val swapResult = for (ind <- arr1.indices) yield {
if (ind % 2 != 0) arr1(ind - 1)
else if (arr1(ind) == arr1.last) arr1(ind)
else if (ind % 2 == 0) arr1(ind + 1)
}
I have a 3d array defined like so:
val 3dArray = new Array[Array[Array[Int]]](512, 8, 8)
In Javascript I would do the following to assign each element to 1:
for (i = 0; i < 512; i++)
{
3dArray[i] = [];
for (j = 0; j < 8; j++)
{
3dArray[i][j] = [];
for (k = 0; k < 8; k++)
{
3dArray[i][j][k] = 1;
}
}
}
What's the most elegant approach to doing the same?
Not sure there's a particularly elegant way to do it, but here's one way (I use suffix s to indicate dimension, i.e. xss is a two-dimensional array).
val xsss = Array.ofDim[Int](512, 8, 8)
for (xss <- xsss; xs <- xss; i <- 0 until 8)
xs(i) = 1
Or, using transform it gets even shorter:
for (xss <- xsss; xs <- xss)
xs transform (_ => 1)
for {
i <- a.indices
j <- a(i).indices
k <- a(i)(j).indices
} a(i)(j)(k) = 1
or
for {
e <- a
ee <- e
i <- ee.indices
} ee(i) = 1
See: http://www.scala-lang.org/api/current/index.html#scala.Array$
You have Array.fill to initialize an array of 1 to 5 dimension to some given value, and Array.tabulate to initialize an array of 1 to 5 dimension given the current indexes:
scala> Array.fill(2,1,1)(42)
res1: Array[Array[Array[Int]]] = Array(Array(Array(42)), Array(Array(42)))
enter code here
scala> Array.tabulate(3,2,1){ (x,y,z) => x+y+z }
res2: Array[Array[Array[Int]]] = Array(Array(Array(0), Array(1)), Array(Array(1), Array(2)), Array(Array(2), Array(3)))