Code:
object Permutations extends App
{
val ar=Array(1,2,3).combinations(2).foreach(println(_))
}
Output:
[I#378fd1ac
[I#49097b5d
[I#6e2c634b
I am trying to execute this but I am getting some other values.
How to print array values in Scala? Can any one help to print?
Use mkString
object Permutations extends App {
Array(1,2,3).combinations(2).foreach(x => println(x.mkString(", ")))
}
Scala REPL
scala> Array(1,2,3).combinations(2).foreach(x => println(x.mkString(", ")))
1, 2
1, 3
2, 3
When array instance is directly used for inside println. The toString method of array gets called and results in output like [I#49097b5d. So, use mkString for converting array instance to string.
Scala REPL
scala> println(Array(1, 2))
[I#2aadeb31
scala> Array(1, 2).mkString
res12: String = 12
scala> Array(1, 2).mkString(" ")
res13: String = 1 2
scala>
You can't print array directly, If you will try to print it will print the reference of that array.
You are almost there, Just iterate over array of array and then on individual array and display the elements like below
Array(1,2,3).combinations(2).foreach(_.foreach(println))
Or Just convert each array to string and display like below
Array(1,2,3).combinations(2).foreach(x=>println(x.mkString(" ")))
I hope this will help you
Related
This question already has answers here:
Printing array in Scala
(7 answers)
Closed 6 years ago.
I am very new to scala programming , hence asking this question
My scala program is below
object FirstMain{
def main(args: Array[String]): Unit={
val myArray = Array(1,2,3,4)
val k =myArray.map(x => x*10)
println(k)
}
}
Why I am getting output as [I#14991ad
I want the array elements to be read and want each element to be multiplied with 10 and print the result
Also I would like to know what is the return type of map() in scala
Array is a Scala's way of representing Java arrays. Invoking Array.toString() method (which you do implicitly in println(k)) invokes toString method defined in Java's Object class (which by default prints hexadecimal representation of the hash code of the object). You should try using Scala's List type, which overrides toString() method and outputs pretty contents of the list.
val list = List(1, 2, 3, 4)
println(list) // Prints: `List(1, 2, 3, 4)`
You can also easily convert an Array to a List:
val list = Array(1, 2, 3, 4).toList
For your second question: mapping an Array returns another Array.
map builds a new collection by applying a function to all elements of this array.
To print an Array you can do the following :
val myArray = Array(1, 2, 3, 4)
// myArray: Array[Int] = Array(1, 2, 3, 4)
val k = myArray.map(x => x * 10)
// k: Array[Int] = Array(10, 20, 30, 40)
println(k.mkString("[", ",", "]"))
// [10,20,30,40]
Hence, you can read here what does actually [I#14991ad means. (Array of Integer with hashcode of the Array object 14991ad)
Array.map will return another Array
println(k) <=> println(String.valueOf(k)) <=> println(k.toString)
the Array does not override the method of toString
if you want to print your array like this Array(10, 20, 30, 40), you can use the following code
println(k.deep)
deep not in Scala 2.13 see SO answer about Array deep documentation.
Good luck with you
Something like:
val arr : Array[Array[Double]] = new Array(featureSize)
sc.parallelize(arr, 100).saveAsTextFile(args(1))
Then Spark will store data type into HDFS.
Array in Scala exactly corresponds to Java Arrays - in particular, it's a mutable type, and its toString method will return a reference to the Array. When you save this RDD as textFile, it's invoking toString method on each element of the RDD and therefore giving you gibberish. If you want to output actual elements of the Array, you first have to stringify the Array, for example by applying mkString(",") method to each array. Example from Spark shell:
scala> Array(1,2,3).toString
res11: String = [I#31cba915
scala> Array(1,2,3).mkString(",")
res12: String = 1,2,3
For double arrays:
scala> sc.parallelize(Array( Array(1,2,3), Array(4,5,6), Array(7,8,9) )).collect.mkString("\n")
res15: String =
[I#41ff41b0
[I#5d31aba9
[I#67fd140b
scala> sc.parallelize(Array( Array(1,2,3), Array(4,5,6), Array(7,8,9) ).map(_.mkString(","))).collect.mkString("\n")
res16: String =
1,2,3
4,5,6
7,8,9
So, your code should be:
sc.parallelize(arr.map(_.mkString(",")), 100).saveAsTextFile(args(1))
or
sc.parallelize(arr), 100).map(_.mkString(",")).saveAsTextFile(args(1))
When I try to print the contents of my RDD, it prints something like displayed below, how can I print the content?
Thanks!
scala> lines
res15: org.apache.spark.rdd.RDD[Array[String]] = MapPartitionsRDD[3] at filter at <console>:23
scala> lines.take(5).foreach(println)
[Ljava.lang.String;#6d3db5d1
[Ljava.lang.String;#6e6be45e
[Ljava.lang.String;#6d5e0ff4
[Ljava.lang.String;#3a699444
[Ljava.lang.String;#69851a51
This is because it uses the toString implementation for the given object. In this case Array prints out the type and hash. If you convert it to a List then it will be a prettier output due to List's toString implementation
scala>println(Array("foo"))
[Ljava.lang.String;HASH
scala>println(Array("foo").toList)
List(foo)
Depending on how you want to print them out you can change your line that prints the elements to:
scala> lines.take(5).foreach(indvArray => indvArray.foreach(println))
I have two lists that I zip and go through the zipped result and call a function. This function returns a List of Strings as response. I now want to collect all the responses that I get and I do not want to have some sort of buffer that would collect the responses for each iteration.
seq1.zip(seq2).foreach((x: (Obj1, Obj1)) => {
callMethod(x._1, x._2) // This method returns a Seq of String when called
}
What I want to avoid is to create a ListBuffer and keep collecting it. Any clues to do it functionally?
Why not use map() to transform each input into a corresponding output ? Here's map() operating in a simple scenario:
scala> val l = List(1,2,3,4,5)
scala> l.map( x => x*2 )
res60: List[Int] = List(2, 4, 6, 8, 10)
so in your case it would look something like:
seq1.zip(seq2).map((x: (Obj1, Obj1)) => callMethod(x._1, x._2))
Given that your function returns a Seq of Strings, you could use flatMap() to flatten the results into one sequence.
How in the world do you get just an element at index i from the List in scala?
I tried get(i), and [i] - nothing works. Googling only returns how to "find" an element in the list. But I already know the index of the element!
Here is the code that does not compile:
def buildTree(data: List[Data2D]):Node ={
if(data.length == 1){
var point:Data2D = data[0] //Nope - does not work
}
return null
}
Looking at the List api does not help, as my eyes just cross.
Use parentheses:
data(2)
But you don't really want to do that with lists very often, since linked lists take time to traverse. If you want to index into a collection, use Vector (immutable) or ArrayBuffer (mutable) or possibly Array (which is just a Java array, except again you index into it with (i) instead of [i]).
Safer is to use lift so you can extract the value if it exists and fail gracefully if it does not.
data.lift(2)
This will return None if the list isn't long enough to provide that element, and Some(value) if it is.
scala> val l = List("a", "b", "c")
scala> l.lift(1)
Some("b")
scala> l.lift(5)
None
Whenever you're performing an operation that may fail in this way it's great to use an Option and get the type system to help make sure you are handling the case where the element doesn't exist.
Explanation:
This works because List's apply (which sugars to just parentheses, e.g. l(index)) is like a partial function that is defined wherever the list has an element. The List.lift method turns the partial apply function (a function that is only defined for some inputs) into a normal function (defined for any input) by basically wrapping the result in an Option.
Why parentheses?
Here is the quote from the book programming in scala.
Another important idea illustrated by this example will give you insight into why arrays are accessed with parentheses in Scala. Scala has fewer special cases than Java. Arrays are simply instances of classes like any other class in Scala. When you apply parentheses surrounding one or more values to a variable, Scala will transform the code into an invocation of a method named apply on that variable. So greetStrings(i) gets transformed into greetStrings.apply(i). Thus accessing an element of an array in Scala is simply a method call like any other. This principle is not restricted to arrays: any application of an object to some arguments in parentheses will be transformed to an apply method call. Of course this will compile only if that type of object actually defines an apply method. So it's not a special case; it's a general rule.
Here are a few examples how to pull certain element (first elem in this case) using functional programming style.
// Create a multdimension Array
scala> val a = Array.ofDim[String](2, 3)
a: Array[Array[String]] = Array(Array(null, null, null), Array(null, null, null))
scala> a(0) = Array("1","2","3")
scala> a(1) = Array("4", "5", "6")
scala> a
Array[Array[String]] = Array(Array(1, 2, 3), Array(4, 5, 6))
// 1. paratheses
scala> a.map(_(0))
Array[String] = Array(1, 4)
// 2. apply
scala> a.map(_.apply(0))
Array[String] = Array(1, 4)
// 3. function literal
scala> a.map(a => a(0))
Array[String] = Array(1, 4)
// 4. lift
scala> a.map(_.lift(0))
Array[Option[String]] = Array(Some(1), Some(4))
// 5. head or last
scala> a.map(_.head)
Array[String] = Array(1, 4)
Please use parentheses () to access the list of elements, as shown below.
list_name(index)