printing elements in list using stream - scala

Question

Why does the following code prints only 1 and not the rest of the list elements?
scala> val l: List [Int] = List(1,2,3)
l: List[Int] = List(1, 2, 3)
scala> l.toStream.map(x => print(x))
1res3: scala.collection.immutable.Stream[Unit] = Stream((), ?)
What is the correct way to write this code?

Answer 1

I'll divide my answer to two:
1. The map method in Scala:
you're using map, which expects a function with no side-effects (printing is a side effect). What you're looking for is:
l.toStream.foreach(x => print(x))
Basically, the general idea is that map takes something and converts it to something else (for example, increasing its value). while foreach is performing some action on that value that isn't supposed to have a return value.
scala> l.toStream.foreach(x => print(x))
123
2. Stream in Scala:
Streams are lazy, so Scala only computes the values it needs. Try this:
scala> l.toStream.map(x => x+1)
res2: scala.collection.immutable.Stream[Int] = Stream(2, ?)
You can see it computed the first value, and the question marks states that it has no idea what comes after it, because it didn't compute it yet. In you're example the first value is nothing, as the print returns no value.

Answer 2

Stream is on demand data structure which means not all the values will be evaluated until you need them.
example,
scala> val stream = (1 to 10000).toStream
stream: scala.collection.immutable.Stream[Int] = Stream(1, ?)
Now if you access head and tail, stream will be evaluated upto 2nd index.
scala> stream.head
res13: Int = 1
scala> stream.tail
res14: scala.collection.immutable.Stream[Int] = Stream(2, ?)
scala> stream
res15: scala.collection.immutable.Stream[Int] = Stream(1, 2, ?)
If you access index 99,
scala> stream(99)
res16: Int = 100
Now if you print stream, stream will be evaluated upto 99th index,
scala> stream
res17: scala.collection.immutable.Stream[Int] = Stream(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, ?)
It's always good to process only those in stream, which you need. you can use take() for that.
scala> stream.take(50).foreach(x => print(x + " "))
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
So, answer to your question can be,
scala> List(1,2,3).toStream.take(3).foreach(x => print(x + " "))
1 2 3
Reference
https://www.coursera.org/learn/progfun2/home/week/2

Answer 3

to print complete stream use
l.toStream.print
Output: 1, 2, 3, empty
to print first n values, you may use take(n)
l.toStream.take(2).print
prints output: 1, 2, empty

Answer 4

You can print it with
l.toStream.foreach(println)
But generally speaking is not a good idea trying to print or even processing without being careful a whole Stream since it may be infinite and cause an error while doing so.
More info about Streams here

Answer 5

Streams in Scala are lazy data structures which means that they tend to perform only the as needed work.
scala> val stream1 = Stream.range(1, 10)
// stream1: scala.collection.immutable.Stream[Int] = Stream(1, ?)
In this case, only the first element is computed. The stream knows how to compute rest of the elements and will compute them only when it actually needs them. for example ("consumers"),
scala> val list1 = stream1.toList
// list1: List[Int] = List(1, 2, 3, 4, 5, 6, 7, 8, 9)
scala> stream1.foreach(print)
// 123456789
But when faced with "transformers", Streams will just keep the new transformation with them but will not apply to whole Stream. The map method is supposed to provide transforms
scala> val stream2 = stream1.map(i => i + 5)
// stream2: scala.collection.immutable.Stream[Int] = Stream(6, ?)
So, it just know that it has to apply this i => i + 5 function to respective elements of stream1 to get element of stream2. And will do that when required (facing any consumer).
Lets consider something similar to your example,
scala> val stream3 = stream1.map(i => println("element :: " + i))
// element :: 1
// stream3: scala.collection.immutable.Stream[Unit] = Stream((), ?)
Here your "transform" function takes an element Int, prints it in line and returns nothing which is called Unit or () in Scala. Out lazy stream here, will compute this transform for first element and will not do for rest. And this computation here will result in that element :: 1 being printed.
Now, lets see what happens when we apply some consumer to it,
scala> val list3 = stream3.toList
// element :: 2
// element :: 3
// element :: 4
// element :: 5
// element :: 6
// element :: 7
// element :: 8
// element :: 9
// list3: List[Unit] = List((), (), (), (), (), (), (), (), ())
Which will look wrong to most people. All I wanted to convert my stream to list but why are all these lines getting printed.
Which is why, when you are using map, you should provide a pure function.
What is a pure function? The simple answer is that a pure function only does the things it is supposed to do and nothing else. It does not cause any change out of its scope. It just takes something and give something else back.
All of the following are pure functions,
scala> val pf1 = (i: Int) => i + 1
// pf1: Int => Int = $$Lambda$1485/1538411140#6fdc53db
scala> val pf2 = (i: Int) => {
| val x = 100
| val xy = 200
| xy + i
| }
// pf2: Int => Int = $$Lambda$1487/14070792#7bf770ba
scala> val pf3 = (i: Int) => ()
// pf3: Int => Unit = $$Lambda$1486/1145379385#336cd7d5
Where as following is not a pure function,
val npf1 = (i: Int) => println(i)
// npf1: Int => Unit = $$Lambda$1488/1736134005#7ac97ba6
Because it causes a "magical" change in the environment.
Why "magical"? Because, it claims to be a function of type Int => Unit which means it should just be transforming an Int to an Unit. But it also printed something on our console, which was outside of its environment.
A real world example of this magic will be that - whenever you put a bread in your toaster it causes a rain storm on the Hulk's current location. And nobody wants the Hulk to come looking for their toaster.

Answer 6

In general, the bottom line is, that you should not use side effects in .map. When you do foo.map(bar) that just returns another collection, that contains element, generated by applying bar to the original collection. It may or may not be lazy. The point is, you should treat the elements of any collection as undefined until something looks at them.
If you want side effects, use foreach: Seq(1,2,3).toStream.foreach(println)