Kotlin's Iterable and Sequence look exactly same. Why are two types required?

KotlinIterableLazy Sequences

Kotlin Problem Overview


Both of these interfaces define only one method

public operator fun iterator(): Iterator<T>

Documentation says Sequence is meant to be lazy. But isn't Iterable lazy too (unless backed by a Collection)?

Kotlin Solutions


Solution 1 - Kotlin

The key difference lies in the semantics and the implementation of the stdlib extension functions for Iterable<T> and Sequence<T>.

  • For Sequence<T>, the extension functions perform lazily where possible, similarly to Java Streams intermediate operations. For example, Sequence<T>.map { ... } returns another Sequence<R> and does not actually process the items until a terminal operation like toList or fold is called.

    Consider this code:

      val seq = sequenceOf(1, 2)
      val seqMapped: Sequence<Int> = seq.map { print("$it "); it * it } // intermediate
      print("before sum ")
      val sum = seqMapped.sum() // terminal
    

    It prints:

      before sum 1 2
    

    Sequence<T> is intended for lazy usage and efficient pipelining when you want to reduce the work done in terminal operations as much as possible, same to Java Streams. However, laziness introduces some overhead, which is undesirable for common simple transformations of smaller collections and makes them less performant.

In general, there is no good way to determine when it is needed, so in Kotlin stdlib laziness is made explicit and extracted to the Sequence<T> interface to avoid using it on all the Iterables by default.

  • For Iterable<T>, on contrary, the extension functions with intermediate operation semantics work eagerly, process the items right away and return another Iterable. For example, Iterable<T>.map { ... } returns a List<R> with the mapping results in it.

    The equivalent code for Iterable:

      val lst = listOf(1, 2)
      val lstMapped: List<Int> = lst.map { print("$it "); it * it }
      print("before sum ")
      val sum = lstMapped.sum()
    

    This prints out:

      1 2 before sum
    

    As said above, Iterable<T> is non-lazy by default, and this solution shows itself well: in most cases it has good locality of reference thus taking advantage of CPU cache, prediction, prefetching etc. so that even multiple copying of a collection still works good enough and performs better in simple cases with small collections.

    If you need more control over the evaluation pipeline, there is an explicit conversion to a lazy sequence with Iterable<T>.asSequence() function.

Solution 2 - Kotlin

Completing hotkey's answer:

It is important to notice how Sequence and Iterable iterates throughout your elements:

Sequence example:

list.asSequence().filter { field ->
    Log.d("Filter", "filter")
    field.value > 0
}.map {
    Log.d("Map", "Map")
}.forEach {
    Log.d("Each", "Each")
}

Log result:

filter - Map - Each; filter - Map - Each

Iterable example:

list.filter { field ->
    Log.d("Filter", "filter")
    field.value > 0
}.map {
    Log.d("Map", "Map")
}.forEach {
    Log.d("Each", "Each")
}

filter - filter - Map - Map - Each - Each

Solution 3 - Kotlin

> Iterable is mapped to the java.lang.Iterable interface on the > JVM, and is implemented by commonly used collections, like List or > Set. The collection extension functions on these are evaluated > eagerly, which means they all immediately process all elements in > their input and return a new collection containing the result. > > Here’s a simple example of using the collection functions to get the > names of the first five people in a list whose age is at least 21: > > val people: List = getPeople() > val allowedEntrance = people > .filter { it.age >= 21 } > .map { it.name } > .take(5) > > Target platform: JVMRunning on kotlin v. 1.3.61 First, the age check > is done for every single Person in the list, with the result put in a > brand new list. Then, the mapping to their names is done for every > Person who remained after the filter operator, ending up in yet > another new list (this is now a List<String>). Finally, there’s one > last new list created to contain the first five elements of the > previous list. > > In contrast, Sequence is a new concept in Kotlin to represent a lazily > evaluated collection of values. The same collection extensions are > available for the Sequence interface, but these immediately return > Sequence instances that represent a processed state of the date, but > without actually processing any elements. To start processing, the > Sequence has to be terminated with a terminal operator, these are > basically a request to the Sequence to materialize the data it > represents in some concrete form. Examples include toList, toSet, > and sum, to mention just a few. When these are called, only the > minimum required number of elements will be processed to produce the > demanded result. > > Transforming an existing collection to a Sequence is pretty > straightfoward, you just need to use the asSequence extension. As > mentioned above, you also need to add a terminal operator, otherwise > the Sequence will never do any processing (again, lazy!). > > val people: List = getPeople() > val allowedEntrance = people.asSequence() > .filter { it.age >= 21 } > .map { it.name } > .take(5) > .toList() > > Target platform: JVMRunning on kotlin v. 1.3.61 In this case, the > Person instances in the Sequence are each checked for their age, if > they pass, they have their name extracted, and then added to the > result list. This is repeated for each person in the original list > until there are five people found. At this point, the toList function > returns a list, and the rest of the people in the Sequence are not > processed. > > There’s also something extra a Sequence is capable of: it can contain > an infinite number of items. With this in perspective, it makes sense > that operators work the way they do - an operator on an infinite > sequence could never return if it did its work eagerly. > > As an example, here’s a sequence that will generate as many powers of > 2 as required by its terminal operator (ignoring the fact that this > would quickly overflow): > > generateSequence(1) { n -> n * 2 } > .take(20) > .forEach(::println)

You can find more here.

Solution 4 - Kotlin

Iterable is good enough for most use cases, the way iteration is performed on them it works very well with caches because of the spatial locality. But the issue with them is that whole collection must pass through first intermediate operation before it moves to second and so on.

In sequence each item passes through the full pipeline before the next is handled.

This property can be determental to the performance of your code especially when iterating over large data set. so, if your terminal operation is very likely to terminate early then sequence should be preferred choice because you save by not performing unnecessary operations. for example

sequence.filter { getFilterPredicate() }   
        .map    { getTransformation()  } 
        .first  { getSelector() }

In above case if first item satisfies the filter predicate and after map transformation meets the selection criteria then filter, map and first are invoked only once.

In case of iterable whole collection must first be filtered then mapped and then first selection starts

Attributions

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionVenkata RajuView Question on Stackoverflow
Solution 1 - KotlinhotkeyView Answer on Stackoverflow
Solution 2 - KotlinLeandro Borges FerreiraView Answer on Stackoverflow
Solution 3 - KotlinSazzad Hissain KhanView Answer on Stackoverflow
Solution 4 - KotlinmightyWOZView Answer on Stackoverflow