How do I infinitely repeat a sequence in Kotlin?

Question

I want to infinitely repeat T elements in a Sequence<T>. This can't be done using kotlin.collections.asSequence. For example:

val intArray = intArrayOf(1, 2, 3)
val finiteIntSequence = intArray.asSequence()
val many = 10
finiteIntSequence.take(many).forEach(::print)
// 123

This is not what I want. I expected some kind of kotlin.collections.repeat function to exist, but there isn't, so I implemented one myself (e.g. for this IntArray):

var i = 0
val infiniteIntSequence = generateSequence { intArray[i++ % intArray.size] }
infiniteIntSequence.take(many).forEach(::print)
// 1231231231

This is quite imperative, so I feel there must be a more functional and less verbose way to do this. If it exists, what is/are Kotlin's standard way(s) to repeat collections / arrays a(n) (in)finite amount of times?

Erik · Accepted Answer · 2018-12-04 06:26:57Z

22

Update: coroutines are no longer experimental as of Kotlin 1.3! Use them as much as you like :)

If you allow the use of coroutines you can do this in a pretty clean way using sequence:

an infinite amount of times

fun <T> Sequence<T>.repeat() = sequence { while (true) yieldAll(this@repeat) }

Note the use of a qualified this expression this@repeat - simply using this would refer to the lambda's receiver, a SequenceScope.

then you can do

val intArray = intArrayOf(1, 2, 3)
val finiteIntSequence = intArray.asSequence()
val infiniteIntSequence = finiteIntSequence.repeat()

println(infiniteIntSequence.take(10).toList())
// ^ [1, 2, 3, 1, 2, 3, 1, 2, 3, 1]

a finite amount of times

fun <T> Sequence<T>.repeat(n: Int) = sequence { repeat(n) { yieldAll(this@repeat) } }

edited Dec 4, 2018 at 6:26

Erik

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answered Dec 28, 2017 at 12:31

Salem

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4 Comments

voddan Over a year ago

Or repeat(n) { yieldAll(...) }

Salem Over a year ago

@voddan Thanks, this is definitely cleaner and arguably more readable :)

Erik Over a year ago

I'm accepting this as coroutines are the way to go since their stable release, although they do require an extra dependency. Edit: not sure if an extra dependency is needed, though, because sequence comes from here: kotlinlang.org/api/latest/jvm/stdlib/kotlin.sequences/…

Salem Over a year ago

@Erik yep, sequence builders are part of the standard library now. IIRC kotlinx.coroutines mostly contains utilities (such as await extensions for various types of futures).

Erik · Accepted Answer · 2017-12-30 16:16:00Z

To avoid using the experimental coroutines, use:

generateSequence { setOf("foo", 'b', 'a', 'r') }
  .flatten() // Put the Iterables' contents into one Sequence
  .take(5) // Take 5 elements
  .joinToString(", ")

// Result: "foo, b, a, r, foo"

or alternatively, if you want to repeat the entire collection a number of times, just take before flattening:

generateSequence { setOf("foo", 'b', 'a', 'r') }
  .take(5) // Take the entire Iterable 5 times
  .flatten() // Put the Iterables' contents into one Sequence
  .joinToString(", ")

// Result: "foo, b, a, r, foo, b, a, r, foo, b, a, r, foo, b, a, r, foo, b, a, r"

For the original question's IntArray, the array first must be converted to an Iterable<Int> (otherwise flatten() isn't available):

val intArray = intArrayOf(1, 2, 3)

generateSequence { intArray.asIterable() }
  .flatten()
  .take(10)
  .joinToString(", ")

// Result: "1, 2, 3, 1, 2, 3, 1, 2, 3, 1"

Furthermore, other types of Array, e.g. ByteArray or LongArray, as well as Map are not Iterable, but they all implement the asIterable() method like IntArray in the example above.

s1m0nw1 · Accepted Answer · 2017-12-28 12:28:37Z

3

I think this is pretty clear:

generateSequence(0) { (it + 1) % intArray.size }
        .map { intArray[it] }
        .forEach { println(it) }

answered Dec 28, 2017 at 12:28

s1m0nw1

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4 Comments

Erik Over a year ago

Your answer can be used to generate a sequence of Int values, but I'm interested in repeating an arbitrary collection or array of any type, e.g. Collection<Person>. By default, there is no Person.plus(other: Number), so this is not the answer I'm looking for. This is what I intended to say with the first sentence of my question: I want to repeat a collection or array of any type T infinitely.

Erik Over a year ago

Ah, I'm sorry I misinterpreted your answer! You're right, this is generic and doesn't require coroutines, which is a great alternative for the Mango's answer. Thanks, have my upvote.

Erik Over a year ago

Still, I think the Kotlin standard lib should contain some kind of repeat() and repeat(times: Int) methods.

s1m0nw1 Over a year ago

Added another more generic answer

suhtai · Accepted Answer · 2019-05-09 07:45:14Z

3

I'm unsure if this is due to API changes in Kotlin, but it's possible to do the following:

fun <T> Sequence<T>.repeatForever() =
  generateSequence(this) { it }.flatten()

Live example: https://pl.kotl.in/W-h1dnCFx

edited May 9, 2019 at 7:45

answered May 8, 2019 at 22:08

suhtai

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4 Comments

Erik Over a year ago

If I'm not mistaken this is basically a less verbose version of this answer, but using an extension and using flatMap instead of flatten. Example: pl.kotl.in/EbpEdBT6W

suhtai Over a year ago

Good point, I forgot about flatten, I'll update. However, using this creates a GeneratorSequence instead of relying on an infinite loop and coroutines for it which I'm not sure about efficiency but at the very least, having a while(true) sounds some alarms on my head (probably unfounded though)

Erik Over a year ago

Note that I referred to a different answer than the one containing infinite loops!

Erik Over a year ago

Also note that using infinite loops in sequence generation where the loop contains yield or yieldAll is not necessarily alarming, because these functions are suspending and will therefore only loop forever if polled forever. Unless the infinite loops contain other 'dangerous' (e.g. computationally heavy) code, they are safe to use.

s1m0nw1 · Accepted Answer · 2017-12-28 14:31:57Z

2

A generic solution would be to reuse the proposal from this answer with extension functions:

fun <T> Array<T>.asRepeatedSequence() =
        generateSequence(0) {
            (it + 1) % this.size
        }.map(::get)

fun <T> List<T>.asRepeatedSequence() =
        generateSequence(0) {
            (it + 1) % this.size
        }.map(::get)

Called like this:

intArray.asRepeatedSequence().forEach(::println)

edited Dec 28, 2017 at 14:31

answered Dec 28, 2017 at 13:50

s1m0nw1

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Comments

Daniel K. · Accepted Answer · 2022-09-21 12:39:08Z

0

If you happen to have Guava on your classpath, you can do the following:

val intArray = intArrayOf(1, 2, 3)
val cyclingSequence = Iterators.cycle(intArray.asList()).asSequence()

// prints 1,2,3,1,2,3,1,2,3,1
println(cyclingSequence.take(10).joinToString(","))

answered Sep 21, 2022 at 12:39

Daniel K.

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