How do I get a list of two random numbers 6 times? This doesn't work: ```sequence<Int> { yield...

# getting-started

[dave08]

How do I get a list of two random numbers 6 times? This doesn't work:

sequence<Int> { yield(Random.nextInt(0..15)) }
    .chunked(2)
    .take(6)

[dave08]

My goal is something like listOf(listOf(0, 3), listOf(5, 15), ...) with six inner lists

[Sam]

List(6) { List(2) { Random.nextInt() } }

[Sam]

Your example doesn't work because the sequence will only yield once. I think you could fix it by adding an infinite loop.

[Vampire]

Or use generateSequence

[Casey Brooks]

sequence<Int> { yield(Random.nextInt(0..15)) } only ever emits a single element, and then quits. Sequences aren’t automatic generator functions, you’ll need to do the iteration within sequence { } yourself.

Try this, with manual iteration: https://pl.kotl.in/R_efr6krJ
Or with `generateSequence`: https://pl.kotl.in/tudLVa1wA

[Vampire]

generateSequence { Random.nextInt(15) }
    .chunked(2)
    .take(6)

[dave08]

Thanks, that was a bit confusing at first, now it's clear!

[CLOVIS]

sequence {
    repeat(6) {
        yield(listOf(Random.nextInt(15), Random.nextInt(15)))
    }
}

is more readable IMO

[Vampire]

More readable than

sequence {
    repeat(6) {
        yield(List(2) { Random.nextInt(15) })
    }
}

or

List(6) { List(2) { Random.nextInt(15) } }

? o_O

[CLOVIS]

Ahah no, I meant than the generateSequence one. Indeed that double List factory one is great

[CLOVIS]

It's not lazy, but it doesn't look like that was the question anyway

[Vampire]

Yeah, just wanted to stay near to original code to show how to do it properly that way in case it was oversimplified. 🙂

[dave08]

Yeah there are a few chained operators after that sequence generation, so a list would just recreate another list for each operation... That's why maybe the sequence version might be better here... Thanks all for the great ideas 😁, different techniques are always useful for different contexts