is there a list building method that can reference previousl kotlinlang #getting-started

is there a list building method that can reference previously constructed elements of the list while building it that i'm not seeing? Or does the collection not exist until after all its elements are determined, so that's not really possible in as a List constructor

Luke

05/10/2019, 8:55 PM

If you start with a collection or some sort of iterable, maybe you can use

.fold

. It uses an accumulator that is passed in the lambda

Hullaballoonatic

05/10/2019, 9:30 PM

yeah, i was thinking fold, but wouldn't that be crazy expensive?

oh, well i could fold into a mutable list

karelpeeters

05/10/2019, 9:34 PM

Or you know, sometimes a for loop is fine as well.

Hullaballoonatic

05/10/2019, 9:34 PM

yeah, i just figured i could make it all pretty

i think if i did it as a sequence then i could pull elements from early parts of the list while i built it

this feels clever but obnoxious...

val (u: List<Vector>, e: List<Vector>) = cols.foldIndexed(ArrayList<Vector>(width) to ArrayList<Vector>(width)) { k, (u, e), a ->
            u[k] = when(k) {
                0 -> a
                1 -> a - a * e[0] * e[0]
                else -> a - e.map { a * it * it }.sum()
            }
            e[k] = u[k] / u[k].magnitude
            u to e
        }

it's a translation of this:

i think i can simplify it all actually

karelpeeters

05/10/2019, 9:55 PM

I think a for loop with a sum in it is way clearer.

Hullaballoonatic

05/10/2019, 9:55 PM

yep

you're right

karelpeeters

05/10/2019, 9:57 PM

val U = mutableListOf<Vector>()
for (a in A)
    U += (a - U.sumBy { it * a * a }).unit

Hullaballoonatic

05/10/2019, 9:58 PM

i'll make it even better actually. i'll give you the result

when i'm done

karelpeeters

05/10/2019, 9:59 PM

Now I'm curious 😒imple_smile:

Hullaballoonatic

05/10/2019, 10:39 PM

val e = arrayListOf<Vector>()
        cols.forEachIndexed { k, a ->
            e[k] = when (k) {
                0 -> a
                1 -> a - a * e[0] * e[0]
                else -> a - e.map { a * it * it }.sum()
            }
        }

karelpeeters

05/10/2019, 10:40 PM

You don't need special cases for 0 and 1 though, right?

Hullaballoonatic

05/10/2019, 10:41 PM

i don't know why it pastes all skewed like that

karelpeeters

05/10/2019, 10:41 PM

Because you didn't select the indent of the first line.

Hullaballoonatic

05/10/2019, 10:41 PM

but yeah, you're right that i don't have to, although it might speed it up a tad, and also it helps for clarity

ah, you're absolutely correct

anyways, u only exists to define e

in the algorithm, so i cut out the middleman

karelpeeters

05/10/2019, 10:42 PM

It's the other way around no?

Hullaballoonatic

05/10/2019, 10:43 PM

that's what it seems to be reading top to bottom

but this is qr decomposition, and neither q nor r are defined by u

only e

karelpeeters

05/10/2019, 10:44 PM

Ah I thought you just wanted to do Gram–Schmidt.

Hullaballoonatic

05/10/2019, 10:44 PM

this is the gram-schmidt qr decomp algorithm

i just posted the excerpt concerning the block of code i wrote for context

karelpeeters

05/10/2019, 10:44 PM

I see.

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