This might even be more obvious:
println ([[x:1, y:100], [x:2, y:1], [x: 2, y:500]].sort{[it.x,
it.y]}) // broken: [[x:2, y:1], [x:1, y:100], [x:2, y:500]]
println ([[x:1, y:100], [x:2, y:1], [x: 2, y:500]].sort{ a, b -> a.x
== b.x ? a.y <=> b.y : a.x <=> b.x }) // works
On Tue, May 21, 2024 at 12:52 AM Paul King <pa...@asert.com.au> wrote:
>
> If you have only two dimensions, you'll get away with your solution
> for small integer coordinate values. Here is a counter example with
> integers:
>
> println ([[x: 1, y: 69273666], [x: 69273666, y: 1]].sort{[it.x,
> it.y]}) // broken
> println ([[x: 1, y: 69273666], [x: 69273666, y: 1]].sort{ a, b ->
> a.x == b.x ? a.y <=> b.y : a.x <=> b.x }) // works
>
> On Tue, May 21, 2024 at 12:25 AM M.v.Gulik <mvgu...@gmail.com> wrote:
> >
> >
> > On Mon, 20 May 2024 at 15:40, Paul King <pa...@asert.com.au> wrote:
> >>
> >> What sort result are you trying to achieve? There should be no need to
> >> perform any recursion.
> >
> >
> > Main target was reordering some set of randomly ordered x,y coordinates
> > into (for example*) a x,y(left to right, top to bottom) order (*:where any
> > of the actual sort directions could be easily flipped).
> >
> > So far "*.sort{[it.y, it.x]}" seems the best and easiest way to do this
> > (despite its Float/Double caveat).
