If you mean you don't want to call it twice at each iteration:
function build_vec(n,x) =
n== 0 ? [] // assume you meant ==
: let(fx=f(x))
concat([[n,fx], build_vec(n-1,fx)]); // assume close square bracket on
the right side
If you mean you want to call it once over all then
function build_vec(n,fx)
and pass f(x) in as an argument. You can use a caller function to calculate
f(x).
I may have inserted the brackets in the wrong place to make your code valid.
As I did above it makes ever deeper nested vectors. If what you actually
want is a table all at the same nesting depth, you might be able to make it
using a list comprehension.
GeoffHorton wrote
Sorry for the vague title, but I couldn't come up with a better one.
I have a recursive function like this:
function build_vec(n, x) =
n = 0
? []
: concat([[n, f(x)], build_vec(n - 1, f(x)));
where f(x) is a multi-level iteration over vectors that takes a long time
to
run. Life would be a lot more efficient if there were a way only to call
f(x) once, but I can't figure out how to do it. Is there a way?
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If you mean you don't want to call it twice at each iteration:
function build_vec(n,x) =
n== 0 ? [] // assume you meant ==
: let(fx=f(x))
concat([[n,fx], build_vec(n-1,fx)]); // assume close square bracket on
the right side
If you mean you want to call it once over all then
function build_vec(n,fx)
and pass f(x) in as an argument. You can use a caller function to calculate
f(x).
I may have inserted the brackets in the wrong place to make your code valid.
As I did above it makes ever deeper nested vectors. If what you actually
want is a table all at the same nesting depth, you might be able to make it
using a list comprehension.
GeoffHorton wrote
> Sorry for the vague title, but I couldn't come up with a better one.
>
> I have a recursive function like this:
>
> function build_vec(n, x) =
> n = 0
> ? []
> : concat([[n, f(x)], build_vec(n - 1, f(x)));
>
> where f(x) is a multi-level iteration over vectors that takes a long time
> to
> run. Life would be a lot more efficient if there were a way only to call
> f(x) once, but I can't figure out how to do it. Is there a way?
>
>
>
> --
> Sent from: http://forum.openscad.org/
> _______________________________________________
> OpenSCAD mailing list
> To unsubscribe send an email to
> discuss-leave@.openscad
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