[Closed] mini-challenge #5

You’ll have to define some logic, won’t you? Is 7 bigger than red? What is your intended output?

Agreed, what’s the sample output? Should the original values be kept or is the goal a sorted array of the extracted numbers, types unchanged (int, float etc.)? In that case, here’s the obvious solution (I guess the hint is they are all iterables but I can’t think of any universal approach):

mapped fn flatten arr buffer = case of
(
	(isKindOf arr Number) : append buffer arr
	(isKindOf arr BitArray) : join buffer arr
	(isKindOf arr Array) : flatten arr buffer
	(isKindOf arr Color) : join buffer #(arr.r, arr.g, arr.b, arr.a)
	(isProperty arr #x) :
	(
		join buffer #(arr.x, arr.y)
		if isProperty arr #z do append buffer arr.z 
		if isProperty arr #w do append buffer arr.w
	)
)

fn sortNumbers arr buffer:#() =
(
	flatten arr buffer
	makeUniqueArray (sort buffer)
)

sortNumbers #(7, point3 7 1.6 12, red, x_axis, #{87..100}, [1,17], #(#(#(1, "Hi!"))))

so from #(red, 7) you want to get #(0,7,255)?

I honestly have no idea what you’re looking to achieve. You’ll need to post sample output and explain it.

the final output for the sample is

 #(0.0, 1.0, 1.6, 7, 12.0, 17.0, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100)
 

red, and “Hi!” were wiped because one doesn’t have get method, and another returns not number

 red[1] -- not works
 "Hi!"[1] == "H" -- not number
 

for example (matrix3 1) is good:

 (matrix3 1)[1] == [1,0,0] -- OK
 ([1,0,0])[1] == 1.0 -- OK   
 --so 
 (matrix3 1) => 1.0,0.0,0.0,0.0,1.0,0.0,0.0,0.0,1.0,0.0,0.0,0.0 => #(0.0,1.0)
 

where has this task came from …
there are several methods that returns vertices as Point3, or bitarray, or array, or single index… you can handle every case individually but… maybe there is a universal solution

I’m still confused as to why you would need to sort this type of array in the first place?