Partition the Path?

[Asitaka]

In short, I wonder if there’s a way to partition the list with desire numbers?

ex.
(0;0) > (0;0;0) (25;50;1) (50;100;2)
(0;1) > (0;1;0) (25;51;1) (50;101;2)
(1;2) > (1;2;0) (26;52;1) (51;102;2)
(1;3) > (1;3;0) (26;53;1) (51;103;2)
(2;4) > (2;4;0) (27;54;1) (52;104;2)
.
.
(74;149) > (24;49;0) (49;99;1) (75;149;2)

Long story is, I try to do Random Reduce at the end but as the images showing below I need each 1/3 circle regions to have the even amount of the reducing. In other words my current definitions only allow me to randomly reduce from the total amount of the geometries despite the location Note that 1/3 and 150 just numbers for temporary place holder.

Thanks in advance for any replies.

[DavidRutten]

You can measure the angle from each point to the centre of the disc (Vector Angle and be sure to include the plane). Then you generate three domains and figure out in which domain each angle sits using Find Domain. You can then use the found domain index to Sift your data into three outputs.

If you’re working in degrees, the domains you need are (0 \rightarrow 120), (120 \rightarrow 240) and (240 \rightarrow 360). In radians, the domains are (0 \rightarrow \frac{2}{3}\pi), (\frac{2}{3}\pi \rightarrow \frac{4}{3}\pi) and (\frac{4}{3}\pi \rightarrow 2\pi).

[anikolo (Aris)]

David, is this what you’re talking about?
equalReduce.gh (7.2 KB)

[DavidRutten]

Not what I said, but I think a better solution. It doesn’t break up the points into three different sets.

Here’s what I meant: domaincategory.gh (13.2 KB)

[anikolo (Aris)]

pure, condensed knowledge