algorithmic modeling for Rhino
Hi there !
I hit a roadblock in my definition.
I have a tree containing lists of integers.
I need to merge the lists that contain at least one identical integer.
Since the number of lists can vary depending on upstream components, I don't see how I can use the components in the "Sets" tab to reach my goal...
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Does sound complicated. To be sure I understand, what you have is:
{44, 9}, {42, 5}, {4, 42}, {41, 11}, {0, 10}, {9, 6}, {12, 14}, {45, 25}, {42, 21}, {20, 43}, {41, 27}, {16, 26}, {25, 22}, {28, 30}, {36, 46}, {38, 49}, {19, 3}, {17, 1}
and what you're after is:
{44, 9, 9, 6}, {42, 5, 4, 42, 42, 21}, {41, 11, 41, 27}, {0, 10}, {12, 14}, {45, 25, 25, 22}, {20, 43}, {16, 26}, {28, 30}, {36, 46}, {38, 49}, {19, 3}, {17, 1}
correct?
Hi David !
Yes, that's what I want.
A bit of background : I'm using the "Topology" component on closed polysurfaces to perform sheet metal flattening.
I need to find the lists of faces which are tangentially connected ; these are generally the "extrados" and "intrados" of my bent sheet metal parts, excluding all the sides.
In the case "01", I can join all the faces that are tangentially connected, and I will end up with two polysurfaces which I can then unfold.
In the case "02", I have a set of "edge" faces which are also tangentially connected, and after "join", it will join outer and inner skin together, and I can't unfold that.
I really need to get a tree with lists of tangentially connected faces.
There might be a simpler way to achieve this, but still, I thought it would be possible to use a couple of "Set" tools to process my tree and get the right lists.
Cheers,
Thanks David !
You can hit the peanuts now :)
Actually, no :(
Kick your friends out and back to work ;)
In the attached example, in the merged tree, there are two separate branches containing common items...
Making the merged tree go through a second C# component provides the expected result though...
Oh I finally see. The 47 doesn't get added to Branch[1] until after Branch[39] has already decided that it has nothing in common with Branch[1].
Tricky, a whole other approach may be needed or some recursion needs to be added.
Yep ! it seems to work.
At least it produces the same (expected) result as passing the tree two time through the first version of the component.
Thanks David !
So you only want to merge branches that have a path like {0;?} and perform a completely different merge on branches that look like {1:?}
Because the script right now doesn't take that into account at all, it just iterates over all branches in the tree, regardless of path overlap.
Anyway, I sent you my definition.
The "Topology" component seemed to be the elegant solution here, but if I can't process the information it gives me, I'm stuck.
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