algorithmic modeling for Rhino

Hello, I wanted to switch them number of paths and indices of each path branch by using path mapper.


I have a series of points which have 10 path and each path is composed by 5 points.


When i connect path mapper directly from point param, path mapper parameter turned to red color which has an error.


I do not understand why.


Does anyone know how to switch path and number of indices?




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Hi Seung

The problem is you need the same number of letters as numbers in the path and then to pick the letter with set of changing numbers in the other side. So if it says.. {0;0;1;0} the input into the mapper would be {A;B;C;D} and as your list of numbers changes in the 3rd row the output letter would be (C)
so would look like this.... {A;B;C;D} (i) > {i} (C)

Hi Matt.
The Path Mapper component is something i haven't been able to tackle yet.
Here's an image trying to follow the logic of your comment.
But not working yet

Hi Mario

I understand what you have tried to do, i guess what i was calling paths/branches has not been understood. when i speak of adding the same number of letters as numbers in paths/branches i ment not the number of paths in the list i.e in your case 4 but the number in each path/branch so for yours it would be {0;1} so 2 numbers thus.. 2 letters {A:B} say. the second half of what you have is fine. so should work.

P.s This thread should help you both better understand the path mapper component. See first reply to question.

hi mario

so first if you want to use pathmapper you have to understand the data tree logic
so as you read in your param viewer you have a data tree structure that has 4 path's
where each path has a unique name like in your example {0;0},{0;1},{0;2},{0;3} where each path contains 11 elements(curves,points,values,etc)
so if you want to restructure the structure first right click on path mapper and take create null mapping which reads your current path structure into the component
after that you can additional write on the left side to the existing path (i) as variable for the elements and then on the right side type in the structure you want to achieve
maybe {A;i}(B) so after that you have 11 paths where each have 3 elements
hope that helps
Hey Matt.
Thanks for the quick reply!!!
So when you talk about "paths/branches" i should be looking at the panel component not the parameter viewer.
Thanks for the link thread, i have looked at it already, but your explanation was clearer to me
BTW, I still don't get why sometimes you can use
{A;B;C;D} (i) > {i} (C)
and sometimes
{A;B;C;D} (i) > {C} (i)
Thanks again!
Hi mario
No no you sould be looking at the parameter viewer just not at the top number that reads the number of paths in the list but at each path..

got it!
Thanks again to]
I'll try that!
Hi, Matt

I have another question.

I want to connect each points to create triangle.

What I made two seperate data tree.

One tree has points which will be bottom line of trangle.

The other one has middle points.

How can I connect and create triangle?

hi seung so if you want to have triangles for all of your points just flatten your points and use mesh/delaunay edges
is you want t construct it on your own you can use a series in a series

I want to add some more.

I guess sincedata tree has different degree, it would not work by using line command.

One has two degree ({A,0})
The other one has three degree({A,0,0})

Do you know how to simplify?

I used simplify but it only delete one in front.

Like that {0,0,A} {A} - simplify command.






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