Grasshopper

algorithmic modeling for Rhino

problem with: expected IEnumerable[LineCurve], got LineCurve

Hi everyone, I think I have a little problem with something I thought very easy but unfortunately I cannot find a way to overcome it.

I did a component in python in gh to get out of a surface tesselation the points of each panel. Using .Branch(i) I can, in a loop, retrieve the element from the tree I have input. 

My problem is when I want to OUTPUT a tree, I got stuck with the ArgymentTypeException telling me that the tree type expected and what I am feeding into it are not consistent: 

Runtime error (ArgumentTypeException): expected IEnumerable[LineCurve], got LineCurve

So I can manage to export the elements flatten into a list but not as a tree. How can I do in this scenario (and in general with different types of geometries)? What is IEnumerable about?

Thanks a Lot!


V.

This is the script:

import rhinoscriptsyntax as rs
import clr
clr.AddReference("Grasshopper")
from Grasshopper.Kernel.Data import GH_Path
from Grasshopper import DataTree
import Rhino


b = []

tree = DataTree[Rhino.Geometry.LineCurve]()


for i in range(0,y):
    a = (x.Branch(i))
    branchList = x.Branch(i)
    branchPath = x.Path(i)
    line = rs.AddLine(a[0],a[2])
    b.append(line)

    tree.AddRange(rs.coercecurve(line),branchPath)

Views: 1005

Replies to This Discussion

Permalink Reply by Mostapha Sadeghipour Roudsari on May 6, 2013 at 1:28pm

Hi Vincenzo.

You should use AddRange when you want to add a list.

In this case just use Add to add a single line.

tree.Add(rs.coercecurve(line),branchPath) should work.

-Mostapha

Permalink Reply by Vincenzo Reale on May 6, 2013 at 1:46pm

Thank you a lot Mostapha!!!

Another question then, what if I would access / create a tree which has element in more than one level (like branches out of branches).

Is there a x.Branch(i)(j) command?

Or x.Branch(i).Branch(j)?

If it is not clear I'll try to make an example.

Permalink Reply by Mostapha Sadeghipour Roudsari on May 6, 2013 at 1:57pm

You can use (i,j). Something like this should work:

import clr

clr.AddReference("Grasshopper")

from Grasshopper.Kernel.Data import GH_Path

from Grasshopper import DataTree

from System import Object

a = DataTree[Object]()

branches = range(10)
paths = range(5)

for i in branches:

    for j in paths:
        p = GH_Path(i,j)
        a.Add(i, p)

RSS

About

Scott Davidson created this Ning Network.

Translate

Search

Photos

  • Add Photos
  • View All

Videos

  • Add Videos
  • View All

© 2024   Created by Scott Davidson.   Powered by

Badges  |  Report an Issue  |  Terms of Service