etting when I merge the three trees, but what I would like to get is:
essentially a tree with 27 branches, each with a single list of either 11 or 21 points.
{0} (N=11)
{1} (N=11)
...
{10} (N=21)
{11} (N=21)
...
{17} (N=11)
{18) (N=11)
{27} (N=11)
Any help would be greatly appreciated.
All the best,
Matt
…
Added by Matt Schmid at 3:06pm on December 4, 2010
5 8, and then the following values are obtain as the last one (8) plus 3, then this last one (11) plus 5, and then this last one (16) plus 8, and then it starts again: 24+3, 27+5, 32+8...
Thanks
…
Added by Jesus Galvez at 5:17am on November 27, 2012
nts me this:
[[0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7]]
this is what I wanted but how to convert this to tree in grasshopper?
In grasshopper I just get:
8x IronPython.Runtime.List…
to install parallels and Windows 7 (I've read to get Windows 7 for now not 8) and use my Rhino 4 (got this a while ago and haven't used it yet due to I'm on a mac). The macbook has 16gb RAM so I could allocate 8 to Windows side and load Rhino 4 or upgrade to Rhino 5 but want to be sure Grasshopper will run or I'll do the Bootcamp thing and sacrifice the restart every time I use Rhino/Grasshopper. I'm just starting this Grasshopper road so zero experience. First thing is get the system right (for me who uses mac.s). I've heard Rhino 4 and 5 will run okay on parallels with Windows 7.
Thanks…
)
3. KeyError(1417,)
4. KeyError(1417,)
5. KeyError(1417,)
6. KeyError(1417,)
7. KeyError(1417,)
8. KeyError(1417,)
9. KeyError(1417,)
10. KeyError(1417,)
11.......
i tried different weather file but also same result. it seems i have same problem. the file am working on is the radiation file i took from the examples . whats seems to be the problem?
thank you for your time…
grouped together. In this case, 5 layers... 7 curves for each.
I had thought FlipMatrix would work here... but it doesn't.
Suggestions? I could write my own contour object in this case by intersecting planes, but is there an easier option?
…
Added by Matt Lavoie at 2:16pm on February 25, 2015
e! I do not have good ideas today!
The end result of the list would be:
5, 10, 15, 20, 21, (21 + 5), (21 + 10), (21 + 15), (21 + 20), (21 + 21), (42 + 5), (42 + 10), (42 + 15), (42 + 20), (42 + 21), etc …
So the distance between each pair of closest points must be exactly 5, 7, or 15? How about the distance between a point and it's 2nd closest neighbour? 3rd closest? ...
Added by David Rutten at 9:41am on November 10, 2017