t item (see the image), is it possible to do this in another way (quickly) ?
Is it possible to divide that curve into 2 separate curves using a point that i've used for the division?
Thanks…
Added by luca.pavarin at 4:08pm on January 7, 2010
ches it with the first branch in Tree B (and then the first branch in Tree C if more than two trees are involved).
I'm planning to add better branch matching logic, but I'm not going to touch it until I have a good idea about what's needed and how it can be accomplished without breaking existing files.
So, the branch "address" is only used to sort the branches in a single tree. Thus, a tree with the following branches is always sorted in the exact same way:
{0;0}
{0;1}
{0;2}
{0;3;0}
{0;3;1}
{1;6}
If you have another tree with different branches:
{0}
{1}
{2}
{3}
{4}
{5}
Then the matching will be:
{0;0} -> {0}
{0;1} -> {1}
{0;2} -> {2}
{0;3;0} -> {3}
{0;3;1} -> {4}
{1;6} -> {5}
As long as people adhere to your advice: "it is best for the addresses of each tree branch to be in the same format", there will be no problem. But it is at the moment extremely difficult to perform complex matchings.
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 9:25am on August 11, 2010
eep track of the path names yourself.
You can use the Replace Branches component to rename.
For instance:
(Paths = 2)
{0;1}
{0;2}
and
(Paths = 3)
{0;1} rename {0;3}
{0;2} rename {0;4}
{0;3} rename {0;5}
Does this make enough sense to get you started?
-taz…
}
X*
{0;2;0}{0;2;1}{0;2;2}{0;2;3}
{0;3;0}{0;3;1}{0;3;2}{0;3;3}
And now we want to insert at the place marked with X* another list formated like:
{0;0}{0;1}{0;2}{0;3}
So we want it resulted like:
{0;0;0}{0;0;1}{0;0;2}{0;0;3}
{0;1;0}{0;1;1}{0;1;2}{0;1;3}
*{0;2;0}{0;2;1}{0;2;2}{0;2;3}* - the inserted list
{0;3;0}{0;3;1}{0;3;2}{0;3;3}
{0;4;0}{0;4;1}{0;4;2}{0;4;3}
Since we plug the list formated with only {A;B} into the place with has {A;B;C} then that list has to be reformated in the same manner and every lists which are next to it have to be reformated too by adding 1 to B so it's {A;B+1;C}.
Param viewer with the data tree diagram is a great tool for visualising data structure. It seems to me that it would be easier to play with lists in the same way as we do with connecting components together. So if we have list of points and we want to insert them at some certain place in the tree then we don't need to play with Patch Mapper, Insert List and others but we just plug them on one go into the tree and format will adopt itself automaticly according to the choicen position on the tree.
Same with OUTPUT. We can pick some elements from the tree and connect it to the component which will receive every element from that branch. (example):There is a list of points with complex data structure. we pick node {0;0;3} and move it out from param viewer to connect it with point component which will receive all the sublists with elements which are under it:
{0;0;3;0}0 - pt1 - pt2 - pt{0;0;3;1}0 -pt1 - pt2 - pt3 - pt...and so on...
I don't know if this solution make better sense then other solutions of this case. Maybe there are easier ways to do it without such complication and I have no idea about it :)
greets!
Adam
…
4}
{0;2;0}
{0;2;1}
{0;2;2}
{0;2;3}
{0;2;4}
You cannot flip this because this is more complex than a rectangular matrix. You're going to have to do the mapping yourself. Try a Path Mapper with the following masks:
{A;B;C}(i) -> {A;B;i}(C)
Which should give you a structure that results in 3 lofts.
--
David Rutten
david@mcneel.com
Poprad, Slovakia
…
Added by David Rutten at 3:18pm on November 27, 2011