branches in each A's list of B's, or remove its ends etcso that if I want to remove the last B in every A{0;1},{0;2},{0;3},{0;4},{0;5},{0;6}{1;1},{1;2},{1;3},{1;4}{2;1},{2;2},{2;3},{2;4},{2;5}would become{0;1},{0;2},{0;3},{0;4},{0;5}
{1;1},{1;2},{1;3}
{2;1},{2;2},{2;3},{2;4}I guess the question is do I need to figure out the cull pattern- each B may have different lengths...…
}
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
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Is it like this:
If a beam is connected from nod 0 to 1 and from 1 to 4. Another from 2 to 3 and from 3 to 5.
Node 1 and 3 have the same coordinates, but are they rigidly connected or not?