ep is to understan the logics of what you want to do, in your case, build 4 point surfaces (u also need to know the right direction to build the surfaces). Then you can write an hipotetic list (by hand in a paper) of what you want. In your case the list was (0, 1, 3, 2) (2, 3, 5, 4) (4, 5, 7, 6), etc... if you can imagine building 2 lists, each one with the sequences (0, 2, 4, 6, etcc) and (1, 3, 5, 7, etc..) then you can manage with shift and graft to finally have four lists. A( 0 1 2 3 ...) B (1 3 5 etc..) C(3 5 7 etc..) D (2 4 6 etc..). And to achieve the 2 first lists, you need to get the odd and the pair numbers. The cull pattern does that amazingy well. With a pattern True-False you get de pair numbers, and with the False-True pattern you get de odd numbers.
Hope it was clear enough…
Added by Pep Tornabell at 5:32am on November 19, 2009
For example.
If you have two lists of points.
List A List B
{0;0;0}(0) {0;0}(0)
{0;0;1}(0) {0;1}(0)
{0;2}(0)
{0;3}(0)
{0;4}(0)
And you want to merge the two lists so that the two points in list A are the end points.
Merge Lists Results:
{0;0}(0)
{0;0;0}(0)
{0;0;1}(0)
{0;1}(0)
{0;2}(0)
{0;3}(0)
{0;4}(0)
Because of their path structures the order is wrong from a simple merge so Flattening now is out of the question.
Path Mapper
{A;B} --> {A;B+1}
{A;B;C} --> {A;C*6}
---------------------
Results:
{0;0} --> {0;0+1} = {0;1}
{0;1} --> {0;1+1} = {0;2}
{0;2} --> {0;2+1} = {0;3}
{0;3} --> {0;3+1} = {0;4}
{0;4} --> {0;4+1} = {0;5}
{0;0;0} --> {0;0*6} = {0;0}
{0;0;1} --> {0;1*6} = {0;6}
Now with the Path Structures similar when they are re-ordered the results will have the two points of list A as the end points.
Question 2
why did the curve-line intersection lose the path structure? Both trees had 38 branches.
Both trees had 38 Paths but Tree A had more Items, 147 compared to 38 in Tree B.
So you get this happening:
{0;0;0;0;0;0}(0) compared to {0;0;0;0}(0) results: Null {0;0;0;0;0;0}(0)
Base result paths on longest
{0;0;1;0;0;0}(0) compared to {0;0;0;1}(0) results: Null {0;0;1;0;0;0}(0)
{0;0;2;0;0;0}(0) compared to {0;0;0;2}(0) results: Yes {0;0;2;0;0;0;0}(0)
Add a branch to contain result
{0;0;3;0;0;0}(0) compared to {0;0;0;3}(0) results: Yes {0;0;3;0;0;0;0}(0)
{0;0;3;0;0;0}(1) compared to {0;0;0;3}(0) results: No {0;0;3;0;0;0;1}(0)
{0;0;4;0;0;0}(0) compared to {0;0;0;4}(0) results: Yes {0;0;4;0;0;0;0}(0)
{0;0;4;0;0;0}(1) compared to {0;0;0;4}(0) results: Yes {0;0;4;0;0;0;1}(0)
{0;0;5;0;0;0}(0) compared to {0;0;0;5}(0) results: Yes {0;0;5;0;0;0;0}(0)
{0;0;5;0;0;0}(1) compared to {0;0;0;5}(0) results: Yes {0;0;5;0;0;0;1}(0)
{0;0;5;0;0;0}(2) compared to {0;0;0;5}(0) results: Yes {0;0;5;0;0;0;2}(0)
...... etc
…
s points"
(the set didnt change from the picture above)
so param viewers show me (with 3 columns points)
1 list with 3 branches : {0;0}/{0;1}/{0;2}
an other with 4 branches : {0;0;0}/{0;0;1}/{0;0;2}/{0;0;3}
I tried to merge those 2 and then input the list in a path mapper but could'nt manage to make it work to get :
{0;0;0}/{0;0}/{0;0;1}/{0;1}/{0;0;2}/{0;2}/{0;0;3}
Am I using the right component?…
ok, very similar to what i am looking for, but the thing is i have a list not just like (0,n), its like (0;0)...(0;3),(1;0)...(1;3), (2;0)...(2;3), etc
se
4 False
(0;1)
0 False
1 False2 False
3 True
4 False
into a branch taking all the True values at their original coresponding lines:(0;0)
0 False
1 True2 False
3 True
4 False
I hope it's understandable what I'm asking, thanks!…
Added by Martin Oravec at 2:42am on January 15, 2017