s 8, 4, 2, 10, 1, 3, 8, 4, 2, 0. But then for the end result to maintain all numbers above 5 but replace all numbers below with a defined number..Let's say zero. So then the list would read...8, 0, 0, 10, 0, 0, 8, 0, 0.…
another example could be:
index 3 value 6
index 4 value 6
index 5 value 6
flipped and branched:
branch 6 index 0 value 3
branch 6 index 1 value 4
branch 6 index 2 value 5
Added by Ante Ljubas at 12:50pm on October 22, 2010
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
…
owing:
{0}0. d1. e2. f
3. g4. h5. i
{1}0. a1. b2. c
3. g4. h5. i
{2}0. a1. b2. c
3. e
4. f
5. g
Thought maybe I could use relative Item but I cant figure out how to do an offset that includes multiples.
…