- C
{2;0} (N=61) - D
{2;1} (N=60) - E
{2;2} (N=61) - F
group 2:
{0;0} (N=10) - U
{0;1} (N=10) - V
{0;2} (N=10) - W
{0;3} (N=10) - X
{0;4} (N=10) - Y
{0;5} (N=10) - Z
the idea case is I can merge those date sets in a pattern of A-U-B-V-C-W-D-X-E-Y-F-Z...so on
therefore I am thinking how could I modify the path on group 2 and make them becomes things like:
{0;0} (N=10) - U
{0;1} (N=10) - V
{0;2} (N=10) - W
{1;0} (N=10) - X
{1;1} (N=10) - Y
{1;2} (N=10) - Z
but I have no idea how could I modify the path in that way....
can anyone show me how to?…
Added by Preston Chan at 8:34pm on October 26, 2010
in the desired order.
0 = 0
1 = 1
2 = 6
3 = 7
4 = 8
5 = 9
6 = 12
7 = 13
8 = 2
9 = 3
10 = 4
11 = 5
12 = 10
13 = 11
Where the first number is the index and the second number is the actual sorting key. Then you sort these keys while sorting your curves in parallel using the A input of the Sort component.
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
{2:2} {2:3}
Each branch has 10 points. I'd like all the points in {0} to draw lines to each sub-branch {0,#}, {1} to {1:#}, etc. I am simply confounded, how would this be accomplished?
Thanks!
…
ee 3)
{5}
0 15
{6}
0 16
And I want to place points at every possible combination of these coordinates, treating Tree 1 as X coordinates, Tree 2 as Y coordinates, and Tree 3 as Z coordinates. Also, I would like the list of points to be a tree with paths corresponding to the coordinates. Wouldn't it be nice if I could plug these trees into a Point XYZ, with a new "branch cross reference" method, and get the following result?
{0:3:5}
0 {10.0, 13.0, 15.0}
{0:3:6}
0 {10.0, 13.0, 16.0}
{0:4:5}
0 {10.0, 14.0, 15.0}
{0:4:6}
0 {10.0, 14.0, 16.0}
{1:3:5}
0 {11.0, 13.0, 15.0}
{1:3:6}
0 {11.0, 13.0, 16.0}
{1:4:5}
0 {11.0, 14.0, 15.0}
{1:4:6}
0 {11.0, 14.0, 16.0}
{2:3:5}
0 {12.0, 13.0, 15.0}
{2:3:6}
0 {12.0, 13.0, 16.0}
{2:4:5}
0 {12.0, 14.0, 15.0}
{2:4:6}
0 {12.0, 14.0, 16.0}
In this form of cross referencing, every combination of individual branches from the different lists is used as separate input, and the output for each combination is put onto a branch in the result whose path is the concatenation of the input branch paths used.…
Added by Andy Edwards at 7:03pm on November 3, 2009
ards to the number before the start number...
i.e. 9, 0, 1, 2, 3, 4, 5, 6, 7, 8
then it will need to repeat this pattern (continuing to count upwards) and the repeat number is based on a slider (for example 3 in the case illustrated below):
9, 0, 1, 2, 3, 4, 5, 6, 7, 8
19, 10, 11, 12, 13, 14, 15, 16, 17, 18,
29, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29,
if anyone has any ideas on how to create this series it would be great
M.…
. In this new branches list ,{2} contain the same items as {0} ,{3} contains the same items as {1} .
I hope I have made myself clear enough .Thanks for any help!!
…
he same order of the list. for example i have a list with 4 different lenght of curve like this:
0= 10
1= 12
2= 8 (minimum)
3= 17 (maximum)
and wont to make a ranking that the longest curve gets the value 4 and the smallest the value 1, like this
0= 2
1= 3
2= 1
3= 4
i tried the sort list function, but it dosn`t work
can anybody help me!
thx a lot…
number of divisions on that curve as in the defintion (i.e. by 4). The offset in the def is slightly different and should cull two or three more curves as in the lists that show my aim below.
Basically I want to look into each branch of the groups of points from each closed curve . Marking in a list whether it contains a one or a zero (0= outside 1 = coincidents).
{0;0}0. 21. 22. 23. 2 {0;1} 0. 01. 22. 03. 2 {0;2}0. 01. 02. 03. 0 {0;3}0. 21. 22. 23. 2 {0;4}0. 21. 22. 23. 2 {0;5}0. 21. 22. 23. 2 {0;6}0. 01. 22. 23. 1 {0;7}0. 21. 22. 03. 0 {0;8}0. 21. 22. 23. 2 {0;9}0. 21. 22. 23. 2 {0;10}0. 21. 22. 23. 2 {0;11}0. 21. 22. 23. 2 {0;12}0. 21. 22. 23. 2 {0;13}0. 01. 22. 23. 0 {0;14}0. 21. 22. 23. 2
I want to create a list from these points. That marks each curve that pokes out, in a cull pattern as such:
20022210222202
Using a 1 where there are co-incidents in the curve points and the boundary. A 2 for true (outside points) and a 0 for containment. So I might be able to use the 1 in future developments - however if a true false list is easiest I can live with that.
So could I use F(x) function? - to look for 0 or 1's in each bunch of points and thus list as such for a cull pattern? or will Path mapper help me here? Or can I rely on simply grafting and splitting??
I am usure of the neatest solution and would love to learn. Hope you can direct me.rgrds
J.…