rve
10 curve
11 curve
12 curve
13 curve
...and I'd like to rearrange the order in which the curve are listed, to something like this:
{0,0,0}
0 curve
1 curve
8 curve
9 curve
10 curve
11 curve
2 curve
3 curve
4 curve
5 curve
12 curve
13 curve
6 curve
7 curve
I hope this makes sense.
Thank in advance for any advice,
John…
vector * number
8. number * point
9. point * number
10. complex * complex
11. colour * colour
12. colour * number
13. number * colour
--
David Rutten
david@mcneel.com
Seattle, WA…
Added by David Rutten at 10:39pm on November 12, 2010
Tree:
{0;0;0} N = 2
{0;0;1} N = 1
{0;0;2} N = 3
{0;1;0} N = 5
{0;1;1} N = 8
{0;1;2} N = 10
If we apply the aforementioned mapping to this tree, we'll end up with the following result:
{0;0} N = 6
{0;1} N = 23
Basically {0;0;0}, {0;0;1} and {0;0;2} are combined into a single path {0;0} as we disregard the third index because "C" is no longer present in the target mapping.
Because we only use the Mapper to modify paths, we do not lose any data items, though we might lose some of the paths.
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 1:03pm on August 25, 2010
=1), {1,1}(N=1)...{1,15}(N=1). While Y is made of {0}(N=6), {1}(N=15).
What I need to do, is to cycle through all the branched items of the X lists with every single item of the corresponding list of Y.
So:
{0,0}(N=1) with 1st item of Y{0},
{0,1}(N=1) with 1st item of Y{0}
...
{0,15}(N=1) with 1st item of Y{0} (and this is the first cycle).
Then (cycle 2)
{0,0}(N=1) with 2nd item of Y{0},
{0,1}(N=1) with 2nd item of Y{0}
...
{0,15}(N=1) with 2nd item of Y{0}
Then do the same thing with the X list {1,0} .. {1,15} with the corresponding Y{1}
As a result I need the following list structure:
{0,0}N=6, {0,1}N=6, {0,2}N=6 ... {0,15}N=6,
{1,0}N=15, {1,1}N=15, {1,2}N=15 ... {0,15}N=15.
If I use the component you suggested "shift path" it will match the items per item-order instead of cycling through all the items of the first sublist...
Hope this makes sense.
Any suggestion?…
I have this :
list 3 : 0 1 2 3 4 5 6
list 2 : 0 1 2 3 4 5 6
list 1 : 0 1 2 3 4 5 6
list 0 : 0 1 2 3 4 5 6
and I want to group the points of index 0 in a branch, the points of index 1 in another branch and so on.
I attached a file in which I generated the points.
Thank you in advance for your help !
Regards
Red…
ents will do or which components will be available.
My problem arises because I want to obtain a list such as the following:
{{6, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6}, {5, 6, 5, 4, 3, 2, 1, 2, 3, 4, 5}, {4, 5, 6, 5, 4, 3, 2, 1, 2, 3, 4}, {3, 4, 5, 6, 5, 4, 3, 2, 1, 2, 3}, {2, 3, 4, 5, 6, 5, 4, 3, 2, 1, 2}, {1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1}}
Which displayed as a matrix is:
If it were possible to combine GH operations (series, shift list, replace string...) with matrices I think it would be quite powerful. A matrix to list component like those available on scientific calculators, would then translate the matrix to list.
For me, matrices come in handy when dealing with surface patterns.
…
Added by Jesus Galvez at 6:46am on November 26, 2012
se the cull pattern, so I wanted to make the pattern using a function component. x=y. x= the original list and y= the interval i wanted to remove. So the pattern should be:
0: false
1:false
2:false
3:false
4:true
5:true
6:true
7:true
8:false
9:false
10:false
etc...…
Added by Rasmus Holst at 3:32am on November 17, 2009
e
7. True
8. True <-- this one
9. True
10. False
11. True
12. False
13. True
14. True <-- this one
15. True
16. False
17. True
18. False
19. True
20. True <-- this one
21. True
22. False
23. True
24. False
25. True
26. True <-- this one
27. True
28. False
29. True
30. False
31. True
32. True <-- this one
33. True
Any idea how I can solve this?
Thanks!…