the one-but-last list [4]. After running out of the n- items avalaible it should continue with the second item of list 0 and so on for all items on all the lists.
Intput, six lists of 30 items each
[0] (n=30)
[1] (n=30)
[2] (n=30)
[3] (n=30)
[4] (n=30)
[5] (n=30)
Output, 18 lists of 10 items each
[0],i=0;[5],i=4; [4],i=7;...
[0],i=1;[5],i=5; [4],i=8;...
...
[5],i=1;[4],i=5; [3],i=0;...
I thought perhaps the weave component or the relative tree item component but didn't manage to figure out how to compose the mask. I couldn't find much on how to use these. I guess it should wrap the lists, but not the items.
Any help would be greatly appreciated.…
Added by Thorsten Lang at 2:27am on January 24, 2011
ggle A
7. Toggle A
8. Toggle 2
9. Toggle 3
10. Toggle A
11. Toggle A
12. Toggle 3
I was thinking to use somehow slider and animate option....but without luck
Any idea would be appreciated…
cture, Rhino treats them as a single flat list. For example a surface can have 10 rows and 6 columns of control-points, resulting in a list of 60 points.
But 10 times 6 isn't the only way to get to 60. If you want to make a surface out of a list of 60 points, you'll also have to tell Rhino how those 60 points should be interpreted in terms of a grid. It could be 2*30, 3*20, 4*15, 5*12, 6*10, and all of the aforementioned products the other way around.
Sometimes there's only one way for a number of points to fit into a rectangular grid. For example if you provide 49 points, then 7*7 is the only way to make it work, but these cases are rare so we always demand you give us all the information required to actually make a rectangular grid of control-points from a linear collection.
As for "Why is it, sometimes we need to attach additional value into it?", this is usually because when you divide a domain or a curve into N segments, you end up with N+1 points. For example take the domain {0 to 5}, and divide it into 5 equal subdomains. You end up with {0 to 1}, {1 to 2}, {2 to 3}, {3 to 4} and {4 to 5}. However there are six numbers that mark the transitions between these domains 0, 1, 2, 3, 4 and 5. This is why you often have to add 1 to the UCount, because the number that controls the UCount often results in N+1 actual points.…
Added by David Rutten at 8:30am on December 25, 2014
I extract the first two with a "Redim Preserve t(1)" command.
In the first case, the redim is correct, Line 7 = Line 2 and Line 8 = Line 3. It just kept the first two values like it is supposed to be.
But, for the second curve starting Line 9, some t values are messed up after the Redim. Line 16 = Line 17 despite Line 11 was different from Line 12. That's what is creating a problem later in the Split.
Weird.
…
53 → 53 → 63 → 74 → 74 → 84 → 9
As you can see from the above list the connection sequence comes in waves of three, where each group of similar indices on the left is associated with a group of three incrementing indices on the right.
Some combination of Series components will probably generate this list, but it'll only work for the first ring, the second one will need a different connection pattern. It is perhaps better to just encode the integer pairs by hand. But then you cannot change your mind about the number of sides later.…
Added by David Rutten at 10:39am on October 21, 2015
shift. I realize I can use 'replace branch' but I do not have an available mask to utilize. I have simplified the problem to its simplest form so my question is understandable, however, the tree I am trying perform this operation on is a much larger 3 digit path address.
{1;3;2}
{2;3;4}
{3;5;4}
{4;3;7}
Change the above list to the list below.
{0;3;2}
{1;3;4}
{2;5:4}
{3;3;7}
I wish for a more robust arsenal of branch manipulation components. Most of the things I need to do are possible with the existing components, however, many operations take several components to perform even simple manipulations. Since branch/path manipulation is so integral to using GH successfully, it seems the GH community would be well served by enhancing the available path manipulation components.
Thanks,
Stan
…
a follow up question... how do I wrap a list onto itself at a certain frequency?
i.e. I want the list {1;2;3;4;5;6;7;8;9}
to become {1,4,7; 2,6,8; 3,6,9} wrapped every 3rd item
Added by Joshua Jordan at 5:30pm on November 17, 2012