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
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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
where each branch contains all the points generated by dividing each curve, so if you divide into 10 segments, you'll get:
{0;0}(N = 11)
{0;1}(N = 11)
{0;2}(N = 11)
{0;3}(N = 11)
{0;4}(N = 11)
Where the second integer in the curly brackets refers back to the index of the curve in the original list.
Another way to look at this data is to see it as a table. It's got 5 rows (one for each original curve) and 11 columns, where every column contains a specific division point.
--
David Rutten
david@mcneel.com
Poprad, Slovakia…