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dont get you, i am saying sleect numbers in range 1 to 10, starting from 1 with a step of 2.
1 to 10 by 3 = 1 4 7 10
1 to 10 by 5 = 1 6
1 to 10 by 1 = 1 to 10 = 1 2 3 4 5 6 7 8 9 10
Added by Steve Lewis at 3:15pm on November 11, 2013
hen the each element of the corresponding X branch list will be repeated 6 times >> basically the output list would have the same structure of the input Xlist but instead of having only one item per branch I would have 6 items per branch >>
{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.
I made a quick example of my problem, here attached.
…
nts me this:
[[0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7]]
this is what I wanted but how to convert this to tree in grasshopper?
In grasshopper I just get:
8x IronPython.Runtime.List…
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
points 0, X-1, (2*x)-1, (3*X)-1, (4*X)-1, (5*X)-1 and then
1, X, (2*x), (3*X), (4*X), (5*X)
2, X+1, (2*x)+1, (3*X)+1, (4*X)+1, (5*X)+1
and so on till
5, X+4, (2*x)+4, (3*X)+4, (4*X)+4, (5*X)+4
How can I do this best?
Thanks,
Niels…