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
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.
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Added by Jesus Galvez at 6:46am on November 26, 2012
it,
[3] the upper limit, [4] the slider position.
What do you think?
Matt
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Added by Matt Trimble at 7:54pm on November 17, 2009
t 2^15 are being listed. Later I will use this info to cull;
all instances where important nodes are not present,
geometric unstable trusses using Grubler's criterion,
the remainder will go to FEM for further analysis,
I can use this method but i require to separate each binary boolean
so they are stored separately like so;
{0} (0) 0 (1) 0 (2) 0 (3) 0 (4) 0 . . . . . . . . (15) 0
{1} (0) 1 (1) 0 (2) 0 (3) 0 (4) 0 . . . . . . . . (15) 0
{2} (0) 0 (1) 1 (2) 0 (3) 0 (4) 0 . . . . . . . . (15) 0 (I know there must be an easier scripted way how to do it) or else is there a way to decompose them directly from the method you sent me using standard Grasshopper components?. Any method would work.
Best,
Kane…