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.
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David Rutten
david@mcneel.com
Poprad, Slovakia…
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…
lues. What I want to do is combine them so that the structure would be something like:
{4;0}
{4;1}
{4;2}
{4;3}
{5;0}
{5;1}
{5;2}
{5;3}
I tried the method here, but it didn't give me what I wanted, it was just tacking the new values onto the end, and not maintaining their paths. Any help would be appreciated. Thanks!…
Added by Dennis Goff at 8:13am on February 10, 2016
etc.
Group 2 - 1, 6, 11, 16, 21 etc.
Group 3 - 2, 7, 12, 17, 22 etc.
Group 4 - 3, 8, 13, 18, 23 etc.
Group 5 - 4, 9, 14, 19, 24 etc. "
except in data, the branches start at 0, so 'group 1' is branch 0
as for the order of your points, that depends on the input prior sorting...
yrs …