0;3} (N = 2)
{0;0;0;4} (N = 2)
{0;0;1;0} (N = 2)
{0;0;1;1} (N = 2)
{0;0;1;2} (N = 2)
{0;0;1;3} (N = 2)
{0;0;1;4} (N = 2)
Flattening this structure using the Flatten component would result in:
{0} (N = 20)
However, using a Path Mapper with the following masks will flatten is somewhat more intelligently:
{A;B;C;D} -> {A;B;C}
Now, you get:
{0;0;0} (N = 10)
{0;0;1} (N = 10)
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 3:19am on December 14, 2009
of 400 interlocked rings in a 20 X 20 grid.
V1 - A single 'suLoop' component doing 400 'SUnion' operations (20 X 20): 11.6 minutes
V2 - Two phases: 5 X 10 in phase one and 2 X 4 in phase 2, 58 'SUnions' total: ~88 seconds combined
V3 - Two phases: 4 X 5 in phase one and 4 X 5 in phase 2, 40 'SUnions' total: ~104 seconds combined
Again, these Profiler benchmarks don't reflect the whole picture, and might be affected by other things I was doing on the laptop while the code was running.…
Added by Joseph Oster at 12:29pm on March 23, 2017
13;2} ... 20.{13;12}
21. {21;0}22. {21;1}23. {21;2} ... 41. {21;20}
42. {34;0}43. {34;1}44. {34;2} ... 75. {34;33}
76. {55;0}77. {55;1} ... ....
I want to grab the first 8 [0-7], the next 13[8-20], the next 21[21-42] etc
so i have the (known fibonacci seq) list of numbers on the left here:
C S
8 0
13 8
21 21
34 42
55 76
89 131
144 220
233 364
and i need the list on the right, so that i can select items using a Series (N=1 and S and C from the list above) and a List Item component.
the simple question is:
is there a component that can take a list and accumulate it in this way that I need?
if not, is there anyone that can point me to a simple relevant VB example so i could easily adapt it?
many thanks,
gotjosh…
I would now like to do is "combine" (only count once) the panels that are stacked vertically between the domain. (I've only got 3 surfaces selected to cut down on the confusion)The domain is the horizontal lines on the object to the ground plane. so Level 1 is 0-8'-7 3/4". Level 2 is 8'-7 3/4" - 15'-3 3/4". ect.
The approach is that the panels that are stacked on top of each other have the same x,y coordinate so they can be separated and counted as 1 instance in the final count.
I started trying to deconstruct the points and create sets of each x and y value but I've got no idea where to go from there. Any have any ideas?
Excel File being referenced looks like this:
Ground
0
1
8.6458
2
15.0625
3
22.375
4
28.5
5
35.333
6
42.2708
…
it is my machine is not letting me install Rhino 5 either.
FYI my machine is running with an Intel(R) Xeon(R) CPU x5675 @ 3.07GHz 3.07GHz GHz (2 processors)
http://www.cpu-world.com/CPUs/Xeon/Intel-Xeon%20X5675%20-%20AT80614006696AA%20(BX80614X5675).html
This is a server chip, I dont think that is the problem but I wanted to throw that out there. If anyone has any advice or has had similar problems please let me know. Thanks much.
-Charles
…
ll these 12500 points.
Group 1 would represent the point located at 0, 5, 10, 15, 20 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.
I can create the pattern but the selection of points are all the points in row 0 and then all the points in row 5 and so on.
I would like the selection of points to start at the bottom left, and sequentially continue to the right and then continue on the 2nd row (left to right & bottom to top). i am hoping the pattern i am trying to achieve is more understood with the quick screen capture I uploaded.
the end goal is to be able to select all the points in the grid that are in each pattern.
Thanks in advance for any guidance with this. …
Added by Alyne Rankin at 6:53am on October 11, 2017