nput parameter and then set the named values on the second?
protected override void BeforeSolveInstance() { Param_Integer param0 = Params.Input[0] as Param_Integer; Param_Integer param1 = Params.Input[1] as Param_Integer; param1.ClearNamedValues();
GH_Structure<GH_Integer> data = param0.VolatileData as GH_Structure<GH_Integer>; if (data.IsEmpty) return; foreach (GH_Integer value in data.AllData(true)) { switch (value.Value) { case 1: param1.AddNamedValue("First option for 1", 11); param1.AddNamedValue("Second option for 1", 12); param1.AddNamedValue("Third option for 1", 13); break;
case 2: param1.AddNamedValue("First option for 2", 21); param1.AddNamedValue("Second option for 2", 22); param1.AddNamedValue("Third option for 2", 23); break;
case 3: param1.AddNamedValue("First option for 3", 31); param1.AddNamedValue("Second option for 3", 32); param1.AddNamedValue("Third option for 3", 33); break; } return; } }
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David Rutten
david@mcneel.com…
Added by David Rutten at 1:56am on December 18, 2013
byte-accuracy red, green, blue channels) = 27 bytes. More likely 28 bytes as colours are probably stored as 32-bit integers, allowing for an unused alpha channel.
28 * 800,000 equals roughly 22 megabytes, which is way down from 9 gigabytes. That's a 400 fold memory overhead, which is pretty hefty.
Grasshopper stores points as instances of classes, so on 64-bit systems it actually takes 64+64+3*8 = 152 bytes per point*, which adds up to 122MB, still way less than 9GB. It would be interesting to know where all the memory goes...
* Grasshopper points also store reference data, in case they come from the Rhino document. This data will not exist, but even so it will require 64-bits of storage.…
Added by David Rutten at 4:13pm on December 11, 2014
s 8, 4, 2, 10, 1, 3, 8, 4, 2, 0. But then for the end result to maintain all numbers above 5 but replace all numbers below with a defined number..Let's say zero. So then the list would read...8, 0, 0, 10, 0, 0, 8, 0, 0.…
the map? For example in one list I want curves 5, 20, 21, 22, 23, 60. In another I want curves 1, 37, 40. In another maybe 70-80. And in the last, all curves that aren't specified in those three lists. Is there a way to partition the lists as such?…
I'm trying to retrieve data from a tag heuer timing system through rs 232. Firefly ports available says com ports 1 and 3 are available. Is one of those two the rs 232 9 pin? Thanks for your help.
ems in the same way. Lofting was particularly difficult, you had to have a separate loft component for every lofted surface that you wanted to generate because the component would/could only see one large list of inputs. Then came along the data structures in GH v0.6 which allowed for the segregation of multiple input sets.
If you go to Section 8: The Garden of Forking Paths of the Grasshopper Primer 2nd Edition you will find the image above describing the storing of data.
Here you will notice a similarity between the path {0;0;0;0}(N=6) and the pathmapper Mask {A;B;C;D}(i). A is a placeholder for all of the first Branch structures (in this case just 0). B is a place holder for all the second branch structures possibly either 0, 1 or 2 in this case. And so forth.
(i) is a place holder for the index of N. If you think of it like a for loop the i plays the same role. For the example {A;B;C;D}(i) --> {i\3}
{0;0;0;0}(0) --> {0\3} = {0}
{0;0;0;0}(1) --> {1\3} = {0}
{0;0;0;0}(2) --> {2\3} = {0}
{0;0;0;0}(3) --> {3\3} = {1}
{0;0;0;0}(4) --> {4\3} = {1}
{0;0;0;0}(5) --> {5\3} = {1}
{0;0;0;1}(0) --> {0\3} = {0}
{0;0;0;1}(1) --> {1\3} = {0}
{0;0;0;1}(2) --> {2\3} = {0}
{0;0;0;1}(3) --> {3\3} = {1}
{0;0;0;1}(4) --> {4\3} = {1}
{0;0;0;1}(5) --> {5\3} = {1}
{0;0;0;1}(6) --> {6\3} = {2}
{0;0;0;1}(7) --> {7\3} = {2}
{0;0;0;1}(8) --> {8\3} = {2}
...
{0;2;1;1}(8) --> {8\3} = {2}
I'm not entirely sure why you want to do this particular exercise but it goes some way towards describing the process.
The reason for the tidy up: every time the data stream passes through a component that influences the path structure it adds a branch. This can get very unwieldy if you let it go to far. some times I've ended up with structures like {0;0;1;0;0;0;3;0;0;0;14}(N=1) and by remapping the structure to {A;B;C} you get {0;0;1}(N=15) and is much neater to deal with.
If you ever need to see what the structure is there is a component called Param Viewer on the first Tab Param>Special Icon is a tree. It has two modes text and visual double click to switch between the two.
Have a look at this example of three scenarios in three situations to see how the data structure changes depending on what components are doing.
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