)
3. KeyError(1417,)
4. KeyError(1417,)
5. KeyError(1417,)
6. KeyError(1417,)
7. KeyError(1417,)
8. KeyError(1417,)
9. KeyError(1417,)
10. KeyError(1417,)
11.......
i tried different weather file but also same result. it seems i have same problem. the file am working on is the radiation file i took from the examples . whats seems to be the problem?
thank you for your time…
would like to group the paths based on their item count (n) values resulting in a tree which should look something like this:
{0;0} (3)
{0;1} (2)
{0;2} (2)
{0;3} (1)
in other words, all paths with 2 items are under one path, all with 6 items in another, and so on.
I feel that the pathmapper should be able to do this very easily but cannot figure out what the expression should be... I have tried searching the forum but have not had much luck!
Any ideas? Thanks a ton!…
cture, Rhino treats them as a single flat list. For example a surface can have 10 rows and 6 columns of control-points, resulting in a list of 60 points.
But 10 times 6 isn't the only way to get to 60. If you want to make a surface out of a list of 60 points, you'll also have to tell Rhino how those 60 points should be interpreted in terms of a grid. It could be 2*30, 3*20, 4*15, 5*12, 6*10, and all of the aforementioned products the other way around.
Sometimes there's only one way for a number of points to fit into a rectangular grid. For example if you provide 49 points, then 7*7 is the only way to make it work, but these cases are rare so we always demand you give us all the information required to actually make a rectangular grid of control-points from a linear collection.
As for "Why is it, sometimes we need to attach additional value into it?", this is usually because when you divide a domain or a curve into N segments, you end up with N+1 points. For example take the domain {0 to 5}, and divide it into 5 equal subdomains. You end up with {0 to 1}, {1 to 2}, {2 to 3}, {3 to 4} and {4 to 5}. However there are six numbers that mark the transitions between these domains 0, 1, 2, 3, 4 and 5. This is why you often have to add 1 to the UCount, because the number that controls the UCount often results in N+1 actual points.…
Added by David Rutten at 8:30am on December 25, 2014
simplified in the bake name attribute and on the bake object component. I was counting on maintaining that structure.
{0;4}
{0;4}
{1;2}
{1;2}
Easily fixed with an additional attribute. Just curious on the behavior. Thanks.
…
Added by japhy to EleFront at 3:18pm on October 9, 2017
13 5 15 6 17 7 ... …
But it seems I have not been clear:
the input values (3,5,7,9,11,13) should be presented in the way of "number slider" parameter. So by increasing the value of the slider from minimum 3, to 5, 7, 9, 13, 15, 17... I should get an output numbers of 0,1,3,4,5,6,7...
So I if choose value 3 on the input "number slider", on the other end, I want 0 as an output.
Or if I choose value 5 on the input "number slider", I want 1 as an output, and so on.
Is this possible?…
up, 6 in the sixth group through the 31st group, 4 in the 32nd group, 2 in the 33rd group.
Side note to David... Since data matching is such a key aspect as definitions become more sophisticated it seems like there should be a much, more comprehensive and intuitive methods to manipulate tree structures. …
Added by Stan Carroll at 12:43pm on April 14, 2011
That is correct. In reality there are 3 major versions of .NET 1, 2, and 4 (different versions of mscorlib.dll)
Rhino 4 uses .NET 2 which includes 3 and 3.5.
Rhino 5 uses .NET 4
dimension of matrices must be identical) and division is the same as multiplication (dimension must be in the order of A(mxn)*\/B(nxk) where n is the common dimension): to divide one element by another you just multiply it by 1/value (part or all of the elements can multiply while part or all of the elements divide):
so for example matrix addition of matrices A(2x2): {2,-1}{1,2} and B(2,2): {3,-5}{4,-2} will result in matrix C(2x2):{5,-6}{5,0}. subtraction of those matrices will result in D(2x2): {-1,4}{-3,4}
Division of matrices A(2x2): {2,0.5}{2,4} and B(2x1) :{2}{2} will result in matrix C(2x1): {1+0.25}{1+2}={1.25,3}. Multiplication of those matrices will result in D(2x1):{4+1}{4+8}={5,12}.…