The best way is to use a C# or a VB component to transpose these
lists. I think in C# you can use transpose directly. You can ask this
on the VB/C# forum on our new website, www.grasshopper3d.com
- Scott
On May 27, 3:56 am, Tonsgaard wrote:
> Being a long time user of Generative Components trying to use
> grasshopper i miss the "transpose" command.
> I have a point list like this:
>
> 0, 1, 2, 3, 4, 5
> 0, 1, 2, 3, 4, 5
> 0, 1, 2, 3, 4, 5
> 0, 1, 2, 3, 4, 5
> 0, 1, 2, 3, 4, 5
>
> and a want to transpose dimensions to:
>
> 1, 1, 1, 1, 1
> 2, 2, 2, 2, 2
> 3, 3, 3, 3, 3
> 4, 4, 4, 4, 4
> 5, 5, 5, 5, 5
>
> Surely I am not the first in need of this...
> how would i go about and do this...? I suppose its quite easy in VB
> script, but being used to GC's C# like language, I kinda dont know how
> to do this...
>
> thanks...
>
> Tonsgaard…
indexes later and the k value is just 5. The goal is to output all combinations of length 5 from the list of integers 0 to 19.
I placed the list in brackets in order to make it iterable (I don't really know what this means.) Somehow I need to get the combinations out of the itertools.combination output.…
nch, xno items in one list)2 divide the list lenght value by the numer of items per branch needed3A generate a list with the series component: the step equal to the target numer of items per branch; the no of items equals the number of target branches
3B generate a list with the series component: the first number of the series equals to the number of items needed (-1 to account for the 0 index); the step size again equal to the target number of itmes per branch as 3A4 feed 3A & 3B to a domain component thus identifying the start -3A- and end -3B- of the domains by which the list will be subdivided5 use a subset component with the domains above thus creating 19 branches with lists having 5 items eachfor lists which are subdivided into branches when the target number of branches is not a multiple of the number of items contained in the list:6 identify if the target number of branches is a multiple of the list by using the modulus component fed by the list lenght -1- and the target number of branches7 identify last index in the 3B series with the item component (reversed to take the last value fed)8 add 6+7 above which dill define the start of the domain that will pick up the remanent items not accommodated in 59 add (+1) to 7 above to define the end of the domain that will pick up the the remanent items not accommodated in 510 feed 8 & 9 to a domain component11 include 10 as part of the subset in 5I'm now trying to understand the components mentioned by Michael...
sn
…
each circle's border, let us say 1.0
3) So, the curve will end up with 5 points, in each point will have a circle, each circle will have a different Radius, but the distance in between the borders of each circle is always the same = 1.0 in this case.
4) The end result list here would be like this to evaluate a curve with these values and find the points on the curve:
List = 1, 5, 11, 19 etc If I use these values to eval a line, I will get the perfect points where I can draw the circles.
…
s
(distance from Spine to Profile)
4) Create Circles at each floor
5) Rotate Each Floor by twisted amount
(according to height of floor)
6) Divide each floor by number of Flutes
7) Flip Matrix of Flute Points (version 0.7)
(if using v0.6 then search flip matrix on site for method)
(Rows to Columns)
8) Interpolate Curve through Flute Points
9) Mirror Flute Curves
10) Create display grid
11) Make a vector 2Pt from the floor centre
to the corresponding display point…
ital; aproveche esta oportunidad y vea la demostración en vivo: Lugar: Escuela Digital - Mexico D.F. Hora: 7:00 - 9:00 pm Con el apoyo de: ESCUELA DIGITAL - Metzli Valle metzli@escueladigital.com.mx…