in the desired order.
0 = 0
1 = 1
2 = 6
3 = 7
4 = 8
5 = 9
6 = 12
7 = 13
8 = 2
9 = 3
10 = 4
11 = 5
12 = 10
13 = 11
Where the first number is the index and the second number is the actual sorting key. Then you sort these keys while sorting your curves in parallel using the A input of the Sort component.
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
rated by "<" symbols. Examples: "2<10", "2<4<10", "Pow(2, 1)<5*Sin(3)<10".
The entered text contains 2 or 3 segments separated by two or more consecutive dots. Examples "2..10", "2..4..10", "Pow(2, 1)....5*Sin(3)..10".
If only two segments are provided, then the initial value will be the same as the minimum value. If a bounds number or a default value is written as a simple number, then the number of decimal places will be harvested. I.e. "2..4..10" is not the same as "2..4..10.00" as the former will result in an integer slider and the latter in a slider with two decimal places.
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 10:08am on February 15, 2013
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…
≈ 4.8 " as " x= 4.8 ± a ", do you know what is the min and max for "a"?
and second, i had tried the "round" function, but i faced problem with it too! for example:
if the input is a series as {0.0, 0.5, 1, 1.5, 2, 2.5, ...}
the output for Round(x, 0.5) is : {0, 0, 1, 2, 2, 2, 3, 4, 4, 4, 5, 6, 6, 6, ... }
and for Round(x, 2) the output is : {0.0, 0.5, 1, 1.5, 2, 2.5, ... }
i can't understand the logic that lies behind this function, i think
for Round(x, 0.5) the output must be {0.0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, ... }
and for Round(x, 2) it must be {0.0, 0.0, 2, 2, 2, 2, 4, 4, 4, 4, ... }
so, is there any problem with it, or I misendestood the logic ?…
is created for each point (25 paths, N=1 for each) which is feed into [Pull Point] for the pull geometry [G].
Correspondingly for the 4 source points a branch is created for each point and duplicated 25 times (4 paths, N=25 for each). This tree then needs to be inverted with [Path Mapper] so the structure will correspond to the format of the pull geometry. The mapping {A;B;C}(i) > {i}(B) produces (25 paths, N=4 for each) the structure to feed into the search point [P].
The [Pull Point] boolean toggle [C] needs to get set to False to obtain all the distances between all search and pull points (4 x 25 = 100 values).
Simultaneously there is also an index being created to correspond to the list of the 4 source points. This index is the integers 0 to 3 which are branched and inverted similar to the source points (25 paths, N=4 for each).
The distance output [D] from [Pull Point] is then sorted synchronously with the source point index for each branch. From the following screengrab branch {0;0} corresponds to a point in the 5 x 5 grid and the shortest distance between that point and a referenced source point index is 5.261. The index of the referenced source point is 3.
For each following sorted branch the first sorted index value will correspond to the closest source point (first [List Item] shown). This index value is then used to select from the original list of duplicated and inverted points and this is done for each of the 25 branches (second List Item shown).
Draw a line or whatever an away we go!…
ep is to understan the logics of what you want to do, in your case, build 4 point surfaces (u also need to know the right direction to build the surfaces). Then you can write an hipotetic list (by hand in a paper) of what you want. In your case the list was (0, 1, 3, 2) (2, 3, 5, 4) (4, 5, 7, 6), etc... if you can imagine building 2 lists, each one with the sequences (0, 2, 4, 6, etcc) and (1, 3, 5, 7, etc..) then you can manage with shift and graft to finally have four lists. A( 0 1 2 3 ...) B (1 3 5 etc..) C(3 5 7 etc..) D (2 4 6 etc..). And to achieve the 2 first lists, you need to get the odd and the pair numbers. The cull pattern does that amazingy well. With a pattern True-False you get de pair numbers, and with the False-True pattern you get de odd numbers.
Hope it was clear enough…
Added by Pep Tornabell at 5:32am on November 19, 2009