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?…
e possible to change the component definition making possible to customize the number of outputs.Now Dispatch moves "true" values to A and "False" values to B
INPUT:
L (List to work on) -> 1, 2, 3, 4, 5, 6, 7, 8
D (Dispatch Pattern) -> True, False
OUTPUT:
A (List) -> 1, 3, 5, 7
B (List) -> 2, 4, 6, 8
Could it be possible/useful to modify it so it could dispatch items to several outputs, like:
INPUT:
L (List to work on) -> 1, 2, 3, 4, 5, 6, 7, 8, 9, 0
D (Dispatch Pattern) -> A, B, C
OUTPUT:
A (List) -> 1, 4, 7, 0
B (List) -> 2, 5, 8
C (List) -> 3, 6, 9
maybe I'm missing something and there's already a component with this function... I have been searching on the forum for half afternoon, but can't find anything about it!
Thank you!…
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
branches in each A's list of B's, or remove its ends etcso that if I want to remove the last B in every A{0;1},{0;2},{0;3},{0;4},{0;5},{0;6}{1;1},{1;2},{1;3},{1;4}{2;1},{2;2},{2;3},{2;4},{2;5}would become{0;1},{0;2},{0;3},{0;4},{0;5}
{1;1},{1;2},{1;3}
{2;1},{2;2},{2;3},{2;4}I guess the question is do I need to figure out the cull pattern- each B may have different lengths...…
≈ 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 ?…
rve
10 curve
11 curve
12 curve
13 curve
...and I'd like to rearrange the order in which the curve are listed, to something like this:
{0,0,0}
0 curve
1 curve
8 curve
9 curve
10 curve
11 curve
2 curve
3 curve
4 curve
5 curve
12 curve
13 curve
6 curve
7 curve
I hope this makes sense.
Thank in advance for any advice,
John…