4}
{0;2;0}
{0;2;1}
{0;2;2}
{0;2;3}
{0;2;4}
You cannot flip this because this is more complex than a rectangular matrix. You're going to have to do the mapping yourself. Try a Path Mapper with the following masks:
{A;B;C}(i) -> {A;B;i}(C)
Which should give you a structure that results in 3 lofts.
--
David Rutten
david@mcneel.com
Poprad, Slovakia
…
Added by David Rutten at 3:18pm on November 27, 2011
;-2;-1;0;1;2;3) and beside that i don't want to have 0 as a result.
So at the End I would like to have only 6 numbers: -3;-2;-1;1;2;3
I already came to an result with integer numbers but it actually only rounded up the random results. when one of the results was -0.473 it ended up as 0.
I'm sure it's pretty easy to solve, I'm just missing something.
thx for help…
i-branches is removing similar branches. This will only be removing 0's.
e.g.
{1;1;1;3;0}
{1;1;1;4;0}
{1;1;1;5;0}
would end up after Simplify as:
{3}
{4}
{5}
But the single branch (remove zeros algorithm, as summarised above) would give:
{1;1;1;3} …
nce to graft:
{A}(i) --> {A;i}
becomes
{0}(0) --> {0;0}
{0}(1) --> {0;1}
{0}(2) --> {0;2}
...
{0}(n) --> {0;n}
So now to apply this to any complex situation that the Path Mapper might be able to do. A recent post here on the Forum asked about duplicating every branch to the next branch. For example:
{0;0} = 1
{0;1} = 2
{0;2} = 3
needs to be
{0;0} = 1
{0;1} = 1
{0;2} = 2
{0;3} = 2
{0;4} = 3
{0;5} = 3
First we need to make a gap for the duplicated data to be slotted into.
{A;B} --> {A;2*B}
becomes
{0;0} --> {0;2*0} = {0;0}
{0;1} --> {0;2*1} = {0;2}
{0;2} --> {0;2*2} = {0;4}
Then we need to map the same branches into the next branch
{A;B} --> {A;2*B+1}
becomes
{0;0} --> {0;2*0+1} = {0;1}
{0;1} --> {0;2*1+1} = {0;3}
{0;2} --> {0;2*2+1} = {0;5}
Now when these two paths structures are combined you will get the desired results.
So it comes down to how you wish to map your paths and coming up with a formula to do so.
…
Added by Danny Boyes at 2:50pm on October 20, 2011
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.
3} N=2
....
{527} N=4
What I want to do is figure out the same the item in group A and retrieve their branch index so that I can get the corresponding item in group A. Is there any way to make it?…
3)
0:1 (N=3)
Is there a way to do this with the current tree utilities or is there a planned component such as the path mapper that has been previously discussed?
Best regards
Danny…
Added by Danny Boyes at 3:59am on November 13, 2009