- C
{2;0} (N=61) - D
{2;1} (N=60) - E
{2;2} (N=61) - F
group 2:
{0;0} (N=10) - U
{0;1} (N=10) - V
{0;2} (N=10) - W
{0;3} (N=10) - X
{0;4} (N=10) - Y
{0;5} (N=10) - Z
the idea case is I can merge those date sets in a pattern of A-U-B-V-C-W-D-X-E-Y-F-Z...so on
therefore I am thinking how could I modify the path on group 2 and make them becomes things like:
{0;0} (N=10) - U
{0;1} (N=10) - V
{0;2} (N=10) - W
{1;0} (N=10) - X
{1;1} (N=10) - Y
{1;2} (N=10) - Z
but I have no idea how could I modify the path in that way....
can anyone show me how to?…
Added by Preston Chan at 8:34pm on October 26, 2010
ards to the number before the start number...
i.e. 9, 0, 1, 2, 3, 4, 5, 6, 7, 8
then it will need to repeat this pattern (continuing to count upwards) and the repeat number is based on a slider (for example 3 in the case illustrated below):
9, 0, 1, 2, 3, 4, 5, 6, 7, 8
19, 10, 11, 12, 13, 14, 15, 16, 17, 18,
29, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29,
if anyone has any ideas on how to create this series it would be great
M.…
etc.
Group 2 - 1, 6, 11, 16, 21 etc.
Group 3 - 2, 7, 12, 17, 22 etc.
Group 4 - 3, 8, 13, 18, 23 etc.
Group 5 - 4, 9, 14, 19, 24 etc. "
except in data, the branches start at 0, so 'group 1' is branch 0
as for the order of your points, that depends on the input prior sorting...
yrs …
e! I do not have good ideas today!
The end result of the list would be:
5, 10, 15, 20, 21, (21 + 5), (21 + 10), (21 + 15), (21 + 20), (21 + 21), (42 + 5), (42 + 10), (42 + 15), (42 + 20), (42 + 21), etc …
icture you can see what I mean, this is just with randome reduce, it should be like this:
->If the last object is (1) then is a 50% chance that the next one is also (1)
-->If the last two objects are (1) then is a 40% chance that the next one is also (1)
--->If the last three objects are (1) then is a 30% chance that the next one is also (1)
---->If the last four objects are (1) then the next object must be (2)
->If the last object is (2) then is a 50% chance that the next one is also (2)
-->If the last two objects are (2) then is a 30% chance that the next one is also (2)
--->If the last three objects are (2) then is a 10% chance that the next one is also (2)
---->If the last four objects are (2) then the next object must be (1)
It would be nice if it is possible to change the number how many objects should be tested and the percent values.
Has anybody an idea how I can do this?
Best Regards
…
0, 5, 10, 15, 20
1, 6, 11, 16, 21
2, 7, 12, 17, 22
3, 8, 13, 18, 23
4, 9, 14, 19, 24
and if i'm here is because i'm not able... :)
can you help me?
thank you
…
an = True
For j As Integer = i + 1 To x.Count - 1
If round((x(i).x * 10 ^ 8 + x(i).y * 10 ^ 4 + x(i).z), 2) = round((x(j).x * 10 ^ 8 + x(j).y * 10 ^ 4 + x(j).z), 2) Then bol = False
Next
If bol Then ptlist.add(x(i))
Next
a = ptlist
I think someone posted a more appropiate way of doing it, similar to how the "seldup" command works. You can also run the seldup command in a script using app.RunScript("-seldup"), but its a bit messy since you have to bake the geometry first and select the resulting geometry all within the script.…