1+2+3+4+5+6 = 21
1+2+3+4+5+6+7 = 28
1+2+3+4+5+6+7+8 = 36
1+2+3+4+5+6+7+8+9 = 45
Is there a tool, that can do me that job?
How do I get this List {1,3,6,10,15,21,28,36,45}?…
Added by Ahmed Hossam at 2:19pm on September 22, 2013
en 3 of them, and one poolyline between two of them.
It would also be very nice if i could control it so that only the successive ones can be connected
so if {0:0:0} has 8 points and {0:0:1} has 8, as do {0:0:5} and {0:0:6} i would like to have this as two polylines, not one continoous that would in this case jump three branches (or curves that are shorter).
Does this make any sense?…
Added by Dusan Bosnjak at 2:08pm on September 28, 2009
{0;0;0;0 -1, 2... 19 ) N=5
i know is basic but im still trying to get my head around lists/trees and how to manipulate them properly
thanks for your time
sn…
{1;1;4}{1;1;5}{1;2;0}{1;2;1}{1;2;2}{1;2;3}{1;2;4}{1;2;5}{1;3;0}{1;3;1}{1;3;2}{1;3;3}{1;3;4}{1;3;5}
etc...
and I want to format as a text it so it replaces the innermost branch with a letter so {0;0;1} would read A-0-1. I am able to replace all the symbols using replace text but am no sure if there's a way to convert a number to a letter.…
Added by Ryan Whitby at 12:40pm on February 3, 2015
For example.
If you have two lists of points.
List A List B
{0;0;0}(0) {0;0}(0)
{0;0;1}(0) {0;1}(0)
{0;2}(0)
{0;3}(0)
{0;4}(0)
And you want to merge the two lists so that the two points in list A are the end points.
Merge Lists Results:
{0;0}(0)
{0;0;0}(0)
{0;0;1}(0)
{0;1}(0)
{0;2}(0)
{0;3}(0)
{0;4}(0)
Because of their path structures the order is wrong from a simple merge so Flattening now is out of the question.
Path Mapper
{A;B} --> {A;B+1}
{A;B;C} --> {A;C*6}
---------------------
Results:
{0;0} --> {0;0+1} = {0;1}
{0;1} --> {0;1+1} = {0;2}
{0;2} --> {0;2+1} = {0;3}
{0;3} --> {0;3+1} = {0;4}
{0;4} --> {0;4+1} = {0;5}
{0;0;0} --> {0;0*6} = {0;0}
{0;0;1} --> {0;1*6} = {0;6}
Now with the Path Structures similar when they are re-ordered the results will have the two points of list A as the end points.
Question 2
why did the curve-line intersection lose the path structure? Both trees had 38 branches.
Both trees had 38 Paths but Tree A had more Items, 147 compared to 38 in Tree B.
So you get this happening:
{0;0;0;0;0;0}(0) compared to {0;0;0;0}(0) results: Null {0;0;0;0;0;0}(0)
Base result paths on longest
{0;0;1;0;0;0}(0) compared to {0;0;0;1}(0) results: Null {0;0;1;0;0;0}(0)
{0;0;2;0;0;0}(0) compared to {0;0;0;2}(0) results: Yes {0;0;2;0;0;0;0}(0)
Add a branch to contain result
{0;0;3;0;0;0}(0) compared to {0;0;0;3}(0) results: Yes {0;0;3;0;0;0;0}(0)
{0;0;3;0;0;0}(1) compared to {0;0;0;3}(0) results: No {0;0;3;0;0;0;1}(0)
{0;0;4;0;0;0}(0) compared to {0;0;0;4}(0) results: Yes {0;0;4;0;0;0;0}(0)
{0;0;4;0;0;0}(1) compared to {0;0;0;4}(0) results: Yes {0;0;4;0;0;0;1}(0)
{0;0;5;0;0;0}(0) compared to {0;0;0;5}(0) results: Yes {0;0;5;0;0;0;0}(0)
{0;0;5;0;0;0}(1) compared to {0;0;0;5}(0) results: Yes {0;0;5;0;0;0;1}(0)
{0;0;5;0;0;0}(2) compared to {0;0;0;5}(0) results: Yes {0;0;5;0;0;0;2}(0)
...... etc
…
cture, Rhino treats them as a single flat list. For example a surface can have 10 rows and 6 columns of control-points, resulting in a list of 60 points.
But 10 times 6 isn't the only way to get to 60. If you want to make a surface out of a list of 60 points, you'll also have to tell Rhino how those 60 points should be interpreted in terms of a grid. It could be 2*30, 3*20, 4*15, 5*12, 6*10, and all of the aforementioned products the other way around.
Sometimes there's only one way for a number of points to fit into a rectangular grid. For example if you provide 49 points, then 7*7 is the only way to make it work, but these cases are rare so we always demand you give us all the information required to actually make a rectangular grid of control-points from a linear collection.
As for "Why is it, sometimes we need to attach additional value into it?", this is usually because when you divide a domain or a curve into N segments, you end up with N+1 points. For example take the domain {0 to 5}, and divide it into 5 equal subdomains. You end up with {0 to 1}, {1 to 2}, {2 to 3}, {3 to 4} and {4 to 5}. However there are six numbers that mark the transitions between these domains 0, 1, 2, 3, 4 and 5. This is why you often have to add 1 to the UCount, because the number that controls the UCount often results in N+1 actual points.…
Added by David Rutten at 8:30am on December 25, 2014