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
Hi I want to eliminate every 11 and 13 numbers (list A) from list B. I used equals and dispatch and I removed the 13 but not the 11. JPG attached. Suggestions?
THX
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
onsecutive points at the same height then your 'Break at discontinuities' component eliminates the middle point completely and then the 'Interpolate Curve' component gives a much bigger bump in the wrong direction. This was enough to get curves to meet from opposite sides.
I fixed this by changing the heights to 1.1 or 2.9, rather than 1.0 and 3.0, but it took a little while to work it out! Sigh.
I attach a new version. But I actually preferred it as it was before. See what you think!
Bob
p.s. in the first list, elements 11, 12, 23 and 24 go from 1 to 3; elements 17 and 18 go from 3 to 1. In the second list, elements 6, 17, 18 and 29 go from 1 to 3; elements 12 and 23 go from 3 to 1. Given the above fix, these can be easily seen.…
Added by Bob Mackay at 10:40pm on November 24, 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
…
a specific domain, for example:
0.) 0 to 1 -----> 11 random values from 0 to 1 (0.245,0.678,0.36,0.78,.28,0.18........)
1.) 1 to 2 -----> 11 random values from 1 to 2 (1.26,1.36,1.01,1.68,1.26,1.96.........)
3.) 2 to 3 -----> 11 random values from 2 to 3 (2.96,2.45,2.78,2.56,2.98,2.10..........)
4.) 3 to 4 and so on where I have a data set containing 11 paths with 11 values and the values fall within the specific domain.
Like my post above I have the correct path but I need to feed it the correct seed to get different values for each number. I tried grafting a series similar to the last post but it scrambles my data. Thanks so much for the help!
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