rated by "<" symbols. Examples: "2<10", "2<4<10", "Pow(2, 1)<5*Sin(3)<10".
The entered text contains 2 or 3 segments separated by two or more consecutive dots. Examples "2..10", "2..4..10", "Pow(2, 1)....5*Sin(3)..10".
If only two segments are provided, then the initial value will be the same as the minimum value. If a bounds number or a default value is written as a simple number, then the number of decimal places will be harvested. I.e. "2..4..10" is not the same as "2..4..10.00" as the former will result in an integer slider and the latter in a slider with two decimal places.
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David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 10:08am on February 15, 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
13;2} ... 20.{13;12}
21. {21;0}22. {21;1}23. {21;2} ... 41. {21;20}
42. {34;0}43. {34;1}44. {34;2} ... 75. {34;33}
76. {55;0}77. {55;1} ... ....
I want to grab the first 8 [0-7], the next 13[8-20], the next 21[21-42] etc
so i have the (known fibonacci seq) list of numbers on the left here:
C S
8 0
13 8
21 21
34 42
55 76
89 131
144 220
233 364
and i need the list on the right, so that i can select items using a Series (N=1 and S and C from the list above) and a List Item component.
the simple question is:
is there a component that can take a list and accumulate it in this way that I need?
if not, is there anyone that can point me to a simple relevant VB example so i could easily adapt it?
many thanks,
gotjosh…
where each branch contains all the points generated by dividing each curve, so if you divide into 10 segments, you'll get:
{0;0}(N = 11)
{0;1}(N = 11)
{0;2}(N = 11)
{0;3}(N = 11)
{0;4}(N = 11)
Where the second integer in the curly brackets refers back to the index of the curve in the original list.
Another way to look at this data is to see it as a table. It's got 5 rows (one for each original curve) and 11 columns, where every column contains a specific division point.
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David Rutten
david@mcneel.com
Poprad, Slovakia…
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...…