a follow up question... how do I wrap a list onto itself at a certain frequency?
i.e. I want the list {1;2;3;4;5;6;7;8;9}
to become {1,4,7; 2,6,8; 3,6,9} wrapped every 3rd item
Added by Joshua Jordan at 5:30pm on November 17, 2012
s 8, 4, 2, 10, 1, 3, 8, 4, 2, 0. But then for the end result to maintain all numbers above 5 but replace all numbers below with a defined number..Let's say zero. So then the list would read...8, 0, 0, 10, 0, 0, 8, 0, 0.…
t, let's talk about randomness. Randomness is a problem in computing because digital computers are deterministic. If you give them the exact same instructions they always end up with the exact same result. It turns out to be mathematically impossible to generate true random numbers using a digital computer, but it is fairly easy to generate pseudo-random numbers. This is actually not bad news as pseudo-random numbers -unlike real random numbers- can be generated again and again and you'll end up with the same random numbers every time. Being able to get the same random numbers on demand increases the reliability of these number sequences which in turn makes them easier to use.
Pseudo-random numbers are numbers that have certain characteristics. Note that when we talk about random numbers we are really talking about numbers. Plural. It's easy to generate only a single one, as xkcd so eloquently put it:
So what are these characteristics that define pseudo-randomness? Without being actually correct, I can sum them up as follows:
The sequence of generated numbers should never repeat itself*
The numbers in the sequence ought to be spread evenly across the numeric domain**
There are a lot of different algorithms out there, some better than others, some faster than others, some solving very specific problems while others are more generic. The generator used in Grasshopper is the standard Microsoft .NET Random, based on Donald Knuth's subtractive algorithm.
So let's imagine we want random integers between 0 and 10. What would a bad random sequence look like?
3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 (about as bad as it gets)
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 (not random at all)
1 3 2 5 3 9 1 2 4 2 5 1 1 2 8 1 5 2 3 4 (too many low numbers)
2 8 4 6 0 9 8 2 4 8 6 4 2 2 5 1 4 8 6 2 (too many even numbers)
So what about good sequences? Well, here's a few:
6 9 1 2 0 4 2 8 5 7 2 9 1 9 2 5 3 1 9 2 (sure, why not)
6 2 5 3 4 1 9 7 8 0 2 1 6 4 5 8 9 5 0 9 (looks about right)
1 8 5 2 3 4 5 7 9 5 2 1 0 2 1 0 9 7 6 4 (I suppose)
9 0 6 4 8 3 1 5 2 7 6 1 4 6 0 1 9 7 5 6 (whatever)
There are a lot of valid pseudo-random sequences. (Seriously, loads). So even if we have a good pseudo-random generator we may be given a random sequence that isn't entirely to our liking. The shorter the sequence we need, the more likely it is that statistical aberrations invalidate that particular sequence for us. What we need is some control over the generator so we don't just get a repeatable sequence, but a repeatable sequence we actually like.
Enter seed values. The random generator requires a seed value before it can generate a random sequence. These seed values are always integers, and they can be any valid 32-bit integer. Every unique seed value results in the same sequence. Every time.
Unfortunately there is no clear relationship between seeds and sequences. Changing the seed value from 5 to 6 will result in a completely difference random sequence, and two sequences that are very similar may well have to wildly different seeds. There is therefore no way to guess a good seed value, it is completely trial-and-error. Also because of this extremely discontinuous nature, you cannot use tools like Galapagos to optimize a seed value.
If you are looking for a pseudo-random sequence which has custom characteristics, you may well end up having to write your own generator algorithm. Ask questions about this on the Grasshopper main forum or the VB/C# forum.
Conclusion: Seed values are integers that define the exact sequence of pseudo-random numbers, but there's no way of knowing ahead of time what sequence it will be and there's no way of tweaking a sequence by slightly changing the seed. Even the tiniest change in seed value will result in a radically different random sequence.
--
David Rutten
david@mcneel.com
Poprad, Slovakia
* This is not actually possible. A finite amount of numbers always repeats itself eventually.
** This should only be true for long enough sequences, short sequences are allowed to cluster their values somewhat.
Interesting links for further reading:
Coding Horror: Computers are Louse Random Number Generators
StackOverflow: When do random numbers start repeating?…
Added by David Rutten at 9:52am on October 20, 2012
Integer = 0 To 9
val *= 2
lst.Add(val)
Next
Since val is a ValueType, when we assign it to the list we actually put a copy of val into the list. Thus, the list contains the following memory layout:
[0] = 2
[1] = 4
[2] = 8
[3] = 16
[4] = 32
[5] = 64
[6] = 128
[7] = 256
[8] = 512
[9] = 1024
Now let's assume we do the same, but with OnLines:
Dim ln As New OnLine(A, B)
Dim lst As New List(Of OnLine)
For i As Integer = 0 To 9
ln.Transform(xform)
lst.Add(ln)
Next
When we declare ln on line 1, it is assigned an address in memory, say "24 Bell Ave." Then we modify that one line over and over, and keep on adding the same address to lst. Thus, the memory layout of lst is now:
[0] = "24 Bell Ave."
[1] = "24 Bell Ave."
[2] = "24 Bell Ave."
[3] = "24 Bell Ave."
[4] = "24 Bell Ave."
[5] = "24 Bell Ave."
[6] = "24 Bell Ave."
[7] = "24 Bell Ave."
[8] = "24 Bell Ave."
[9] = "24 Bell Ave."
To do this properly, we need to create a unique line for every element in lst:
Dim lst As New List(Of OnLine)
For i As Integer = 0 To 9
Dim ln As New OnLine(A, B)
ln.Transform(xform)
lst.Add(ln)
Next
Now, ln is constructed not just once, but whenever the loop runs. And every time it is constructed, a new piece of memory is reserved for it and a new address is created. So now the list memory layout is:
[0] = "24 Bell Ave."
[1] = "12 Pike St."
[2] = "377 The Pines"
[3] = "3670 Woodland Park Ave."
[4] = "99 Zoo Ln."
[5] = "13a District Rd."
[6] = "2 Penny Lane"
[7] = "10 Broadway"
[8] = "225 Franklin Ave."
[9] = "420 Paper St."
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 6:26am on September 9, 2010
st with 6 branches the first 3 having 8 items and the last three having 4 items . I really want to say take list 1 and add list 2 onto the bottom of it so i get a list with 9 branches each with 4 items?
As always thanks wonderful forum!…
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 …
Good to hear it helped.
As for the slider you can just change the limits to Min=1 and Max=4, so you'll get the values you want (.2, .4, .6, .8). This works for me fine.