Randomize between a fix collection of numbers

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Permalink Reply by Julian Buhagiar on August 2, 2017 at 4:38am

you can define the angles that you want to select as a list (like a bag of numbers)

then connect this to the L input of a component to a Sets>listItem component

Then connect the random component to I of the listItem component.

Then to generate the number of random angles you need, input that to the N of the random component.

And you'll want the domain of the random select to be 0 To List length (3)...(with Wrap on on the List item component, this takes care of the chance of picking the first or last item as being the same as all in between elements)

You can think of this method as a lucky dip bag with replacement.

Permalink Reply by eminsua on August 2, 2017 at 4:55am

Hi Julian, thanks for the reply.

I must have something in the wrong place, because if I do this, what I get is a tree list of a 100 elements with 3 inside them.

What I'm looking for is a random list of this 100 elements being the 3 diferent numbers on a random disposition.

Let me upload the file and see if you can spot what I' am doing wrong.

cheers

Attachments:

• unnamed.gh, 10 KB
• Capture2.JPG, 95 KB

Permalink Reply by Julian Buhagiar on August 2, 2017 at 5:55am

you were almost there... you just needed a second list length and to ensure you feed it a list of items from the panel by making it multiline.

Permalink Reply by Julian Buhagiar on August 2, 2017 at 6:02am

and for greater accuracy, perhaps use the convert to radians component before the list you select from.

Permalink Reply by Julian Buhagiar on August 2, 2017 at 5:00am

see attached.... the white list is the output... the bit after is just proof that you'll get a uniform distribution because of the wrap=true

i.e

0 To 0.49999 will produce an index of 0, 

0.5 To 1.49999 will produce an index of 1, 

1.5 To 2.49999 will produce an index of 2,

2.5 To 3.0 will produce an index of 3 which transforms to 0, giving 0 a similar likelihood of selection as 1 and 2. (it will have marginally more chance of likelihood) because ideally you want the range to produce the closest number to 3.0 (2.999999999...) and not actually 3

Attachments:

• luckydip.gh, 14 KB

Permalink Reply by Julian Buhagiar on August 2, 2017 at 5:17am

Central Limit Theorem proof of the average for large sample sizes shows uniformity.

Permalink Reply by Julian Buhagiar on August 2, 2017 at 5:35am

Permalink Reply by eminsua on August 2, 2017 at 7:42am

Thank you very much for this Julian , really helpfull stuff.

Now I get How this works

HAVE A LOVELY DAY

Attachments:

• Capture3.JPG, 153 KB