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#1 2013-04-04 15:18:02
Randomizing a list
I am trying to randomize items in a list for a project I am trying out. I am trying to keep tract of a card position in a deck with using a list. I keep running into a problem where there are list omissions and/or duplicates of a single list item. Is is possible to randomize a list without having any of the items in the list being omitted? or is it not even possible to do that in scratch 1.4? Also if using a list to store a set amount of data for use at a later time is not the best way to go that would be useful to know.
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#3 2013-04-05 03:56:01
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- Registered: 2012-09-26
- Posts: 500+
Re: Randomizing a list
You'd need another, new list to be able to shuffle; an empty, unused one. Create a variable called "item", and then you can use this script:
repeat until <(length of [listyouwanttoshuffle v]) = [0]> set [item v] to (pick random (1) to (length of [listyouwanttoshuffle v])) add (item (item) of [listyouwanttoshuffle v]) to [emptyunusedlist v] delete (item) of [listyouwanttoshuffle v] end set [item v] to [0] repeat until <(length of [emptyunusedlist v]) = [0]> change [item v] by (1) add (item (item) of [emptyunusedlist v]) to [listyouwanttoshuffle v] delete (item) of [emptyunusedlist v] endIf you want to keep track of a specific item's position, create a variable called "newpos", another called "oldpos", and another one called "isselected". Then use this code instead:
set [oldpos v] to [1] //Change this to the position of the item in the list you want to keep track of
set [newpos v] to (oldpos)
set [isselected v] to [0]
repeat until <(length of [listyouwanttoshuffle v]) = [0]>
set [item v] to (pick random (1) to (length of [listyouwanttoshuffle v]))
if <(isselected) = [0]>
if <(item) < (newpos)>
change [newpos v] by (-1)
else
if <(item) = (newpos)>
set [newpos v] to ((length of [emptyunusedlist v]) + (1))
set [isselected v] to [1]
end
end
end
add (item (item) of [listyouwanttoshuffle v]) to [emptyunusedlist v]
delete (item) of [listyouwanttoshuffle v]
end
set [item v] to [0]
repeat until <(length of [emptyunusedlist v]) = [0]>
change [item v] by (1)
add (item (item) of [emptyunusedlist v]) to [listyouwanttoshuffle v]
delete (item) of [emptyunusedlist v]
end
"newpos" will contain the new position of the item that was at the position you specified in "oldpos". 
These scripts are off the top of my head, so they may not work.
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#4 2013-04-05 03:59:15
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- Registered: 2012-09-26
- Posts: 500+
Re: Randomizing a list
Actually, erm, yeah, I made a stupid mistake
In those scripts, replace the
set [item v] to [0] repeat until <(length of [emptyunusedlist v]) = [0]> change [item v] by (1) add (item (item) of [emptyunusedlist v]) to [listyouwanttoshuffle v] delete (item) of [emptyunusedlist v] endat the end with
repeat until <(length of [emptyunusedlist v]) = [0]> add (item (1 v) of [emptyunusedlist v]) to [listyouwanttoshuffle v] delete (1 v) of [emptyunusedlist v] end
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