Need help randomly selecting unique values from an array

Discussion in 'Ruby' started by Paul, Aug 30, 2007.

  1. Paul

    Paul Guest

    Hi there,

    I have a simple array that I have to pick one or more unique values
    from and am kind of stumped as to how to do this. Here's the what the
    array looks like:

    foo = [2, 7, 10, 14] # (or could equal a diff't set of 4
    numbers)

    Now for this task, I'm not interested in the first element at all.
    I'd rather just leave it alone and not delete it though.
    What I need to do is pick 1 to 3 elements randomly from the array. I
    can't reuse the numbers.

    So, for example, if I need 1 number then it could be any of the
    elements 1-3 from the array.
    If I need 2 numbers, it could be any of elements 1-3 but I can't pick
    the same number twice. Ditto for when I need 3 numbers.

    I don't need the numbers in any particular order, so I'm just looking
    for the simplest loop/function to pick the unique numbers for me.

    Any suggestions? Thanks.
     
    Paul, Aug 30, 2007
    #1
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  2. Paul

    Kyle Schmitt Guest

    You could port this code over from C...
    http://xkcd.com/221/


    OK ok seriously though.
    rand(foo.length) will randomly pick any one of them, but you want to
    ignore index zero so
    rand(foo.length -1 ) + 1 should do the trick.

    This is a slightly... stupid way of doing it, but considering the size
    of the data, it's a choice of what's more stupid, generating random
    numbers until you get two that aren't the same, or something like
    this..

    number1 = foo[rand(foo.length-1)+1]
    foo2=foo-[number1]
    number2 = foo2[rand(foo2.length-1)+1]
     
    Kyle Schmitt, Aug 30, 2007
    #2
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  3. So you use foo[1..-1]. That will create a new array not containing the first
    item and leave the original array untouched.
    foo[1..-1].sort_by {rand}.first(n)
    where n is a number between 1 and 3.

    HTH,
    Sebastian
     
    Sebastian Hungerecker, Aug 30, 2007
    #3
  4. Paul

    Paul Guest

    Excellent! That's amazingly simple! (and I never would have figured
    that out)
    It's just what I need. Thanks.
     
    Paul, Aug 30, 2007
    #4
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