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Re: majority voting [message #65158 is a reply to message #65079]

Thu, 12 February 2009 04:17

Allan Whiteford
Messages: 117
Registered: June 2006

Senior Member

mort canty wrote:
> Hi all,
>
> Given a 2-D array such as
>
> 0 1 1 2 1
> 0 2 1 1 1
> 1 0 2 2 1
>
> where the entries are labels, the columns represent items and the rows
> are voters, I want a IDL function that returns the majority vote labels.
> So here I should get
>
> 0 ? 1 2 1
>
> as output, where ? = "don't care". There must _not_ be a loop over
> columns. I've got a clumsy solution, but I'm sure there's an elegant one
> somewhere?
>
> Cheers,
>
> Mort

Hi Mort,

It might be less efficient than JD's histogram solution (I didn't check)
but the following also fits the problem specification:

x=[ [0,1,1,2,1],$
[0,2,1,1,1],$
[1,0,2,2,1]]

voters=(size(x,/dim))[1]
items=(size(x,/dim))[0]
max_label=max(x)+1

f=intarr(max_label,items)
++f[max_label*(indgen(voters*items) / voters)+ $
reform(transpose(x),voters*items)]
junk=max(f,idx,dim=1)
print,idx - max_label*findgen(items)

Note that the above solution will also blow up when you end up with
sparse arrays (e.g. if you have someone voting for label 1000000 then f
will end up being an items x 1000000 array even if nobody votes for any
labels between 3 and 1000000).

I think all the discussions on finding the mode (either in 1D or nD)
probably pre-dated the ++ operator. It could be that using the
vectorised ++ operator is a better way to do it - I doubt it though,
normally if histogram can do something then histogram will be the best
way! You'd also need to introduce a clumsy offset to deal with negative
selections (Not an issue for you here but would be if finding the mode
in a more general way).

It would make David's 1D example from his webpage into something like this:

array = [1, 1, 2 , 4, 1, 3, 3, 2, 4, 5, 3, 2, 2, 1, 2, 6,-3]
f=intarr(max(array)-min(array)+1)
f[array-min(array)]++
junk=max(f,idx)
mode=idx + min(array)
print,mode

again, with no idea on what would be more efficient. If you're doing
analysis on measurements (typically non-integers) then you'd need to
invoke histogram anyway to bin them before trying to find the mode.

Thanks,

Allan

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[Message index]

		majority voting By: Mort Canty on Wed, 11 February 2009 08:14
		Re: majority voting By: Allan Whiteford on Fri, 13 February 2009 01:31
		Re: majority voting By: David Fanning on Sat, 14 February 2009 08:44
		Re: majority voting By: JD Smith on Thu, 12 February 2009 14:45
		Re: majority voting By: Foldy Lajos on Thu, 12 February 2009 12:47
		Re: majority voting By: David Fanning on Thu, 12 February 2009 12:11
		Re: majority voting By: pgrigis on Thu, 12 February 2009 11:58
		Re: majority voting By: JD Smith on Thu, 12 February 2009 10:49
		Re: majority voting By: Mort Canty on Thu, 12 February 2009 07:45
		Re: majority voting By: JD Smith on Thu, 12 February 2009 07:08
		Re: majority voting By: David Fanning on Thu, 12 February 2009 05:43
		Re: majority voting By: Allan Whiteford on Thu, 12 February 2009 04:17
		Re: majority voting By: Mort Canty on Thu, 12 February 2009 01:15
		Re: majority voting By: JD Smith on Wed, 11 February 2009 15:20
		Re: majority voting By: ben.bighair on Wed, 11 February 2009 14:03
		Re: majority voting By: Allan Whiteford on Mon, 16 February 2009 04:11

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