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Re: max, mean, min of array [message #29004 is a reply to message #28924]

Thu, 24 January 2002 07:29

Craig Markwardt
Messages: 1869
Registered: November 1996

Senior Member

Alex Schuster <Wonko@planet-interkom.de> writes:
> It's possible without, um, with fewer loops:
>
> zdim = (size( array, /dimension ))[2]
> pixmin = ( pixmax = array[*,*,0] )
> for i = 1, zdim-1 do begin
> pixmax = pixmax > array[*,*,i]
> pixmin = pixmin < array[*,*,i]
> endfor
> pixmean = total( array, 3 ) / zdim
>
> Hey Craig, I think with this method you can get rid of your ho, hum
> comment in cmapply.pro.

Very cool! I think I've been "outvectored" on this one.

Alex, I think *both* solutions should be possible. Consider the
following scenario: instead of a 400x400x12 array, how about a
2x2x1000000 array? Using your technique we would end up doing 1000000
iterations, but with mine it would only be 4.

Thus, the code should contain both solutions, and pick whichever one
takes fewer iterations.

Craig

--
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Craig B. Markwardt, Ph.D. EMAIL: craigmnet@cow.physics.wisc.edu
Astrophysics, IDL, Finance, Derivatives | Remove "net" for better response
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[Message index]

		Re: max, mean, min of array By: dinhnq on Mon, 07 January 2002 02:52
		Re: max, mean, min of array By: Craig Markwardt on Sun, 06 January 2002 12:25
		Re: max, mean, min of array By: Wayne Landsman on Sun, 06 January 2002 11:50
		Re: max, mean, min of array By: David Fanning on Sun, 06 January 2002 07:42
		Re: max, mean, min of array By: Alex Schuster on Wed, 23 January 2002 07:04
		Re: max, mean, min of array By: Craig Markwardt on Thu, 24 January 2002 07:29

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