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Re: find bimodal maximum in each row [message #74867]

Sun, 06 February 2011 12:27

Jeremy Bailin
Messages: 618
Registered: April 2008

Senior Member

That's easily constructed from peakp:

q = [[0,1,2,1,2,1,0], [1,2,0,0,2,2,0], [0,2,2,2,0,1,0]]
nx = (size(q, /dimen))[0]
ny = (size(q, /dimen))[1]

; is this a local maximum? (note that the index starts at element 1,* of q)
peakp = (q[1:nx-2,*] gt q[2:nx-1,*]) and (q[0:nx-3,*] le q[1:nx-2,*])
; pad peakp to make it the same dimensions as q and multiply
peakonly = q * [replicate(0,1,ny), peakp, replicate(0,1,ny)]

IDL> print, peakonly
0 0 2 0 2 0 0
0 2 0 0 0 2 0
0 0 0 2 0 1 0

-Jeremy.

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[Message index]

		Re: find bimodal maximum in each row By: Jeremy Bailin on Sun, 06 February 2011 12:27
		Re: find bimodal maximum in each row By: vijay on Fri, 04 February 2011 19:32
		Re: find bimodal maximum in each row By: Jeremy Bailin on Fri, 04 February 2011 12:57

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