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Re: Help in Image Analysis [message #47744 is a reply to message #47615]

Thu, 02 March 2006 10:30

David Fanning
Messages: 11724
Registered: August 2001

Senior Member

David Fanning writes:

> Any 2D array, which is what you have in front of you,
> can be considered an "image". All you need do to is
> pull out of the data the rows you are interested in
> for further analysis:
>
> interestingStuff = data[*, 280:320]
>
> To display it as an image, you need to byte scale
> it (and perhaps take the absolute value, since your
> data is complex).
>
> TV, BytScl(Abs(interestingStuff))
>
> Now you are ready to perform any analysis that seems
> appropriate.

I'd try to find the row that had the maximum
value in the interesting stuff:

dims = Size(interestingStuff, /Dimension)
s = Scale_Vector(Abs(interestingStuff), 0, 1)
maxVal = Max(s, location)
row = location / dims[0]
Plot, s[*,row]

Now, you can easily find the peak values in this row, etc.

Cheers,

David

--
David Fanning, Ph.D.
Fanning Software Consulting, Inc.
Coyote's Guide to IDL Programming: http://www.dfanning.com/

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[Message index]

		Help in Image Analysis By: Bharani on Tue, 21 February 2006 02:33
		Re: Help in Image Analysis By: Peter Clinch on Thu, 23 February 2006 01:04
		Re: Help in Image Analysis By: Bharani on Thu, 02 March 2006 09:02
		Re: Help in Image Analysis By: Bharani on Wed, 22 February 2006 08:53
		Re: Help in Image Analysis By: Bharani on Wed, 22 February 2006 08:53
		Re: Help in Image Analysis By: btt on Wed, 22 February 2006 08:29
		Re: Help in Image Analysis By: Peter Clinch on Wed, 22 February 2006 08:10
		Re: Help in Image Analysis By: Bharani on Wed, 22 February 2006 07:47
		Re: Help in Image Analysis By: Peter Clinch on Wed, 22 February 2006 06:06
		Re: Help in Image Analysis By: David Fanning on Thu, 02 March 2006 10:30
		Re: Help in Image Analysis By: David Fanning on Thu, 02 March 2006 10:05

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