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Re: Gradient of an Image [message #33720 is a reply to message #33719]

Thu, 23 January 2003 23:05

meron
Messages: 51
Registered: July 1995

Member

In article <3E30D896.7000602@ee.uwa.edu.au>, Thomas Gutzler <tgutzler@ee.uwa.edu.au> writes:
> Hi,
>
> is it true that IDL currently has no function to calculate the gradient
> of an image ?
> Google found this for me:
> http://groups.google.com/groups?q=gradient+image+group:comp. lang.idl-pvwave&hl=en&lr=&ie=UTF-8&oe=UTF-8& amp;selm=3396D128.4CE0%40dlr.de&rnum=1
> and I am about to test, if this is doing the same as "Digital Image
> Processing" by Gonzales/Woods sais on page 418ff.
> Would be interesting to know, if anybody else has diffent (quicker?)
> solutions.
>
If it is just the absolute value of the gradient you're after, then
I've a function like this (written many years ago). And it can easily
be modified to split the components of the gradient. The function is
called ABGRAD and you'll find it in the IDL users contributions page,
in my library (MIDL).

Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"

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[Message index]

		Re: Gradient of an Image By: Thomas Gutzler on Fri, 24 January 2003 00:23
		Re: Gradient of an Image By: meron on Thu, 23 January 2003 23:05
		Re: Gradient of an Image By: David Fanning on Fri, 24 January 2003 06:30
		Re: Gradient of an Image By: meron on Fri, 24 January 2003 00:45

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