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From: Rahul on 28 Sep 2009 15:43
ImageAnalyst <imageanalyst(a)mailinator.com> wrote in news:3d17f33d-e6bf-
478d-8f63-9487df2f177e(a)c3g2000yqd.googlegroups.com:

> Of course using the original image data would be better than using
> some jpg with annotation on it and a colormap applied.
> Hopefully by "z" you mean intensity rather than a coordinate, unless
> your intensity actually corresponds to a height (I'm not really
> familiar with stm images).
>

Yes, the "z" is the "intensity". z was a function of x and y.

Hence the (x,y,z) tuples. Wanted maxima in z=f(x,y). Peaks in the z space
as a function of x and y. or "What is the height (in z units) of a peak and
where (x,y) are these peaks located)

I'm sorry if this wasn't clear earlier.

The jpeg simply has the z coordinate put on a color map.

Bruno: Did my original post confuse you? I'm sorry if it did.

--
Rahul
From: Rahul on 28 Sep 2009 15:45
"Bruno Luong" <b.luong(a)fogale.findmycountry> wrote in news:h9r2vi$qvm$1
@fred.mathworks.com:

> No, as I understand he OP has 3D array I(x,y,z) with
>
> x=1:nx;
> y=1:ny;
> z=1:nz;
>

I guess the array is 2D. z=f(x,y) but the data is still tuples of (x,y,z).
Sorry about the notation confusion.

A(5,5)=40

A(5,4)=30
A(5,6)=30
A(4,5)=30
A(6,5)=30



--
Rahul
From: Bruno Luong on 28 Sep 2009 15:51
Rahul <nospam(a)nospam.invalid> wrote in message <Xns9C9495C5DB00F6650A1FC0D7811DDBC81(a)188.40.43.213>...

>
> Bruno: Did my original post confuse you?

It did obviously.

Bruno
From: Rahul on 28 Sep 2009 16:34
"Bruno Luong" <b.luong(a)fogale.findmycountry> wrote in news:h9r436$dte$1
@fred.mathworks.com:

>> Bruno: Did my original post confuse you?
>
> It did obviously.
>

My bad. Obviously, my terminology was imprecise or I used notation not
common in the community.

--
Rahul
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