Storing points into an array

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Shane
Shane 2014-9-8
编辑： Andy L 2014-9-10
I have a Hyper-spectral image that is 760 X 520 X 361 which is also known as a data cube. I know if i want to find the value of a single pixel I would say Let p be the value/point and img be the image so therefore
p = img('75,104,361');
it should return a 1X1X361 matrix but instead i get a 1x10 what am i doing wrong? My other problem is I have a text file we will call "listp.txt" with 12 different points in it (i.e.75,104,361)( I tried in the below code to call list). How do i create a loop to call this list to p so that they can be stored as an array?
old=cd('C:\Users\Blarg\Desktop\Hal');
%%Get Full Image Data
% ENVI Import Cube and Register images
datafile = 'image_FlatField';
hdrfile = 'image_FlatField.hdr';
%%Read ENVI Files
[im, hdr] = enviread([old '\' datafile],[old '\' hdrfile]);
img = imrotate(im,90, 'nearest');
figure; imshow(img(:,:,361));
%%Reshape .txt into 12x1 list with each coordinate read in a columns as
% | 1 | 2 ...
%1| '342,212,361' |
%2| '313,147,361' |
%3| '295,212,361' |
%4 ...
d = reshape(textread('listp.txt', '%s'),1,12)';
  2 个评论
Andy L
Andy L 2014-9-8
How is your text file laid out - my assumption is each line is a set of co-ordinates? Are these then separated by commas, spaces? etc!
Shane
Shane 2014-9-8
they are laid out in the text file like this with a space between them
232,255,361 334,421,361 ...and so on
I have another .txt laid out as
x y z
232 255 361
334 421 361
... ... ...
I do not know which way I should go with this so any help would be appreciated. Thank you

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Andy L
Andy L 2014-9-8
编辑：Andy L 2014-9-10
For the first question I think that if you change the following line of code
p = img('75,104,361');
to
p = img(75,104,:); % for the full vector at Z.
this should solve your indexing issue. It seems no coincidence to me that your input is 10 characters long and you are getting a 1x10 output...
Based on your response to my comment above, if you have control over how the text file is laid out I would go with the second option (line by line). If you always want the full vector z at (x,y) then the z column in the text file is not necessary either.
[x , y, z] = textread('textExample.txt', '%u %u %u');
Returns column vectors for x y and z assuming the layout below
x y z
232 255 361
334 421 361
It's worth noting that you don't need to type x y and z in the text file - this may confuse things.
You can then use the following to extract the vector z at each coordinate:
for m = 1 : length(x)
p(i,:) = img(x(i),y(i),:); % Store vector z in row i of p.
end
  2 个评论
Shane
Shane 2014-9-8
编辑：Shane 2014-9-8
when i do this it gives me a single cell with a low value of 0.0067. So with the image, I have a x, y and z coordinate and all I want is the data on z vector of all 361 points at location x and y which I figure out I can get this by
p = img(75,104,:);
which gives me the 1X1X361 matrix i am looking for.
Thanks you

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更多回答（1 个）

Iain
Iain 2014-9-8
p = img(75, 104,:);

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