Storing points into an array

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)';

[Andy L]

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]

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

Answer 1

[Andy L]

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

[Shane]

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

[Andy L]

Check updated answer

Answer 2

[Iain]

 p = img(75, 104,:);