spatial coordinates of BW image

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MANI MANI
MANI MANI 2013 年 11 月 6 日
編集済み: Jan 2013 年 11 月 6 日
I have a BW image in which few pixels are white rest all are black.. I want to store the spatial co-ordinates of only white pixels in to 2D matrix having only 2 columns(to represent x y coordinates) and any number of rows.. How do I do that? Pl. help me.

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Image Analyst
Image Analyst 2013 年 11 月 6 日
I don't know why that would be necessary, but if you insist:
[rows, columns] = find(BW); % 2 1-D vectors.
xy = [columns, rows]; % Stitch together to form N by 2 (x,y) array.
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MANI MANI
MANI MANI 2013 年 11 月 6 日
Thanks..it works.. I hv got a circle on pupil for which I need spatial coordinates.
Image Analyst
Image Analyst 2013 年 11 月 6 日
Why? The binary already is the coordinates. You don't need the x,y or row,col coordinates unless you are going to do some kind of inefficient loop instead of the faster vectorized method. Tell me what you want to do and I can help you avoid slowing down your program by looping over x,y or col,row.

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その他の回答 (1 件)

Jan
Jan 2013 年 11 月 6 日
Use find()
e.g.
A = randn(100,100) > .5; % example binary image
[row col] = find( A == 1 );
res = [ row col ];
This should give you the locations of the pixels containing a '1' in the matrix res.
  2 件のコメント
Image Analyst
Image Analyst 2013 年 11 月 6 日
res =[row col] is not (x,y) with your method - it's (y,x) because row = y and col = x.
Jan
Jan 2013 年 11 月 6 日
編集済み: Jan 2013 年 11 月 6 日
right, that should read res=[col, row] and equals your answer.
However, I only saw your answer after I posted mine, so it's futile anyway.

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