Finding row and column number of values in an image matrix

Hi all,

As the picture attached, on line 39, I am trying to use Find to return row and column numbers of a selected area in the image matrix.

However, it returns Col_r and Row_r with unexpected numbers of values (more than 500k values).

I am not sure whether I use the appropriate function to obtain the row and column numbers of values in the image matrix.

Could you please help provide some suggestion for me. Thank you so much.

[Voss]

The syntax and logic seem ok. My guess would be that the 0.415 is wrong. Maybe plot a histogram of Img12 at that point in the code and see if you get a distribution of values you expect:

histogram(Img12(:))

On the other hand, maybe it is not wrong at all. 500k elements represent about 14% of the relevant region in Img12 (rows 921 to 1126, going by the values of the variables shown in the workspace panel on the right). Does that seem reasonable?

By the way you can write that line more concisely as:

[Row_r, Col_r] = find(Img12(fix(0.9*Rows1(1)/2):fix(1.1*Rows1(1)/2),:)>0.415);

[Chutiphon Moranon]

Thanks for your reply. I think it is because of the threshold (0.415). When I see the histogram, I find that there is a number of the values after the threshold. I have got another question.

How can I obtain a value (in histogram) of a specific location in my image? Thank you.

[Voss]

To create a histogram of the region of 2D matrix Img12 containing rows start_row to end_row and columns start_column to end_column, you can say:

histogram(Img12(start_row:end_row,start_column:end_column))

i.e., just take the relevant part of Img12 using indexing when calling histogram().

Answer 1

[KSSV]

If your image is a RGB image, the number of column will be product of second and third dimension if you just use size function.

DEmo

I = rand(10,10,3) ; 
[m,n] = size(I) 

m = 10

n = 30

% You have to use 
[m,n,p] = size(I)

m = 10

n = 10

p = 3

Change your line number 28, so that you get Img12L value right.

[Chutiphon Moranon]

Thank you so much for your reply.