Obtain Submatrix

Hi, need absolutely your help as soon as possible.

I selected a rectangle from an image(that contains values), than trough show m code i obtained the position of rect: [x1 y1 width height]...than i got:

   x2=x1+width;

   y2=y1+height; %(x are columns and y are rows)

[rows columns numberOfColorChannels]=size(NameMatrix);

column1=int32(x1*columns);

column2=int32(x2*columns);

row1=int32(y1*rows);

row2=int32(y2*rows);

submatrix=NameMatrix(row1:row2,column1:column2);

I thought that was the solution but if i do imagesc(submatrix) it appears another part of image and not the area i wanted...why?Is it right or miss something?

[Walter Roberson]

I really have no idea why you are multiplying x1 by columns ?? What is the mechanism you used to obtain the rectangle position, which call?

Answer 1

[Vincenzo]

I multiply for the columns and rows because they give me the number of original matrix just to cover the area i chose...and someone told me this..

http://www.mathworks.it/matlabcentral/answers/13951-how-to-create-image-matrix-via

In any case here there's the explanation of [left right width height]

what should i do?

http://www.mathworks.it/help/techdoc/creating_plots/f5-40107.html

Answer 2

[Vincenzo]

Any idea??

Answer 3

[Image Analyst]

Like Walter said, DON'T multiply x and y by rows or columns. "Someone" told you incorrectly to do that. That's nonsense unless x and y are in the range 0-1 instead of 1-rows and 1-columns, which I doubt. If you got x and y from imrect(), rbbox() or similar, they are already in units of rows and columns so just use them:

x2=x1+width;
y2=y1+height; %(x are columns and y are rows)
[rows columns numberOfColorChannels]=size(NameMatrix);
column1=int32(x1);
column2=int32(x2);
row1=int32(y1);
row2=int32(y2);
submatrix=NameMatrix(row1:row2,column1:column2);

You'll probably also want to write a function to clip x to between 1 and columns, and y to between 1 and rows since, depending on how you got those x and y it might be possible for them to be outside the range of your image.

[Walter Roberson]

That "Someone" was you, I.A. ;-)

imshow -> insert rectangle

The previous discussion does appear to show normalized coordinates, but I never use that particular function so I have no experience with what it returns.

[Image Analyst]

I don't remember that. If I did, then it must have been like I said, using values for x and y that went between 0 and 1.

[Image Analyst]

Oh, I see (http://www.mathworks.com/matlabcentral/answers/18424-sequence-of-submatrix-based-on-the-position-of-a-rectangle) - that was where I was correcting where he swapped rows and columns, but I never addressed what the range of x and y were, and didn't give him the original equation where he multiplied x by the number of columns. I still don't know where the x and y came from or why he apparently has them in the range of 0-1. If they are, then his original code should work, unless he somehow mixed up x&y and rows&columns again.

[Walter Roberson]

It was

http://www.mathworks.com/matlabcentral/answers/13951-how-to-create-image-matrix-via#anwer_19104

[Image Analyst]

Ah, ok, so they *were* normalized values. So then I'm not sure why his code doesn't work. Probably has something to do with trying to do things from code, instead of interactively drawing a rectangle and having the figure generate m-code for him. Probably something about the units property of the figure. He needs to manually check what the values of row1, row2, column1, column2 are and see if they make sense. Turning "axis" on may help with that so he can see what the actual row and column numbers are.

Answer 4

[Walter Roberson]

I would be concerned about indexing going up versus indexing going down. Matrix indexing goes down (that is, the top left corner is said to be (1,1)), but it is common to flip matrices over as they are displayed so that the bottom left corner is (1,1) and y increases upwards.

You have to know which coordinate system the [x1 y1 width height] is using in order to be sure you get the right indices.

[Image Analyst]

Good point. imshow() has a "YData" property where I think you can reverse this, though I haven't experimented around with it. Usually if you call rbbox() or imrect() it seems to know the y direction and return the y coords correctly. I think there's a "ydir reverse" command that could also be worth looking into. But I'd just figure out how to get x and y in pixel coordinates correctly in the first place rather than normalized units.

Answer 5

[Vincenzo]

I did the follow thing:

insert rectangle->Show Property Editor->More Properties->Units->Pixels it change me automatically the position from 0.34.... to 100... but i can't put this numbers because are too high (Indiced Exceed Matrix dimension), so i right click in the area and chose "show m code"...it appears this...

http://imageshack.us/photo/my-images/15/sub2a.jpg/

so i put the values in x1,x2,width and height

PS. x2=x1+height and y2=y1+width;

and that's what appear after doing column1=int32(x1*columns) and so on...till sub=NAME(row1:row2,column1:column2):

http://imageshack.us/photo/my-images/339/sub4s.jpg/

As you can see there is only a part of rectangle ,not all..why?

Answer 6

[Image Analyst]

Vincenzo: If you look at your first figure, you'll see that the units are already in pixels. They're not normalized in the range of 0 to 1. So you don't want to multiply by the number of rows or columns. I'm not sure why, after all the discussion from Walter and me, that you didn't just try what we said, which was to not multiply by the number of rows and columns. It would take just a few seconds to try it. You changed the units to pixels interactively but then you said you still did this: "column1=int32(x1*columns) " which is what we said NOT to do if x1 is already in pixels. Now, drawing rectangles gives you fractional units so if you see things like x1=0.34 then you need to clip x1,x2,y1,y2 to the appropriate range (1 to rows) or (1 to columns) because you can't have a fractional index into your image matrix.

P.S. x corresponds to the width, not height, and y corresponds to the height, not the other way around like you had in your PS.

Answer 7

[Vincenzo]

Sorry but i don't understand what i have to do ...I tried a lot of ways but nothing...reassuming: after i reached the position [x1,y1,width,height] in pixels unit (always in 0.34,0.23,..), i did x2=x1+width; y2=y1+height;...now the column1,column2,row1 and row2? I don't know..please can you write this?

[Walter Roberson]

My approach would be to stop trying to use the insert rectangle tool and instead switch to something like ginput()

[Image Analyst]

Or rbbox().