How display different blocks of the images in one figure?

I have divided image into blocks. How to display it using one figure only?

[Wan Ji]

How many blocks, 4, 5 or 8?

[anu]

Total 5 blocks. 4 blocks along with central block.

[DGM]

Assemble the blocks back into one image by concatenation or indexing. If the central block contains no unique data, then it's redundant. Unless you provide a concrete example of what you did, you'll have to figure out the indexing yourself. Not all image geometries are integer-divisible by 2 or 4, so whatever rounding you did during detiling will need to be taken into account.

Answer 1

[Image Analyst]

First crop out the portions of the image you want then use subplot():

suplot(3, 3, 1);
imshow(upperLeftImage);
suplot(3, 3, 3);
imshow(upperRightImage);
suplot(3, 3, 5);
imshow(middleImage);
suplot(3, 3, 7);
imshow(lowerLeftImage);
suplot(3, 3, 9);
imshow(lowerRightImage);

Is that what you want?

[anu]

Actually central image should display as shown in attached image.

[Image Analyst]

You are showing the quadrants of the image stitched together. Thus that just gives the original image. So you can just simply do

imshow(yourOriginalImage);

If you want lines half way across you can do this:

yourOriginalImage = imread('peppers.png');
imshow(yourOriginalImage);
axis('on', 'image');
[rows, columns, numberOfColorChannels] = size(yourOriginalImage)
xline(columns/2, 'Color', 'y', 'LineWidth', 3);
yline(rows/2, 'Color', 'y', 'LineWidth', 3);
pos = [columns/2 - columns/4, rows/2 - rows/4, columns/2, rows/2];
rectangle('Position', pos, 'EdgeColor', 'y', 'LineWidth', 3, 'LineStyle', '--');

If that's not what you want, then show what you want with an actual image. Mock something up in Photoshop if you have to.

Answer 2

[DGM]

If you want them tiled as shown, then you're going to have to reassemble them into one image.

A = imread('cameraman.tif');

% geometry needs to be integer-divisible by 4
bksz = floor([size(A,1) size(A,2)]/4)

% split image
nw = A(1:2*bksz(1),1:2*bksz(2));
ne = A(1:2*bksz(1),2*bksz(2)+1:4*bksz(2));
sw = A(2*bksz(1)+1:4*bksz(1),1:2*bksz(2));
se = A(2*bksz(1)+1:4*bksz(1),2*bksz(2)+1:4*bksz(2));
md = A(bksz(1)+1:3*bksz(1),bksz(2)+1:3*bksz(2));

% reassemble image
B = [nw ne; sw se];
% if center block is nonunique, it's not needed
% if it is unique, then insert it
B(bksz(1)+1:3*bksz(1),bksz(2)+1:3*bksz(2)) = md;

immse(A,B)
imshow(B)

What I said about being integer-divisible by 4 still applies. If the source image geometry is not integer-divisible by 4, then reconstruction has to take into account how you detiled the image. The above code simply discards excess edge vectors. In the case of an improperly-sized source image, B will be smaller than A, though their NW corners will be aligned.