how to divide a grayscale image into 8*8 blocks and return it from my function?

1 Ansicht (letzte 30 Tage)
i do have a code, but it doesn't work quickly for 8*8. works perfectly for 32 and higher block sizes. it doesn't return the blocks either.
function y=segment1(f)
rgbImage = f;
imshow(rgbImage);
set(gcf, 'units','normalized','outerposition',[0 0 1 1]);
drawnow;
[rows columns numberOfColorBands] = size(rgbImage);
blockSizeR =8;
blockSizeC =8;
wholeBlockRows = floor(rows / blockSizeR);
blockVectorR = [blockSizeR * ones(1, wholeBlockRows), rem(rows, blockSizeR)];
wholeBlockCols = floor(columns / blockSizeC);
blockVectorC = [blockSizeC * ones(1, wholeBlockCols), rem(columns, blockSizeC)];
if numberOfColorBands > 1
ca = mat2cell(rgbImage, blockVectorR, blockVectorC, numberOfColorBands);
else
ca = mat2cell(rgbImage, blockVectorR, blockVectorC);
end
plotIndex = 1;
numPlotsR = size(ca, 1);
numPlotsC = size(ca, 2);
for r = 1 : numPlotsR
for c = 1 : numPlotsC
fprintf('plotindex = %d, c=%d, r=%d\n', plotIndex, c, r);
subplot(numPlotsR, numPlotsC, plotIndex);
rgbBlock=ca{r,c};
imshow(rgbBlock);
[rowsB columnsB numberOfColorBandsB] = size(rgbBlock);
caption = sprintf('Block #%d of %d\n%d rows by %d columns', plotIndex, numPlotsR*numPlotsC, rowsB, columnsB);
drawnow;
plotIndex = plotIndex + 1;
end
end
end

Akzeptierte Antwort

Walter Roberson
Walter Roberson am 30 Mär. 2016
Your code starts
function y=segment1(f)
so it will return whatever is stored in the variable "y" by the function. But you do not store anything into y, so it cannot return the blocks. You need to store the blocks in y.
  4 Kommentare
chaitanya g p
chaitanya g p am 5 Apr. 2016
hey. regarding the storing of value in y. how do i do that? please help me out.
Walter Roberson
Walter Roberson am 5 Apr. 2016
What is your defined output type? What kind of variable do you need to return? What is it that needs to be returned?

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by