Resizing pixels with independent scaling factor?

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JAI PRAKASH 2019-2-15

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/445105-resizing-pixels-with-independent-scaling-factor

评论： JAI PRAKASH 2019-2-18

采纳的回答： Jan

在 MATLAB Online 中打开

How can be pixels resized with independent scaling factor?

Keeping size of final image same as initial image.

e.g.

+

=

Resizing method 'nearest' would be enough.

Right I am using some kind of 'for' loop but want to avoid it for faster processing.

Starting experiment data could be:

%% Creation of starting image
w=8;
im=255*ones(w,w,1, 'uint8');
im(2,2,:)=0;
im(6,6,:)=0;
%% Inputs to resize pixels
row=[2;6]; col=[2;6];
scale=[3;5];            % length of each of these inputs are same, explanation: pixel{2,2} will have a scaling factor 3 and so on..
                        % fractional scaling factor can be entertained as per nearest resizing method.
                        % length of these inputs are large, of the order of number of pixels in the image, thats why for loop takes time.
Any innovative ideas are welcome
Note: During resizing process sometimes pixels may overlap, in this case, new values will overwrite the old values
      i.e., pixels processed due to bottom values of [row, col & scale] can overwrite the previous values.

Thanks

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JAI PRAKASH 2019-2-15

Thanks for consideration @kalyan.

Initial image (SIze is [8x8])

Final image (SIze is [8x8])

Jan 2019-2-15

@JAI: Please post the current working code, if you want us to improve it. Maybe the loop is the most efficient method, when it is optimized.

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Answer 1

Jan 2019-2-15

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/445105-resizing-pixels-with-independent-scaling-factor#answer_361172

编辑：Jan 2019-2-15

在 MATLAB Online 中打开

Let's start with a working loop:

w  = 8;
im = repmat(uint8(255), w, w);
im(2,2,:) = 0;
im(6,6,:) = 0;
row   = [2;6];
col   = [2;6];
scale = [3;5];
for k = 1:numel(row)
    s = (scale(k) - 1) / 2;
    w(row(k) - s:row(k) + s, col(k) - s:col(k) + s) = ...
        w(row(k), col(k));
end

Is this your current code?

Maybe this is slightly faster or nicer:

s    = (scale - 1) / 2;
iRow = row - s;  % Safer: max(1, row - s);
fRow = row + s;  %        min(w, row + s);
iCol = col - s;  %        max(1, col - s);
fCol = col + s;  %        min(w, col + 2);
for k = 1:numel(row)
    w(iRow(k):fRow(k), iCol(k):fCol(k)) = w(row(k), col(k));
end

The assignment inside the loop is vectorized already. If an overlap is possible, a fully vectorized method is hard or impossible. I assume, that the creation of the large index array will take more time than this compact loop.

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Jan 2019-2-18

A parallelization is strange here: The code overwrites elements of the input data and the order of loops matters. Then a parallel version cannot create the same output.

JAI PRAKASH 2019-2-18