Resizing pixels with independent scaling factor?
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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
3 个评论
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.
采纳的回答
Jan
2019-2-15
编辑:Jan
2019-2-15
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.
6 个评论
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.
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