Resizing pixels with independent scaling factor?
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JAI PRAKASH
am 15 Feb. 2019
Kommentiert: JAI PRAKASH
am 18 Feb. 2019
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 Kommentare
Jan
am 15 Feb. 2019
@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.
Akzeptierte Antwort
Jan
am 15 Feb. 2019
Bearbeitet: Jan
am 15 Feb. 2019
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 Kommentare
Jan
am 18 Feb. 2019
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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