Scaling the image using Bilinear Interpolation

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Nick Armas
Nick Armas am 19 Nov. 2017
Kommentiert: Rik am 19 Nov. 2017
I have written a code that reads an image and does scaling of 2 units in x-axis direction. Scaling matrix is filled by values that are read from a text file.
Scaling Matrix looks like
2 0 0
0 1 0
0 0 1
Original Image
Transformed Image (Scaling of 2 units in X-direction)
Code
file = importdata('transform_c.txt');
fileData = file.data;
image = imread('mecca06.pgm');
[row, col] = size(image);
scalingMatrix = zeros(3,3);
scalingMatrix(1,1) = fileData(2);
scalingMatrix(1,2) = fileData(3);
scalingMatrix(1,3) = fileData(4);
scalingMatrix(2,1) = fileData(5);
scalingMatrix(2,2) = fileData(6);
scalingMatrix(2,3) = fileData(7);
scalingMatrix(3,1) = fileData(8);
scalingMatrix(3,2) = fileData(9);
scalingMatrix(3,3) = fileData(10);
m1Inverse = inv(scalingMatrix);
outputImage = applyTransformation(image, row, col, m1Inverse);
figure
imshow(outputImage);
function outImage = applyTransformation(image, row, col, m1Inverse)
points = zeros(3,1);
for i=1:row
for j=1:col
points(1,1) = i;
points(2,1) = j;
points(3,1) = 1;
m2 = m1Inverse * points;
x = m2(1,1);
y = m2(2,1);
xlb = floor(x);
ylb = floor(y);
if(xlb <= 0)
xlb = 1;
end
if(xlb > row)
xlb = row;
end
if(ylb <= 0)
ylb = 1;
end
if(ylb > col)
ylb = col;
end
xub = xlb+1;
yub = ylb+1;
if(xub <= 0)
xub = 1;
end
if(xub > row)
xub = row;
end
if(yub <= 0)
yub = 1;
end
if(yub > col)
yub = col;
end
exub = xub-x;
eyub = yub-y;
exlb = x-xlb;
eylb = y-ylb;
outImage(i,j) = (exub*eyub*image(xlb,ylb))+(exlb*eyub*image(xub,ylb))+(exub*eylb*image(xlb,yub))+(exlb*eylb*image(xub,yub));
end
end
end
Question
When i apply the transformation, image is cropped. My question is how can i modify the above code to get uncropped image ?
I want to get the following image
  1 Kommentar
Rik
Rik am 19 Nov. 2017
You don't pre-allocate the result and you loop only over the size of the original.

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