Changing image aspect ratio of interpolated RGB image. Square to rectangular - matlab

I have some code which takes a fish eye images and converts it to a rectangular image in each RGB channels. I am having trouble with the fact the the output image is square instead of rectangular. (this means that the image is distorted, compressed horizontally.) I have tried changing the output matrix to a more suitable format, without success. Besides this i have also discovered that for the code to work the input image must be square like 500x500. Any idea how to solve this issue? This is the code:
The code is inspired by Prakash Manandhar "Polar To/From Rectangular Transform of Images" file exchange on mathworks.
EDIT. Code now works.
function imP = FISHCOLOR2(imR)
rMin=0.1;
rMax=1;
[Mr, Nr, Dr] = size(imR); % size of rectangular image
xRc = (Mr+1)/2; % co-ordinates of the center of the image
yRc = (Nr+1)/2;
sx = (Mr-1)/2; % scale factors
sy = (Nr-1)/2;
reduced_dim = min(size(imR,1),size(imR,2));
imR = imresize(imR,[reduced_dim reduced_dim]);
M=size(imR,1);N=size(imR,2);
dr = (rMax - rMin)/(M-1);
dth = 2*pi/N;
r=rMin:dr:rMin+(M-1)*dr;
th=(0:dth:(N-1)*dth)';
[r,th]=meshgrid(r,th);
x=r.*cos(th);
y=r.*sin(th);
xR = x*sx + xRc;
yR = y*sy + yRc;
for k=1:Dr % colors
imP(:,:,k) = interp2(imR(:,:,k), xR, yR); % add k channel
end
imP = imresize(imP,[size(imP,1), size(imP,2)/3]);
imP = imrotate(imP,270);
SOLVED
Input image <- Image link
Output image <- Image link

PART A
To remove the requirement of a square input image, you may resize the input image into a square one with this -
%%// Resize the input image to make it square
reduced_dim = min(size(imR,1),size(imR,2));
imR = imresize(imR,[reduced_dim reduced_dim]);
Few points I would like to raise here though about this image-resizing to make it a square image. This was a quick and dirty approach and distorts the image for a non-square image, which you may not want if the image is not too "squarish". In many of those non-squarish images, you would find blackish borders across the boundaries of the image. If you can remove that using some sort of image processing algorithm or just manual photoshoping, then it would be ideal. After that even if the image is not square, imresize could be considered a safe option.
PART B
Now, after doing the main processing of flattening out the fisheye image,
at the end of your code, it seemed like the image has to be rotated
90 degrees clockwise or counter-clockwise depending on if the fisheye
image have objects inwardly or outwardly respectively.
%%// Rotating image
imP = imrotate(imP,-90); %%// When projected inwardly
imP = imrotate(imP,-90); %%// When projected outwardly
Note that the flattened image must have the height equal to the half of the
size of the input square image, that is the radius of the image.
Thus, the final output image must have number of rows as - size(imP,2)/2
Since you are flattening out a fisheye image, I assumed that the width
of the flattened image must be 2*PI times the height of it. So, I tried this -
imP = imresize(imP,[size(imP,2)/2 pi*size(imP,2)]);
But the results looked too flattened out. So, the next logical experimental
value looked like PI times the height, i.e. -
imP = imresize(imP,[size(imP,2)/2 pi*size(imP,2)/2]);
Results in this case looked good.

I'm not very experienced in the finer points of image processing in MATLAB, but depending on the exact operation of the imP fill mechanism, you may get what you're looking for by doing the following. Change:
M = size(imR, 1);
N = size(imR, 2);
To:
verticalScaleFactor = 0.5;
M = size(imR, 1) * verticalScaleFactor;
N = size(imR, 2);
If my hunch is right, you should be able to tune that scale factor to get the image just right. It may, however, break your code. Let me know if it doesn't work, and edit your post to flesh out exactly what each section of code does. Then we should be able to give it another shot. Good luck!

This is the code which works.
function imP = FISHCOLOR2(imR)
rMin=0.1;
rMax=1;
[Mr, Nr, Dr] = size(imR); % size of rectangular image
xRc = (Mr+1)/2; % co-ordinates of the center of the image
yRc = (Nr+1)/2;
sx = (Mr-1)/2; % scale factors
sy = (Nr-1)/2;
reduced_dim = min(size(imR,1),size(imR,2));
imR = imresize(imR,[reduced_dim reduced_dim]);
M=size(imR,1);N=size(imR,2);
dr = (rMax - rMin)/(M-1);
dth = 2*pi/N;
r=rMin:dr:rMin+(M-1)*dr;
th=(0:dth:(N-1)*dth)';
[r,th]=meshgrid(r,th);
x=r.*cos(th);
y=r.*sin(th);
xR = x*sx + xRc;
yR = y*sy + yRc;
for k=1:Dr % colors
imP(:,:,k) = interp2(imR(:,:,k), xR, yR); % add k channel
end
imP = imresize(imP,[size(imP,1), size(imP,2)/3]);
imP1 = imrotate(imP1,270);

Related

How to "disturb" an image with noise, but keep the orginal undisturbed pixel intact, and write to file in imagesc

For simple explanation of the title, imagine you're using a Photoshop layers, where noise is on the top layer.
% load a test image
I = rgb2gray(imread('peppers.png'));
% recreate image
cmap = colormap(); % grab current colormap
ncolors = size(cmap,1);
% do what imagesc does
Iind = double(I) - double(min(I(:)));
Iind = Iind / max(Iind(:));
% quantize image
Iind = round(Iind * ncolors + 0.5);
Iind(Iind > ncolors) = ncolors;
Iind(Iind < 1) = 1;
% convert to RGB from indexed image using cmap as palette
Irgb = ind2rgb(Iind,cmap);
imwrite(Irgb, 'filename.bmp');
These code will scale the colormap and write to file. However, by adding a artificial noise at a certain pixel location after loaded the image:
I(50,50) = rand(1);
This will generate a completely different colormap visually, and the one with the noise will look a little washed out. Top image is the original and bottom is with noise.
EDIT: The below image is the cropped image on the top left corner. Left is original and right is with noise. On the right you can see the color washed out a little bit, and if you look closely you can see a noise (dark blue color, position (50,50)).
Any idea how to still maintain the original after noise was added? Thanks in advance!
The issue is that you're adding noise (between 0 and 1) to the original image which has min = 8 and max = 255. This is reducing the new min to either 0 or 1 (the input image is of type uint8, and it seems MATLAB rounds the random number) which stretches out the colourmap slightly. There are two options to avoid this.
Add random noise in the range of the original image
I(50,50) = randi([min(I(:)), max(I(:))]);
Add random noise after scaling the original image
Irgb(50,50) = rand(1);

Smooth and fit edge of binary images

i am working on a research about the swimming of fishes using analysis of videos, then i need to be carefully with the images (obtained from video frames) with emphasis in the tail.
The images are in High-Resolution and the software that i customize works with binary images, because is easy to use maths operations on this.
For obten this binary images i use 2 methods:
1)Convert the image to gray, invert the colors,later to bw and finally to binary with a treshold that give me images like this, with almost nothing of noise. The images sometimes loss a bit of area and doesn't is very exactly with the tail(now i need more acurracy for determinate the amplitude of tail moves)
image 1
2)i use this code, for cut the border that increase the threshold, this give me a good image of the edge, but i dont know like joint these point and smooth the image, or fitting binary images, the app fitting of matlab 2012Rb doesn't give me a good graph and i don't have access to the toolboxs of matlab.
s4 = imread('arecorte.bmp');
A=[90 90 1110 550]
s5=imcrop(s4,A)
E = edge(s5,'canny',0.59);
image2
My question is that
how i can fit the binary image or joint the points and smooth without disturb the tail?
Or how i can use the edge of the image 2 to increase the acurracy of the image 1?
i will upload a image in the comments that give me the idea of the method 2), because i can't post more links, please remember that i am working with iterations and i can't work frame by frame.
Note: If i ask this is because i am in a dead point and i don't have the resources to pay to someone for do this, until this moment i was able to write the code but in this final problem i can't alone.
I think you should use connected component labling and discard the small labels and than extract the labels boundary to get the pixels of each part
the code:
clear all
% Read image
I = imread('fish.jpg');
% You don't need to do it you haef allready a bw image
Ibw = rgb2gray(I);
Ibw(Ibw < 100) = 0;
% Find size of image
[row,col] = size(Ibw);
% Find connceted components
CC = bwconncomp(Ibw,8);
% Find area of the compoennts
stats = regionprops(CC,'Area','PixelIdxList');
areas = [stats.Area];
% Sort the areas
[val,index] = sort(areas,'descend');
% Take the two largest comonents ids and create filterd image
IbwFilterd = zeros(row,col);
IbwFilterd(stats(index(1,1)).PixelIdxList) = 1;
IbwFilterd(stats(index(1,2)).PixelIdxList) = 1;
imshow(IbwFilterd);
% Find the pixels of the border of the main component and tail
boundries = bwboundaries(IbwFilterd);
yCorrdainteOfMainFishBody = boundries{1}(:,1);
xCorrdainteOfMainFishBody = boundries{1}(:,2);
linearCorrdMainFishBody = sub2ind([row,col],yCorrdainteOfMainFishBody,xCorrdainteOfMainFishBody);
yCorrdainteOfTailFishBody = boundries{2}(:,1);
xCorrdainteOfTailFishBody = boundries{2}(:,2);
linearCorrdTailFishBody = sub2ind([row,col],yCorrdainteOfTailFishBody,xCorrdainteOfTailFishBody);
% For visoulaztion put color for the boundries
IFinal = zeros(row,col,3);
IFinalChannel = zeros(row,col);
IFinal(:,:,1) = IFinalChannel;
IFinalChannel(linearCorrdMainFishBody) = 255;
IFinal(:,:,2) = IFinalChannel;
IFinalChannel = zeros(row,col);
IFinalChannel(linearCorrdTailFishBody) = 125;
IFinal(:,:,3) = IFinalChannel;
imshow(IFinal);
The final image:

Applying temporal median filter to a video

I want to apply Temporal Median Filter to a depth map video to ensure temporal consistency and prevent the flickering effect.
Thus, I am trying to apply the filter on all video frames at once by:
First loading all frames,
%%% Read video sequence
numfrm = 5;
infile_name = 'depth_map_1920x1088_80fps.yuv';
width = 1920; %xdim
height = 1088; %ydim
fid_in = fopen(infile_name, 'rb');
[Yd, Ud, Vd] = yuv_import(infile_name,[width, height],numfrm);
fclose(fid_in);
then creating a 3-D depth matrix (height x width x number-of-frames),
%%% Build a stack of images from the video sequence
stack = zeros(height, width, numfrm);
for i=1:numfrm
RGB = yuv2rgb(Yd{i}, Ud{i}, Vd{i});
RGB = RGB(:, :, 1);
stack(:,:,i) = RGB;
end
and finally applying the 1-D median filter along the third direction (time)
temp = medfilt1(stack);
However, for some reason this is not working. When I try to view each frame, I get white images.
frame1 = temp(:,:,1);
imshow(frame1);
Any help would be appreciated!
My guess is that this is actually working but frame1 is of class double and contains values, e.g. between 0 and 255. As imshow represents double images by default on a [0,1] scale, you obtain a white, saturated image.
I would therefore suggest:
caxis auto
after imshow to fix the display problem.
Best,

Matlab fill shapes by white

As you see, I have shapes and their white boundaries. I want to fill the shapes in white color.
The input is:
I would like to get this output:
Can anybody help me please with this code? it doesn't change the black ellipses to white.
Thanks alot :]]
I = imread('untitled4.bmp');
Ibw = im2bw(I);
CC = bwconncomp(Ibw); %Ibw is my binary image
stats = regionprops(CC,'pixellist');
% pass all over the stats
for i=1:length(stats),
size = length(stats(i).PixelList);
% check only the relevant stats (the black ellipses)
if size >150 && size < 600
% fill the black pixel by white
x = round(mean(stats(i).PixelList(:,2)));
y = round(mean(stats(i).PixelList(:,1)));
Ibw = imfill(Ibw, [x, y]);
end;
end;
imshow(Ibw);
Your code can be improved and simplified as follows. First, negating Ibw and using BWCONNCOMP to find 4-connected components will give you indices for each black region. Second, sorting the connected regions by the number of pixels in them and choosing all but the largest two will give you indices for all the smaller circular regions. Finally, the linear indices of these smaller regions can be collected and used to fill in the regions with white. Here's the code (quite a bit shorter and not requiring any loops):
I = imread('untitled4.bmp');
Ibw = im2bw(I);
CC = bwconncomp(~Ibw, 4);
[~, sortIndex] = sort(cellfun('prodofsize', CC.PixelIdxList));
Ifilled = Ibw;
Ifilled(vertcat(CC.PixelIdxList{sortIndex(1:end-2)})) = true;
imshow(Ifilled);
And here's the resulting image:
If your images are all black&white, and you have the image processing toolkit, then this looks like what you need:
http://www.mathworks.co.uk/help/toolbox/images/ref/imfill.html
Something like:
imfill(image, [startX, startY])
where startX, startY is a pixel in the area that you want to fill.

How to measure the rotation of a image in MATLAB?

I have two images. One is the original, and the another is rotated.
Now, I need to discover the angle that the image was rotated. Until now, I thought about discovering the centroids of each color (as every image I will use has squares with colors in it) and use it to discover how much the image was rotated, but I failed.
I'm using this to discover the centroids and the color in the higher square in the image:
i = rgb2gray(img);
bw = im2bw(i,0.01);
s = regionprops(bw,'Centroid');
centroids = cat(1, s.Centroid);
colors = impixel(img,centroids(1),centroids(2));
top = max(centroids);
topcolor = impixel(img,top(1),top(2));
You can detect the corners of one of the colored rectangles in both the image and the rotated version, and use these as control points to infer the transformation between the two images (like in image registration) using the CP2TFORM function. We can then compute the angle of rotation from the affine transformation matrix:
Here is an example code:
%# read first image (indexed color image)
[I1 map1] = imread('http://i.stack.imgur.com/LwuW3.png');
%# constructed rotated image
deg = -15;
I2 = imrotate(I1, deg, 'bilinear', 'crop');
%# find blue rectangle
BW1 = (I1==2);
BW2 = imrotate(BW1, deg, 'bilinear', 'crop');
%# detect corners in both
p1 = corner(BW1, 'QualityLevel',0.5);
p2 = corner(BW2, 'QualityLevel',0.5);
%# sort corners coordinates in a consistent way (counter-clockwise)
p1 = sortrows(p1,[2 1]);
p2 = sortrows(p2,[2 1]);
idx = convhull(p1(:,1), p1(:,2)); p1 = p1(idx(1:end-1),:);
idx = convhull(p2(:,1), p2(:,2)); p2 = p2(idx(1:end-1),:);
%# make sure we have the same number of corner points
sz = min(size(p1,1),size(p2,1));
p1 = p1(1:sz,:); p2 = p2(1:sz,:);
%# infer transformation from corner points
t = cp2tform(p2,p1,'nonreflective similarity'); %# 'affine'
%# rotate image to match the other
II2 = imtransform(I2, t, 'XData',[1 size(I1,2)], 'YData',[1 size(I1,1)]);
%# recover affine transformation params (translation, rotation, scale)
ss = t.tdata.Tinv(2,1);
sc = t.tdata.Tinv(1,1);
tx = t.tdata.Tinv(3,1);
ty = t.tdata.Tinv(3,2);
translation = [tx ty];
scale = sqrt(ss*ss + sc*sc);
rotation = atan2(ss,sc)*180/pi;
%# plot the results
subplot(311), imshow(I1,map1), title('I1')
hold on, plot(p1(:,1),p1(:,2),'go')
subplot(312), imshow(I2,map1), title('I2')
hold on, plot(p2(:,1),p2(:,2),'go')
subplot(313), imshow(II2,map1)
title(sprintf('recovered angle = %g',rotation))
If you can identify a color corresponding to only one component it is easier to:
Calculate the centroids for each image
Calculate the mean of the centroids (in x and y) for each image. This is the "center" of each image
Get the red component color centroid (in your example) for each image
Subtract the mean of the centroids for each image from the red component color centroid for each image
Calculate the ArcTan2 for each of the vectors calculated in 4), and subtract the angles. That is your result.
If you have more than one figure of each color, you need to calculate all possible combinations for the rotation and then select the one that is compatible with the other possible rotations.
I could post the code in Mathematica, if you think it is useful.
I would take a variant to the above mentioned approach:
% Crude binarization method to knock out background and retain foreground
% features. Note one looses the cube in the middle
im = im > 1
Then I would get the 2D autocorrelation:
acf = normxcorr2(im, im);
From this result, one can easily detect the peaks, and as rotation carries into the autocorrelation function (ACF) domain, one can ascertain the rotation by matching the peaks between the original ACF and the ACF from the rotated image, for example using the so-called Hungarian algorithm.