I am attempting to create a radial gradient image to look like the following using Matlab. The image needs to be of size 640*640*3 as I have to blend it with another image of that size. I have written the following code but the image that prints out is simply a grey circle on a black background with no fading around the edges.
p = zeros(640,640,3);
for i=1:640
for j=1:640
d = sqrt((i-320)^2+(j-320)^2);
if d < 640/3
p(i,j,:) = .5;
elseif d > 1280/3
p(i,j,:) = 0;
else
p(i,j,:) = (1 + cos(3*pi)*(d-640/3))/4;
end
end
end
imshow(p);
Any help would be greatly appreciated as I am new to Matlab.
Change:
p(i,j,:) = (1 + cos(3*pi)*(d-640/3))/4;
to
p(i,j,:) = .5-( (.5-0)*(d-640/3)/(640/3)) ;
This is an example of linear interpolation, where the grey value from the inner rim drops linearly to the background.
You can try other equations to have different kinds of gradient fading!
If you look more closely on your third case (which by the way should be a simple else instead of elseif), you can see that you have
= (1 + cos(3*pi))*...
Since cos(3*pi) = -1, this will always be 0, thus making all pixels within that range black. I assume that you would want a "d" in there somewhere.
Related
after leaf segmentation i got the following binary image:
Is there a way to fill the gaps caused by the similiarity of the veins with the background? I've tried to use imclose, or imdilate etc but it affects teeth shape. I can't find out how to fill these gaps without affecting teeth shape.
You may try bwfill(I, 'hols'), with out without imclose:
I = imbinarize(rgb2gray(imread('leaf.jpg')));
I = I(3:end-4, 1:end-8); %Remove white frame
J = imclose(I, ones(2)); %Minor affect the teeth shape (result looks better with imclose).
K = bwfill(J, 'hols'); %Fill the black hols
Result:
In case you want to fill the "vein gaps", you can try the following approach:
I = imbinarize(rgb2gray(imread('leaf.jpg')));
I = I(3:end-4, 1:end-8); %Remove white frame
I = bwfill(I, 'hols'); %Fill small black hols.
J = imerode(imdilate(I, strel('disk',5)), strel('disk',10)); %Dilate with radius 5 and erode with 10
T = (I == 0) & (J == 1); %Create mask with 1 where I is black and J is white "vein mask".
K = I;
K(T) = 1; %Fill "vein mask" in I with white.
K = bwfill(K, 'hols'); %Fill small black hols (fill tiny holds left).
Result:
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);
Hi all I have a stack of images of fluorescent labeled particles that are moving through time. The imagestack is gray scaled.
I computed a maximum intensity projection by taking the maximum of the image stack in the 3rd dimension.
Example:
ImageStack(x,y,N) where N = 31 image frames.
2DProjection = max(ImageStack,[],3)
Now, since the 2D projection image is black and white, I was hoping to assign a color gradient so that I can get a sense of the flow of particles through time. Is there a way that I can overlay this image with color, so that I will know where a particle started, and where it ended up?
Thanks!
You could use the second output of max to get which frame the particular maximum came from. max returns an index matrix which indicates the index of each maximal value, which in your case will be the particular frame in which it occurred. If you use this with the imagesc function, you will be able to plot how the particles move with time. For instance:
ImageStack(x,y,N) where N = 31 image frames.
[2DProjection,FrameInfo] = max(ImageStack,[],3);
imagesc(FrameInfo);
set(gca,'ydir','normal'); % Otherwise the y-axis would be flipped
You can sum up bright pixels of each image with one another after coloring each image. This way you will have mixed colors on overlapped areas which you will miss using max function. Although I like the previous answer more than mine.
hStep = 1/N;
currentH = 0;
resultImage = uint8(zeros(x,y,3));
for i = 1 : N
rgbColor = hsv2rgb(currentH,1,0.5);
resultImage(:,:,1) = resultImage(:,:,1) + im(:,:,i) * rgbColor(1);
resultImage(:,:,2) = resultImage(:,:,2) + im(:,:,i) * rgbColor(2);
resultImage(:,:,3) = resultImage(:,:,3) + im(:,:,i) * rgbColor(3);
currentH = currentH + hStep;
end
I am trying to rotate the image manually using the following code.
clc;
m1 = imread('owl','pgm'); % a simple gray scale image of order 260 X 200
newImg = zeros(500,500);
newImg = int16(newImg);
rotationMatrix45 = [cos((pi/4)) -sin((pi/4)); sin((pi/4)) cos((pi/4))];
for x = 1:size(m1,1)
for y = 1:size(m1,2)
point =[x;y] ;
product = rotationMatrix45 * point;
product = int16(product);
newx =product(1,1);
newy=product(2,1);
newImg(newx,newy) = m1(x,y);
end
end
imshow(newImg);
Simply I am iterating through every pixel of image m1, multiplying m1(x,y) with rotation matrix, I get x',y', and storing the value of m1(x,y) in to `newImg(x',y')' BUT it is giving the following error
??? Attempted to access newImg(0,1); index must be a positive integer or logical.
Error in ==> at 18
newImg(newx,newy) = m1(x,y);
I don't know what I am doing wrong.
Part of the rotated image will get negative (or zero) newx and newy values since the corners will rotate out of the original image coordinates. You can't assign a value to newImg if newx or newy is nonpositive; those aren't valid matrix indices. One solution would be to check for this situation and skip such pixels (with continue)
Another solution would be to enlarge the newImg sufficiently, but that will require a slightly more complicated transformation.
This is assuming that you can't just use imrotate because this is homework?
The problem is simple, the answer maybe not : Matlab arrays are indexed from one to N (whereas in many programming langages it's from 0 to (N-1) ).
Try newImg( max( min(1,newX), m1.size() ) , max( min(1,newY), m1.size() ) ) maybe (I don't have Matlab at work so I can tell if it's gonna work), but the resulting image will be croped.
this is an old post so I guess it wont help the OP but as I was helped by his attempt I post here my corrected code.
basically some freedom in the implementation regarding to how you deal with unassigned pixels as well as wether you wish to keep the original size of the pic - which will force you to crop areas falling "outside" of it.
the following function rotates the image around its center, leaves unassigned pixels as "burned" and crops the edges.
function [h] = rot(A,ang)
rotMat = [cos((pi.*ang/180)) sin((pi.*ang/180)); -sin((pi.*ang/180)) cos((pi.*ang/180))];
centerW = round(size(A,1)/2);
centerH = round(size(A,2)/2);
h=255.* uint8(ones(size(A)));
for x = 1:size(A,1)
for y = 1:size(A,2)
point =[x-centerW;y-centerH] ;
product = rotMat * point;
product = int16(product);
newx =product(1,1);
newy=product(2,1);
if newx+centerW<=size(A,1)&& newx+centerW > 0 && newy+centerH<=size(A,2)&& newy+centerH > 0
h(newx+centerW,newy+centerH) = A(x,y);
end
end
end
I want to stretch an elliptical object in an image until it forms a circle. My program currently inputs an image with an elliptical object (eg. coin at an angle), thresholds and binarizes it, isolates the region of interest using edge-detect/bwboundaries(), and performs regionprops() to calculate major/minor axis lengths.
Essentially, I want to use the 'MajorAxisLength' as the diameter and stretch the object on the minor axis to form a circle. Any suggestions on how I should approach this would be greatly appreciated. I have appended some code for your perusal (unfortunately I don't have enough reputation to upload an image, the binarized image looks like a white ellipse on a black background).
EDIT: I'd also like to apply this technique to the gray-scale version of the image, to examine what the stretch looks like.
code snippet:
rgbImage = imread(fullFileName);
redChannel = rgbImage(:, :, 1);
binaryImage = redChannel < 90;
labeledImage = bwlabel(binaryImage);
area_measurements = regionprops(labeledImage,'Area');
allAreas = [area_measurements.Area];
biggestBlobIndex = find(allAreas == max(allAreas));
keeperBlobsImage = ismember(labeledImage, biggestBlobIndex);
measurements = regionprops(keeperBlobsImage,'Area','MajorAxisLength','MinorAxisLength')
You know the diameter of the circle and you know the center is the location where the major and minor axes intersect. Thus, just compute the radius r from the diameter, and for every pixel in your image, check to see if that pixel's Euclidean distance from the cirlce's center is less than r. If so, color the pixel white. Otherwise, leave it alone.
[M,N] = size(redChannel);
new_image = zeros(M,N);
for ii=1:M
for jj=1:N
if( sqrt((jj-center_x)^2 + (ii-center_y)^2) <= radius )
new_image(ii,jj) = 1.0;
end
end
end
This can probably be optimzed by using the meshgrid function combined with logical indices to avoid the loops.
I finally managed to figure out the transform required thanks to a lot of help on the matlab forums. I thought I'd post it here, in case anyone else needed it.
stats = regionprops(keeperBlobsImage, 'MajorAxisLength','MinorAxisLength','Centroid','Orientation');
alpha = pi/180 * stats(1).Orientation;
Q = [cos(alpha), -sin(alpha); sin(alpha), cos(alpha)];
x0 = stats(1).Centroid.';
a = stats(1).MajorAxisLength;
b = stats(1).MinorAxisLength;
S = diag([1, a/b]);
C = Q*S*Q';
d = (eye(2) - C)*x0;
tform = maketform('affine', [C d; 0 0 1]');
Im2 = imtransform(redChannel, tform);
subplot(2, 3, 5);
imshow(Im2);