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
Related
I try to remove the white annotations of this image (the numbers and arrows), as well as the black grid, with MATLAB:
I tried to compute, for each pixel, the mode of neighbors, but this process is very slow and I get poor results.
How can I obtain an image like this one?
Thank you for your time.
The general name for such a task is inpainting. If you search for that you will find better methods than what I'm showing here. This is no more than a proof of concept. I'm using DIPimage 3 (because I'm an author and it's easy for me to use).
First we need to create a mask for the regions that we want to remove (inpaint). It is easy to find pixels where all three channels have a high value (white) or a low value (black):
img = readim('https://i.stack.imgur.com/16r9N.png');
% Find a mask for the areas to remove
whitemask = min(img,'tensor') > 50;
blackmask = max(img,'tensor') < 30;
mask = whitemask | blackmask;
This mask doesn't capture all of the black grid, if we increase the threshold we will also remove the dark region of sea off the coast of Spain. And it also captures the white outline of the coasts. We can do a little bit better than this with some additional filtering:
% Find a mask for the areas to remove
whitemask = min(img,'tensor') > 50;
whitemask = whitemask - pathopening(whitemask,50);
blackmask = max(img,'tensor');
blackmask2 = blackmask < 80;
blackmask2 = blackmask2 - areaopening(blackmask2,6);
blackmask = blackmask < 30 | blackmask2;
mask = whitemask | blackmask;
This produces the following mask:
Still far from perfect, but a good start for our proof of concept.
One simple inpainting method uses normalized convolution: using the inverse of the mask we made, convolve the image multiplied by the mask, and convolve the mask separately. The ratio of these two results is a smoothed image that doesn't take the masked pixels into account. Finally, we replace the pixels in the original image under the mask with the values from this normalized convolution:
% Solution 1: normalized convolution
smooth = gaussf(img * ~mask, 2) / gaussf(~mask, 2);
img(mask) = smooth(mask);
An alternative solution applies a closing on the image multiplied by the mask (note that this multiplication makes the pixels we don't want completely black; the closing will spread the surrounding colors over the black areas):
% Solution 2: morphology
smooth = iterate('closing',img * ~mask, 13);
img(mask) = smooth(mask);
I am facing the same problem as mentioned in this post, however, I am not facing it with OpenGL, but simply with MATLAB. Depth as distance to camera plane in GLSL
I have a depth image rendered from the Z-Buffer from 3ds max. I was not able to get an orthographic representation of the z-buffer. For a better understanding, I will use the same sketch as made by the previous post:
* |--*
/ |
/ |
C-----* C-----*
\ |
\ |
* |--*
The 3 asterisks are pixels and the C is the camera. The lines from the
asterisks are the "depth". In the first case, I get the distance from the pixel to the camera. In the second, I wish to get the distance from each pixel to the plane.
The settins of my camera are the following:
WIDTH = 512;
HEIGHT = 424;
FOV = 89.971;
aspect_ratio = WIDTH/HEIGHT;
%clipping planes
near = 500;
far = 5000;
I calulate the frustum settings like the following:
%calculate frustums settings
top = tan((FOV/2)*5000)
bottom = -top
right = top*aspect_ratio
left = -top*aspect_ratio
And set the projection matrix like this:
%Generate matrix
O_p = [2/(right-left) 0 0 -((right+left)/(right-left)); ...
0 2/(top-bottom) 0 -((top+bottom)/(top-bottom));...
0 0 -2/(far-near) -(far+near)/(far-near);...
0 0 0 1];
After this I read in the depth image, which was saved as a 48 bit RGB- image, where each channel is the same, thus only one channel has to be used.
%Read in image
img = imread('KinectImage.png');
%Throw away, except one channel (all hold the same information)
c1 = img(:,:,1);
The pixel values have to be inverted, since the closer the values are to the camera, the brigher they were. If a pixel is 0 (no object to render available) it is set to 2^16, so , that after the bit complementation, the value is still 0.
%Inverse bits that are not zero, so that the z-image has the correct values
c1(c1 == 0) = 2^16
c1_cmp = bitcmp(c1);
To apply the matrix, to each z-Buffer value, I lay out the vector one dimensional and build up a vector like this [0 0 z 1] , over every element.
c1_cmp1d = squeeze(reshape(c1_cmp,[512*424,1]));
converted = double([zeros(WIDTH*HEIGHT,1) zeros(WIDTH*HEIGHT,1) c1_cmp1d zeros(WIDTH*HEIGHT,1)]) * double(O_p);
After that, I pick out the 4th element of the result vector and reshape it to a image
img_con = converted(:,4);
img_con = reshape(img_con,[424,512]);
However, the effect, that the Z-Buffer is not orthographic is still there, so did I get sth wrong? Is my calculation flawed ? Or did I make mistake here?
Depth Image coming from 3ds max
After the computation (the white is still "0" , but the color axis has changed)
It would be great to achieve this with 3ds max, which would resolve this issue, however I was not able to find this setting for the z-buffer. Thus, I want to solve this using Matlab.
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,
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);
I have performed rgb2gray on an image and did a sobel edge detection on the image.
then did
faceEdges = faceNoNoise(:,:) > 50; %binary threshold
so it sets the outline of the image (a picture of a face), to black and white. Values 1 is white pixel, and 0 is black pixel. Someone said I could use this,
mouthsquare = rectangle('position',[recX-mouthBoxBuffer, recY-mouthBoxBuffer, recXDiff*2+mouthBoxBuffer/2, recYDiff*2+mouthBoxBuffer/2],... % see the change in coordinates
'edgecolor','r');
numWhite = sum(sum(mouthsquare));
He said to use two sum()'s because it gets the columns and rows of the contained pixels within the rectangle. numWhite always returns 178 and some decimal numbers.
If you have a 2D matrix M (this being -- for exmple -- an image), the way to count how many elements have the value 1 is:
count_1 = sum(M(:)==1)
or
count_1 = sum(reshape(M,1,[])==1)
If the target values are not exactly 1, but have a Δ-threshold of, let's say, +/- 0.02, then one should ask for:
count_1_pm02 = sum((M(:)>=0.98) & (M(:)<=1.02))
or the equivalent using reshape.