need a working example of how to do this, i have a bmp file showing the shape of a lake, the bmp's size is the rectangular area and known. i need to take this picture and estimate the lake size. so far, i have a script that generates a giant matrix of each pixel, telling me whether or not it's in the lake - but this isn't monte carlo! i need to generate random points and compare them against the shape somehow, this is where i'm getting stuck. i don't understand how to compare here, i don't have an equation for the shape or lines, i only have exact point information - either it is or isn't in the lake. so i guess i have the exact area already, but i need to find a way to compare random points against this.
function Yes = Point_In_Lake(x,y,image_pixel)
[pHeight,pWidth]=size(image_pixel);
%pHeight = Height in pixel
%pWidth = Width in pixel
width = 1000; %width is the actual width of the lake
height = 500; %height is the actual height of the lake
%converting x_value to pixel_value in image_pixel
point_x_pixel = x*pWidth/width;
xl = floor(point_x_pixel)+1;
xu = min(ceil(point_x_pixel)+1,pWidth);
%converting y_value to pixel_value in image_pixel
point_y_pixel = y*pHeight/height;
yl = floor(point_y_pixel)+1;
yu = min(ceil(point_y_pixel)+1,pHeight);
%Finally, perform the check whether the point is in the lake
if (image_pixel(yl,xl)~=0)&&(image_pixel(yl,xu)~=0)&&(image_pixel(yu,xl)~=0)&&(image_pixel(yu,xu)~=0)
Yes=0;
else
Yes=1;
end
Here was the solution:
binaryMap = image_pixel
for i = 1:numel(image_pixel)
xrand = randperm(size(image_pixel,1),1);
yrand = randperm(size(image_pixel,2),1);
Yes(i) = Point_In_Lake(xrand,yrand,binaryMap);
end
PercentLake = length(find(Yes==1))/length(Yes);
LakeArea = (PercentLake * 500000)/43560;
Related
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:
I am working on Simulink to develop my algorithm.
I have a video stream with dimensions 640x360. I am trying to extract region of interest (ROI) from each frame. However, my video turns into grayscale when I use the following code:
MATLAB Function Block which I am using for the ROI extraction:
function y = fcn(u)
%some more code
width = 639;
height = 210;
top = 150;
left = 1;
y = u(top:top+height, left:left+width);
Solution
Change the last line as follows:
y = u(top:top+height, left:left+width,:);
Explanation
The dimensions of an RGB image are actually mxnx3. The m and n are the image height and width, and there are 3 channels: red, green and blue.
when you perform a cropping of an RGB image, it should be performed on each channel separately. You can achieve that by using my code example above.
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