I am currently writing a code in matlab to analyze the optical flow in leach hearts and for some reason, whenever I run this it returns weird complex functions. I'm not sure where they come from and I would love some help on figuring that out.
function [opticalFlow] = opticalflowanalysis(handles,hOpticalflow)
videoReader = vision.VideoFileReader('jun07_0165_segment8to12_20.avi','ImageColorSpace','Intensity','VideoOutputDataType','single');
converter = vision.ImageDataTypeConverter;
opticalFlow = vision.OpticalFlow('OutputValue', 'Horizontal and vertical components in complex form','ReferenceFrameDelay', 6);
shapeInserter = vision.ShapeInserter('Shape','Lines','BorderColor','Custom', 'CustomBorderColor', 255);
videoPlayer = vision.VideoPlayer('Name','Motion Vector');
%Convert the image to single precision, then compute optical flow for the video. Generate coordinate points and draw lines to indicate flow.
i=0;
mm = ones(1080,1920);
%Display results.
while ~isDone(videoReader)
frame = step(videoReader);
im = step(converter, frame);
of = step(opticalFlow, im); %always complex number
aa = size(of)
lines = videooptflowlines(of, 5); %complex number only sometimes - when lines appear?
bb = size(lines)
x = i+ 1;
if(x==2)
mm = of;
end
% show diff bw of and lines matrices
if (x == 2)||(x == 10)
for j=1:1:1080 %gives j = [1 2 ... 720]
for k=1:1:1920 %gives k = [1 2 ... 1280]
of(j,k)
lines(j,k)
if(of(j,k) ~= lines(j,k))
disp(['of[',num2str(j),',',num2str(k),'] = ', num2str(of(j,k)), '...', 'lines[',num2str(j),',',num2str(k),'] = ', num2str(lines(j,k))])
end
end
end
end
if ~isempty(lines)
out = step(shapeInserter, im, lines);
step(videoPlayer, out);
end
end
%Close the video reader and player ,
%handles.output = hObject;
release(videoPlayer);
release(videoReader);
mm
It returns:
aa =
1080 1920
bb =
36465 4
Where do the variables from bb come from?
Thanks,
Jacob
Try putting semi-colons (ie ;) at the ends of the lines in which aa and bb are assigned to
aa = size(of);
...
bb = size(lines);
and see what happens.
Mind you, since neither aa nor bb seems to be used later in the program you could probably safely delete both those lines.
Related
How can I read specific images from TID2013 dataset in MATLAB? I written the following code but it start from first to end of the list. The images are in this format: ixx.yy.z.bmp means xx is the number of image, yy is the model of the noise and z is the level of the noise. I just want to work with models 1,2 in level 4,5 but I don't know how to do it. please someone help me! By the way there are 25 reference images, 24 models of noise and 5 level of each model of noise that I wrote them vertically in info1.txt , info2.txt , info3.txt respectively.
clc; clear; close all;
% read Original images
cd 'C:\Users\Desktop'
for NO1 = 1:25
in1 = fopen('info1.txt');
xx = fgets(in1);
A = imread(strcat('C:\Users\Desktop\reference_images\',xx,'.bmp'));
A = rgb2gray(A);
end
% read distorted images
for NO1 = 1:25
in1 = fopen('info1.txt');
xx = fgets(in1);
for NO2 = 1:24
in2 = fopen('info2.txt');
yy = fgets(in2);
for NO3 = 1:5
in3 = fopen('info3.txt');
z = fgets(in3);
B = imread(strcat('C:\Users\Desktop\distorted_images\',xx,yy,z,'.bmp'));
B = rgb2gray(B);
C = imadjust(B);
% Write restored images
imwrite(C,['C:\Users\Desktop\restored_images\','i',sprintf('%02d',NO1),'_',sprintf('%02d',NO2),'_',num2str(NO3),'.bmp']);
end
end
end
The only thing you need to do is to change values of your for loops. Note that in matlab, in contrast with most other languages, for-loops are not restricted to specific step length and can be any vectors.
imageNames = textread('info1.txt', '%s');
noiseModels = textread('info2.txt', '%s');
noiseLevels = textread('info3.txt', '%s');
imageIndices = 1:25;
modelIndices = [1, 2, 7:8];
levelindices = [4 5];
sourceDir = 'C:\Users\Desktop\distorted_images\';
destDir = 'C:\Users\Desktop\restored_images\';
for ii = imageIndices
name = imageNames{ii};
for jj = modelIndices
model = noiseModels{jj};
for kk = levelindices
level = noiseLevels{kk};
sourcePath = sprintf('%s%s%s%s.bmp', sourceDir, name, model, level)
destPath = sprintf('%si%02d_%02d_%02d.bmp', destDir, ii, jj, kk)
B = imread(sourcePath);
B = rgb2gray(B);
C = imadjust(B);
imwrite(C, destPath);
end
end
end
The following code correctly loads an mp4 file and stores it in a 3D matrix.
r = 1;
fileName = testDummyMP4;
readerobj = VideoReader(fileName, 'tag', 'myreader1');
F = get(readerobj, 'numberOfFrames');
tampon = single(read(readerobj,1));
tampon = imresize(tampon(:,:,1),r);
I = zeros(size(tampon,1),size(tampon,2),F,'uint8');
for k = 1:F
disp(['Open: ' num2str(round(100*k/F)) '%'])
vidFrames = single(read(readerobj,k));
I(:,:,k) = imresize(vidFrames(:,:,2),r);
end;
imagesc((I(:,:,1)));
This is the output
I'm trying to reverse engineer this code so that it produces the same sort of result for but for a .raw 8bit rbg file. Following the answers from this question I tried the following:
You can download the 'M1302000245_1436389857.982603.raw' rbg file here
Or Google Drive version here
Ix = 256;
Iy = 256;
SF = 30; % Sample frequency
RecordingTime = 30;
Iz = SF*RecordingTime
testDummy = 'M1302000245_1436389857.982603.raw'
fin = fopen(testDummy, 'r');
I = fread(fin, Ix*Iy*3*Iz, 'uint8');
fclose(fin);
I = reshape(I, [Ix Iy 3 Iz]); % The rbg should be 256x256x3x900
% I've tried each of the following manipulations before calling imagesc to no avail
% I = flipdim(imrotate(I, -90),2);
% I=impixel(I)
% I=I'
imagesc((I(:,:,1))); % view first slice
This gives:
What am I doing wrong?
Additional info:
Recordings are taken using raspberry pi cameras with the following python code
class BrainCamera:
def __init__(self):
self.video_format = "rgb"
#self.video_quality = 5
# Set up the settings of the camera so that
# Exposure and gains are constant.
self.camera = picamera.PiCamera()
self.camera.resolution = (256,256)
self.camera.framerate = 30
sleep(2.0)
self.camera.shutter_speed = self.camera.exposure_speed
self.camera.exposure_mode = 'off'
g = self.camera.awb_gains
self.camera.awb_mode = 'off'
self.camera.awb_gains = g
self.camera.shutter_speed = 30000
self.camera.awb_gains = (1,1)
def start_recording(self, video_name_path):
self.camera.start_recording(video_name_path, format=self.video_format)
self.camera.start_preview()
def stop_recording(self):
self.camera.stop_recording()
self.camera.stop_preview()
# Destructor
def __del__(self):
print ("Closed Camera")
self.camera.close()
I don't get the output you have, but I get some reasonable image:
%your vode from above, ending with I=fread(...)
I=uint8(I)
I2=reshape(I, 3, Ix ,Iy, []);
I2=permute(I2,[3,2,1,4]);
imagesc(I2(:,:,:,1));
implay(I2);
With Daniel's help I got the following working. Additionally a co-worker found out how to get the number of frames of a .raw rbg video. This we needed since it turns out assuming noFrame = SF*RecordingTime was a bad idea.
testDummy = 'M1302000245_1436389857.982603.raw'
testDummyInfo = dir(testDummy);
noFrames = testDummyInfo.bytes/(256*256*3); % In my question Iz = noFrames
fin = fopen(testDummy, 'r');
I = fread(fin, testDummyInfo.bytes, 'uint8');
fclose(fin);
I=uint8(I);
I2=reshape(I, 3, Ix ,Iy, noFrames);
I2=permute(I2,[3,2,1,4]);
imagesc(I2(:,:,2,1)); % Throw out unnecessarry channels r and b. Only want GFP signal
to produce:
My script is supposed to run Runge-Kutta and then interpolate around the tops using polyfit to calculate the max values of the tops. I seem to get the x-values of the max points correct but the y-values are off for some reason. Have sat with it for 3 days now. The problem should be In the last for-loop when I calculate py?
Function:
function funk = FU(t,u)
L0 = 1;
C = 1*10^-6;
funk = [u(2); 2.*u(1).*u(2).^2./(1+u(1).^2) - u(1).*(1+u(1).^2)./(L0.*C)];
Program:
%Runge kutta
clear all
close all
clc
clf
%Given values
U0 = [240 1200 2400];
L0 = 1;
C = 1*10^-6;
T = 0.003;
h = 0.000001;
W = [];
% Runge-Kutta 4
for i = 1:3
u0 = [0;U0(i)];
u = u0;
U = u;
tt = 0:h:T;
for t=tt(1:end-1)
k1 = FU(t,u);
k2 = FU(t+0.5*h,u+0.5*h*k1);
k3 = FU((t+0.5*h),(u+0.5*h*k2));
k4 = FU((t+h),(u+k3*h));
u = u + (1/6)*(k1+2*k2+2*k3+k4)*h;
U = [U u];
end
W = [W;U];
end
I1 = W(1,:); I2 = W(3,:); I3 = W(5,:);
dI1 = W(2,:); dI2 = W(4,:); dI3 = W(6,:);
I = [I1; I2; I3];
dI = [dI1; dI2; dI3];
%Plot of the currents
figure (1)
plot(tt,I1,'r',tt,I2,'b',tt,I3,'g')
hold on
legend('U0 = 240','U0 = 1200','U0 = 2400')
BB = [];
d = 2;
px = [];
py = [];
format short
for l = 1:3
[H,Index(l)]=max(I(l,:));
Area=[(Index(l)-2:Index(l)+2)*h];
p = polyfit(Area,I(Index(l)-2:Index(l)+2),4);
rotp(1,:) = roots([4*p(1),3*p(2),2*p(3),p(4)]);
B = rotp(1,2);
BB = [BB B];
Imax(l,:)=p(1).*B.^4+p(2).*B.^3+p(3).*B.^2+p(4).*B+p(5);
Tsv(i)=4*rotp(1,l);
%px1 = linspace(h*(Index(l)-d-1),h*(Index(l)+d-2));
px1 = BB;
py1 = polyval(p,px1(1,l));
px = [px px1];
py = [py py1];
end
% Plots the max points
figure(1)
plot(px1(1),py(1),'b*-',px1(2),py(2),'b*-',px1(3),py(3),'b*-')
hold on
disp(Imax)
Your polyfit line should read:
p = polyfit(Area,I(l, Index(l)-2:Index(l)+2),4);
More interestingly, take note of the warnings you get about poor conditioning of that polynomial (I presume you're seeing these). Why? Partly because of numerical precision (your numbers are very small, scaled around 10^-6) and partly because you're asking for a 4th-order fit to five points (which is singular). To do this "better", use more input points (more than 5), or a lower-order polynomial fit (quadratic is usually plenty), and (probably) rescale before you use the polyfit tool.
Having said that, in practice this problem is often solved using three points and a quadratic fit, because it's computationally cheap and gives very nearly the same answers as more complex approaches, but you didn't get that from me (with noiseless data like this, it doesn't much matter anyway).
im working on a video steganography using LSB technique..im using traffic.avi and xylophone.mpg as the cover medium and when im using the licence.txt file (in the attach file) to encode into the video it runs well however when im using a short messsage for the input text it shows an error which is
"The matrix MSG in ENCODE must have K columns." and sometimes when use short text it gives error "msg is too long to encode"
i have no idea what does this 2 set of coding means and how to edit the code to make it possible to encode a short msg...below this is some of the code that i guess relate to this problem
num2add = 80-length(msg); % Number of spaces to add to end of MSG.
if num2add < 0, error('This message is too long to encode.'), end
newmsg = [msg,repmat(' ',1,num2add)]; % 80 chars always encoded.
msgmat = dec2bin(newmsg)-48; % Each row is a bin. rep. of an ascii char.
and also this coding
if m_msg == 1
type_flag = 2; % binary vector
[msg, added] = vec2mat(msg, k);
elseif m_msg ~= k
error('comm:encode:InvalidMatrixColumnSize','The matrix MSG in ENCODE must have K columns.');
BELOW THIS IS THE FULL ENCODE CODING continue after the first part of the above coding!
B = pic1(:,:,1); [piclngth pichght] = size(B); % Choose the first page.
dim1 = piclngth-2; dim2 = pichght-3; keyb = key(end:-1:1);
rows = cumsum(double(key));
columns = cumsum(double(keyb)); % Coord pairs for KEY (rows,columns)
A = zeros(dim1,dim2); % This matrix will house the hiding points.
A = crtmtrx(A,rows,columns,dim1,dim2,key);
idx = find(A==1); % This same index will be used for pic matrix.
for vv = 1:80 % This is the encoder.
for uu = 1:8
if msgmat(vv,uu)==1;
if rem(B(idx(uu+8*(vv-1))),2)==0
if(frame==1)
disp('some pixel value of original frame');
B(idx(uu+8*(vv-1)))
end
B(idx(uu+8*(vv-1))) = B(idx(uu+8*(vv-1)))+1;
if(frame==1)
disp('some pixel value of stegno video frame');
B(idx(uu+8*(vv-1)))
end
end
elseif rem(B(idx(uu+8*(vv-1))),2)==1
if(frame==1)
disp('some pixel value of original frame');
B(idx(uu+8*(vv-1)))
end
B(idx(uu+8*(vv-1))) = B(idx(uu+8*(vv-1)))-1;
if(frame==1)
disp('some pixel value of stegno video frame');
B(idx(uu+8*(vv-1)))
end
end
end
end
global newpic;
newpic = pic1; newpic(:,:,1) = B;
f(frame) = im2frame(newpic);
end
frameRate = get(vidObj,'FrameRate');
movie2avi(f,'stegano_video.avi','compression','None', 'fps', 20);
success = 1;
function A = crtmtrx(A,rows,columns,dim1,dim2,key)
% Creates the matrix used to find the points to hide the message.
jj = 1; idx = 1;
while 640 > length(idx) % Need 560 points to hide 80 characters.
for ii = 1:length(rows)
if rows(ii) < dim1
rows(ii) = rem(dim1,rows(ii))+1;
else
rows(ii) = rem(rows(ii),dim1)+1;
end
if columns(ii) < dim2
columns(ii) = rem(dim2,columns(ii))+1;
else
columns(ii) = rem(columns(ii),dim2)+1;
end
A(rows(ii),columns(ii)) = 1;
end
rows = jj*cumsum(double(columns))+round(dim2/2); % Each pass is diff.
columns = jj*cumsum(double(rows))+round(dim1/2);
if jj > ceil(640/length(key))+2 % Estimate how many iters. needed.
idx = find(A==1);
end
jj = jj+1;
end
this is some of the input text and the right one is the encypted txt
The code that triggers the error is pretty clear:
num2add = 80-length(msg); % Number of spaces to add to end of MSG.
if num2add < 0, error('This message is too long to encode.'), end
So basically you will get the error as soon as there are more than 80 characters in msg. I am not sure whether the 80 is meaningfull, you can try to increase it but that may break something else.
I implemented a method for removing shadows based on invariant color features found in the paper Entropy Minimization for Shadow Removal. My implementation seems to be yielding similar computational results sometimes, but they are always off, and my grayscale image is blocky, maybe as a result of incorrectly taking the geometric mean.
Here is an example plot of the information potential from the horse image in the paper as well as my invariant image. Multiply the x-axis by 3 to get theta(which goes from 0 to 180):
And here is the grayscale Image my code outputs for the correct maximum theta (mine is off by 10):
You can see the blockiness that their image doesn't have:
Here is their information potential:
When dividing by the geometric mean, I have tried using NaN and tresholding the image so the smallest possible value is .01, but it doesn't seem to change my output.
Here is my code:
I = im2double(imread(strname));
[m,n,d] = size(I);
I = max(I, .01);
chrom = zeros(m, n, 3, 'double');
for i = 1:m
for j = 1:n
% if ((I(i,j,1)*I(i,j,2)*I(i,j,3))~= 0)
chrom(i,j, 1) = I(i,j,1)/((I(i,j,1)*I(i,j,2)*I(i,j, 3))^(1/3));
chrom(i,j, 2) = I(i,j,2)/((I(i,j,1)*I(i,j,2)*I(i,j, 3))^(1/3));
chrom(i,j, 3) = I(i,j,3)/((I(i,j,1)*I(i,j,2)*I(i,j, 3))^(1/3));
% else
% chrom(i,j, 1) = 1;
% chrom(i,j, 2) = 1;
% chrom(i,j, 3) = 1;
% end
end
end
p1 = mat2gray(log(chrom(:,:,1)));
p2 = mat2gray(log(chrom(:,:,2)));
p3 = mat2gray(log(chrom(:,:,3)));
X1 = mat2gray(p1*1/(sqrt(2)) - p2*1/(sqrt(2)));
X2 = mat2gray(p1*1/(sqrt(6)) + p2*1/(sqrt(6)) - p3*2/(sqrt(6)));
maxinf = 0;
maxtheta = 0;
data2 = zeros(1, 61);
for theta = 0:3:180
M = X1*cos(theta*pi/180) - X2*sin(theta*pi/180);
s = sqrt(std2(X1)^(2)*cos(theta*pi/180) + std2(X2)^(2)*sin(theta*pi/180));
s = abs(1.06*s*((m*n)^(-1/5)));
[m, n] = size(M);
length = m*n;
sources = zeros(1, length, 'double');
count = 1;
for x=1:m
for y = 1:n
sources(1, count) = M(x , y);
count = count + 1;
end
end
weights = ones(1, length);
sigma = 2*s;
[xc , Ak] = fgt_model(sources , weights , sigma , 10, sqrt(length) , 6 );
sum1 = sum(fgt_predict(sources , xc , Ak , sigma , 10 ));
sum1 = sum1/sqrt(2*pi*2*s*s);
data2(theta/3 + 1) = sum1;
if (sum1 > maxinf)
maxinf = sum1;
maxtheta = theta;
end
end
InvariantImage2 = cos(maxtheta*pi/180)*X1 + sin(maxtheta*pi/180)*X2;
Assume the Fast Gauss Transform is correct.
I don't know whether this makes any difference as it is more than a month now, but the blockiness and different information potential plot is simply caused by compression of the used image. You can't expect to be getting same results using this image as they had, because they have used raw, high resolution uncompressed version of it. I have to say I am fairly impressed with your results, especially with implementing the information potential. That thing went over my head a little.
John.