I write JPEG compression in Scilab (equivalent of MATLAB) using function imdct. In this function is used function DCT from openCV and I don't know which equation is used in dct function.
lenna by imdct
lenna by my_function
You can see lenna by imdct which is internal function and lenna by my_function is my function in scilab.
I add my code in scilab
function vystup = dct_rovnice(vstup)
[M,N] = size(vstup)
for u=1:M
for v=1:N
cos_celkem = 0;
for m=1:M
for n=1:N
pom = double(vstup(m,n));
cos_citatel1 = cos(((2*m) * u * %pi)/(2*M));
cos_citatel2 = cos(((2*n) * v * %pi)/(2*N));
cos_celkem = cos_celkem + (pom * cos_citatel1 * cos_citatel2);
end
end
c_u = 0;
c_v = 0;
if u == 1 then
c_u = 1 / sqrt(2);
else
c_u = 1;
end
if v == 1 then
c_v = 1 / sqrt(2);
else
c_v = 1;
end
vystup(u,v) = (2/sqrt(n*m)) * c_u * c_v * cos_celkem;
end
end
endfunction
function vystup = dct_prevod(vstup)
Y = vstup(:,:,1);
Cb = vstup(:,:,2);
Cr = vstup(:,:,3);
[rows,columns]=size(vstup)
vystup = zeros(rows,columns,3)
for y=1:8:rows-7
for x=1:8:columns-7
blok_Y = Y(y:y+7,x:x+7)
blok_Cb = Cb(y:y+7,x:x+7)
blok_Cr = Cr(y:y+7,x:x+7)
blok_dct_Y = dct_rovnice(blok_Y)
blok_dct_Cb = dct_rovnice(blok_Cb)
blok_dct_Cr = dct_rovnice(blok_Cr)
vystup(y:y+7,x:x+7,1)= blok_dct_Y
vystup(y:y+7,x:x+7,2)= blok_dct_Cb
vystup(y:y+7,x:x+7,3)= blok_dct_Cr
end
end
vystup = uint8(vystup)
endfunction
You can see equation I used
EQUATION
The issue seems to be in the use of different normalization of the resulting coefficients.
The OpenCV library uses this equation for a forward transform (N=8, in your case):
The basis g is defined as
where
(Sorry for the ugly images, but SO does not provide any support for typesetting equations.)
Take care there are several definitions of the dct function (DCT-I, DCT-II, DCT-III and DCT-IV normalized and un-normmalized)
Moreover have you tried the Scilab builtin function dct (from FFTW) which can be applied straightforward to images.
Related
I'm trying to fit a line using quadratic poly, but because the fit results in continuous values, the integer conversion (for CartesianIndex) rounds it off, and I loose data at that pixel.
I tried the method
here. So I get new y values as
using Images, Polynomials, Plots,ImageView
img = load("jTjYb.png")
img = Gray.(img)
img = img[end:-1:1, :]
nodes = findall(img.>0)
xdata = map(p->p[2], nodes)
ydata = map(p->p[1], nodes)
f = fit(xdata, ydata, 2)
ydata_new .= round.(Int, f.(xdata)
new_line_fitted_img=zeros(size(img))
new_line_fitted_img[xdata,ydata_new].=1
imshow(new_line_fitted_img)
which results in chopped line as below
whereas I was expecting it to be continuous line as it was in pre-processing
Do you expect the following:
Raw Image
Fitted Polynomial
Superposition
enter image description here
enter image description here
enter image description here
Code:
using Images, Polynomials
img = load("img.png");
img = Gray.(img)
fx(data, dCoef, cCoef, bCoef, aCoef) = #. data^3 *aCoef + data^2 *bCoef + data*cCoef + dCoef;
function fit_poly(img::Array{<:Gray, 2})
img = img[end:-1:1, :]
nodes = findall(img.>0)
xdata = map(p->p[2], nodes)
ydata = map(p->p[1], nodes)
f = fit(xdata, ydata, 3)
xdt = unique(xdata)
xdt, fx(xdt, f.coeffs...)
end;
function draw_poly!(X, y)
the_min = minimum(y)
if the_min<0
y .-= the_min - 1
end
initialized_img = Gray.(zeros(maximum(X), maximum(y)))
initialized_img[CartesianIndex.(X, y)] .= 1
dif = diff(y)
for i in eachindex(dif)
the_dif = dif[i]
if abs(the_dif) >= 2
segment = the_dif รท 2
initialized_img[i, y[i]:y[i]+segment] .= 1
initialized_img[i+1, y[i]+segment+1:y[i+1]-1] .= 1
end
end
rotl90(initialized_img)
end;
X, y = fit_poly(img);
y = convert(Vector{Int64}, round.(y));
draw_poly!(X, y)
I am learning Machine Learning course from coursera from Andrews Ng. I have written a code for logistic regression in octave. But, it is not working. Can someone help me?
I have taken the dataset from the following link:
Titanic survivors
Here is my code:
pkg load io;
[An, Tn, Ra, limits] = xlsread("~/ML/ML Practice/dataset/train_and_test2.csv", "Sheet2", "A2:H1000");
# As per CSV file we are reading columns from 1 to 7. 8-th column is Survived, which is what we are going to predict
X = [An(:, [1:7])];
Y = [An(:, 8)];
X = horzcat(ones(size(X,1), 1), X);
# Initializing theta values as zero for all
#theta = zeros(size(X,2),1);
theta = [-3;1;1;-3;1;1;1;1];
learningRate = -0.00021;
#learningRate = -0.00011;
# Step 1: Calculate Hypothesis
function g_z = estimateHypothesis(X, theta)
z = theta' * X';
z = z';
e_z = -1 * power(2.72, z);
denominator = 1.+e_z;
g_z = 1./denominator;
endfunction
# Step 2: Calculate Cost function
function cost = estimateCostFunction(hypothesis, Y)
log_1 = log(hypothesis);
log_2 = log(1.-hypothesis);
y1 = Y;
term_1 = y1.*log_1;
y2 = 1.-Y;
term_2 = y2.*log_2;
cost = term_1 + term_2;
cost = sum(cost);
# no.of.rows
m = size(Y, 1);
cost = -1 * (cost/m);
endfunction
# Step 3: Using gradient descent I am updating theta values
function updatedTheta = updateThetaValues(_X, _Y, _theta, _hypothesis, learningRate)
#s1 = _X * _theta;
#s2 = s1 - _Y;
#s3 = _X' * s2;
# no.of.rows
#m = size(_Y, 1);
#s4 = (learningRate * s3)/m;
#updatedTheta = _theta - s4;
s1 = _hypothesis - _Y;
s2 = s1 .* _X;
s3 = sum(s2);
# no.of.rows
m = size(_Y, 1);
s4 = (learningRate * s3)/m;
updatedTheta = _theta .- s4';
endfunction
costVector = [];
iterationVector = [];
for i = 1:1000
# Step 1
hypothesis = estimateHypothesis(X, theta);
#disp("hypothesis");
#disp(hypothesis);
# Step 2
cost = estimateCostFunction(hypothesis, Y);
costVector = vertcat(costVector, cost);
#disp("Cost");
#disp(cost);
# Step 3 - Updating theta values
theta = updateThetaValues(X, Y, theta, hypothesis, learningRate);
iterationVector = vertcat(iterationVector, i);
endfor
function plotGraph(iterationVector, costVector)
plot(iterationVector, costVector);
ylabel('Cost Function');
xlabel('Iteration');
endfunction
plotGraph(iterationVector, costVector);
This is the graph I am getting when I am plotting against no.of.iterations and cost function.
I am tired by adjusting theta values and learning rate. Can someone help me to solve this problem.
Thanks.
I have done a mathematical error. I should have used either power(2.72, -z) or exp(-z). Instead I have used as -1 * power(2.72, z). Now, I'm getting a proper curve.
Thanks.
I am a newbie in the field of CV and IP. I was writing the HoughTransform algorithm for finding line.I am not getting what is wrong with this code in which i m trying to find the accumulator array
numRowsInBW = size(BW,1);
numColsInBW = size(BW,2);
%length of the diagonal of image
D = sqrt((numRowsInBW - 1)^2 + (numColsInBW - 1)^2);
%number of rows in the accumulator array
nrho = 2*(ceil(D/rhoStep)) + 1;
%number of cols in the accumulator array
ntheta = length(theta);
H = zeros(nrho,ntheta);
%this means the particular pixle is white
%i.e the edge pixle
[allrows allcols] = find(BW == 1);
for i = (1 : size(allrows))
y = allrows(i);
x = allcols(i);
for th = (1 : 180)
d = floor(x*cos(th) - y*sin(th));
H(d+floor(nrho/2),th) += 1;
end
end
I m applying this for a simple image
I m getting this result
But this is expected
I am not able to find the mistake.Please help me.Thanks in advance.
There are several issues with your code. The main issue is here:
ntheta = length(theta);
% ...
for i = (1 : size(allrows))
% ...
for th = (1 : 180)
d = floor(x*cos(th) - y*sin(th));
% ...
th seems to be an angle in degrees. cos(th) is meaningless. Instead, use cosd and sind.
Another issue is that th iterates from 1 to 180, but there is no guarantee that ntheta is 180. So, loop as follows instead:
for i = 1 : size(allrows)
% ...
for j = 1 : numel(theta)
th = theta(j);
% ...
and use th as the angle, and j as the index into H.
Finally, given your image and your expected output, you should apply some edge detection first (Canny, for example). Maybe you already did this?
I am getting different results between the below iterative way and a simple vector way in octave(simple regression). What am I doing wrong in the iterative way ?
Iterative version
sum_val = 0;
for m_val = 1:m,
h = X(m_val,:) * theta;
err_sq = power((h - y(m_val)),2);
sum_val = sum_val + err_sq;
end;
J = (1/2*m)*sum_val;
Vector way:
J = (1/(2*m))*sum(power((X*theta - y),2));
In MATLAB, and so I presume in Octave as well, 1/2*m is not the same as 1/(2*m). This is your source of error.
I get what this wiki page says(http://en.wikipedia.org/wiki/Multinomial_logistic_regression), but I don't know how to get the update rules for stochastic gradient descent. Sorry to ask this here(this is really just about machine learning theories instead of actual implementation). Could someone provide a solution with explanation? Thanks in advance!
I happened to write code to implent softmax, I refer most to the page http://ufldl.stanford.edu/wiki/index.php/Softmax_Regression
this is the code I wrote in matlab ,hope it will help
function y = sigmoid_multi(weight,x,class_index)
%% weight feature_dim * class_num
%% x feature_dim * 1
%% class_index scalar
sum = eps;
class_num = size(weight,2);
for i = 1:class_num
sum = sum + exp(weight(:,i)'*x);
end
y = exp(weight(:,class_index)'*x)/sum;
end
function g = gradient(train_patterns,train_labels,weight)
m = size(train_patterns,2);
class_num = size(weight,2);
g = zeros(size(weight));
for j = 1:class_num
for i = 1:m
if(train_labels(i) == j)
g(:,j) = g(:,j) + (1 - log( sigmoid_multi(weight,train_patterns(:,i),j) + eps))*train_patterns(:,i);
end
end
end
g = -(g/m);
end
function J = object_function(train_patterns,train_labels,weight)
m = size(train_patterns,2);
J = 0;
for i = 1:m
J = J + log( sigmoid_multi(weight,train_patterns(:,i),train_labels(i)) + eps);
end
J = -(J/m);
end
function weight = multi_logistic_train(train_patterns,train_labels,alpha)
%% weight feature_dim * class_num
%% train_patterns featur_dim * sample_num
%% train_labels 1 * sample_num
%% alpha scalar
class_num = length(unique(train_labels));
m = size(train_patterns,2); %% sample_number;
n = size(train_patterns,1); % feature_dim;
weight = rand(n,class_num);
for i = 1:40
J = object_function(train_patterns,train_labels,weight);
fprintf('objec function value : %f\n',J);
weight = weight - alpha*gradient(train_patterns,train_labels,weight);
end
end