I write in variable 'O' some values using
for i = 1:size(I,1)
for j = 1:size(1,I)
h = i * j;
O{h} = I(i, j) * theta(h);
end
end
I - double, theta - double.
I need to sum()all 'O' values, but when I do it its give me error: sum: wrong type argument 'cell'.
How can I sum() it?
P.s. when I want to see O(), its give me
O =
{
[1,1] = 0.0079764
[1,2] = 0.0035291
[1,3] = 0.0027539
[1,4] = 0.0034392
[1,5] = 0.017066
[1,6] = 0.0082958
[1,7] = 1.4764e-04
[1,8] = 0.0024597
[1,9] = 1.1155e-04
[1,10] = 0.0010342
[1,11] = 0.0039654
[1,12] = 0.0047713
[1,13] = 0.0054305
[1,14] = 3.3794e-04
[1,15] = 0.014323
[1,16] = 0.0026826
[1,17] = 0.013864
[1,18] = 0.0097778
[1,19] = 0.0058029
[1,20] = 0.0020726
[1,21] = 0.0016430
etc...
The exact answer to your question is to use cell2mat
sum (cell2mat (your_cell_o))
However, this is the very wrong way to solve your problem. The thing is that you should not have created a cell array in first place. You should have created a numeric array:
O = zeros (size (I), class (I));
for i = 1:rows (I)
for j = 1:columns (I)
h = i * j;
O(h) = I(i, j) * theta(h);
endfor
endfor
but even this is just really bad and slow. Octave is a language to vectorize operations. Instead, you should have:
h = (1:rows (I))' .* (1:columns (I)); # automatic broadcasting
O = I .* theta (h);
which assumes your function theta behaves properly and if givena matrix will compute the value for each of the element of h and return something of the same size.
If you get an error about wrong sizes, I will guess you have an old version of Octave that does not perform automatic broadcasting. If so, update Octave. If you really can't, then:
h = bsxfun (#times, (1:rows (I))', 1:columns (I));
Related
I have a vector with a couple of elements and I want to write a function that returns me all combinations of x items from this vector.
The following code produces the right output for the case x=2 or x=3 or x=4.
However, I can not implement a solution for every possible x following this idea.
values = {'A','B','C','D','E'};
n = length(values);
data2 = {}; % case x=2
for i = 1:n
for j = i+1:n
data2{end+1} = {values{i}, values{j}};
fprintf('%s %s\n',values{i}, values{j})
end
end
data3 = {}; % case x=3
for i = 1:n
for j = i+1:n
for k = j+1:n
data3{end+1} = {values{i}, values{j}, values{k}};
fprintf('%s %s %s\n',values{i}, values{j}, values{k})
end
end
end
data4 = {}; % case x=4
for i = 1:n
for j = i+1:n
for k = j+1:n
for l = k+1:n
data4{end+1} = {values{i}, values{j}, values{k}, values{l}};
fprintf('%s %s %s %s\n',values{i}, values{j}, values{k}, values{l})
end
end
end
end
How would a function look like which would be able to return my data variable?
data = getCombinations(values, x) %values is vector with elements, x is integer value
EDIT
The following code comes pretty close:
data = perms(values)
data = data(:,1:x)
data = unique(data,'rows')
but it still produces output like A,B and B,A
EDIT2
This fixed it somehow but it is not very nice to look at and it does not work for text entries in cells but only for numbers
data = perms(values)
data = data(:,1:x)
data = sort(data,2)
data = unique(data,'rows')
EDIT3
This did it but it is not very nice to look at... Maybe there is a better solution?
function [data] = getCombinations(values,x)
i = 1:length(values);
d = perms(i);
d = d(:,1:x);
d = sort(d,2);
d = unique(d,'rows');
data = v(d);
end
If you don't want repetitions (and your example suggests you don't) then try nchoosek as nchoosek(1:n, x) to give indices:
values = {'A','B','C','D','E'};
n = length(values);
x = 3;
C = nchoosek(1:n, x);
data = values(C)
In the above, each row is a unique combination of 3 of the 5 elements of values.
Alternatively pass in the values directly:
data = nchoosek(values, x);
To calculate an enhancement function for an input image I have written the following piece of code:
Ig = rgb2gray(imread('test.png'));
N = numel(Ig);
meanTotal = mean2(Ig);
[row,cal] = size(Ig);
IgTransformed = Ig;
n = 3;
a = 1;
b = 1;
c = 1;
k = 1;
for ii=2:row-1
for jj=2:cal-1
window = Ig(ii-1:ii+1,jj-1:jj+1);
IgTransformed(ii,jj) = ((k*meanTotal)/(std2(window) + b))*abs(Ig(ii,jj)-c*mean2(window)) + mean2(window).^a;
end
end
How can I reduce the calculation time?
Obviously, one of the factors is the small window (3x3) that should be made in the loop each time.
Here you go -
Igd = double(Ig);
std2v = colfilt(Igd, [3 3], 'sliding', #std);
mean2v = conv2(Igd,ones(3),'same')/9;
Ig_out = uint8((k*meanTotal)./(std2v + b).*abs(Igd-cal*mean2v) + mean2v.^a);
This will change the boundary elements too, which if not desired could be set back to the original ones with few additional steps, like so -
Ig_out(:,[1 end]) = Ig(:,[1 end])
Ig_out([1 end],:) = Ig([1 end],:)
I am implementing a fast optimization algorithm using fixed point method in matlab. The goal of that method is that find optimal value of u. Denote u={u_i,i=1..2}. The optimal value of u can be obtained as following steps:
Sorry about my image because I cannot type mathematics equation in here.
To do that task, I tried to find u follows above steps. However, I don't know how to implement the term \sum_{j!=i} (u_j-1) in equation 25. This is my code. Please see it and could you give me some comment or suggestion about my implementation to correct them. Currently, I tried to run that code but it give an incorrect answer.
function u = compute_u_TV(Im0, N_class)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Initialization
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
theta=0.001;
gamma=0.01;
tau=0.1;
sigma=0.1;
N_class=2; % only have u1 and u2
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Iterative segmentation process
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
for i=1:N_class
v(:,:,i) = Im0/max(Im0(:)); % u between 0 and 1.
qxv(:,:,i) = zeros(size(Im0));
qyv(:,:,i) = zeros(size(Im0));
u(:,:,i) = v(:,:,i);
for iteration=1:10000
u_temp=u;
% Update v
Divqi = ( BackwardX(qxv(:,:,i)) + BackwardY(qyv(:,:,i)) );
Term = Divqi - u(:,:,i)/ (theta*gamma);
TermX = ForwardX(Term);
TermY = ForwardY(Term);
Norm = sqrt(TermX.^2 + TermY.^2);
Denom = 1 + tau*Norm;
%Equation 24
qxv(:,:,i) = (qxv(:,:,i) + tau*TermX)./Denom;
qyv(:,:,i) = (qyv(:,:,i) + tau*TermY)./Denom;
v(:,:,i) = u(:,:,i) - theta*gamma* Divqi; %Equation 23
% Update u
u(:,:,i) = (v(:,:,i) - theta* gamma* Divqi -theta*gamma*sigma*(sum(u(:))-u(:,:,i)-1))./(1+theta* gamma*sigma);
u(:,:,i) = max(u(:,:,i),0);
u(:,:,i) = min(u(:,:,i),1);
check=u_temp(:,:,i)-u(:,:,i);
if(abs(sum(check(:)))<=0.1)
break;
end
end
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Sub-functions- X.Berson
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [dx]=BackwardX(u);
[Ny,Nx] = size(u);
dx = u;
dx(2:Ny-1,2:Nx-1)=( u(2:Ny-1,2:Nx-1) - u(2:Ny-1,1:Nx-2) );
dx(:,Nx) = -u(:,Nx-1);
function [dy]=BackwardY(u);
[Ny,Nx] = size(u);
dy = u;
dy(2:Ny-1,2:Nx-1)=( u(2:Ny-1,2:Nx-1) - u(1:Ny-2,2:Nx-1) );
dy(Ny,:) = -u(Ny-1,:);
function [dx]=ForwardX(u);
[Ny,Nx] = size(u);
dx = zeros(Ny,Nx);
dx(1:Ny-1,1:Nx-1)=( u(1:Ny-1,2:Nx) - u(1:Ny-1,1:Nx-1) );
function [dy]=ForwardY(u);
[Ny,Nx] = size(u);
dy = zeros(Ny,Nx);
dy(1:Ny-1,1:Nx-1)=( u(2:Ny,1:Nx-1) - u(1:Ny-1,1:Nx-1) );
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% End of sub-function
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
You should do
u(:,:,i) = (v(:,:,i) - theta* gamma* Divqi -theta*gamma*sigma* ...
(sum(u(:,:,1:size(u,3) ~= i),3) -1))./(1+theta* gamma*sigma);
The part you were searching for is
sum(u(:,:,1:size(u,3) ~= i),3)
Let's decompose this :
1:size(u,3) ~= i
is a vector containing all values from 1 to the max size of u on the third dimension except i.
Then
u(:,:,1:size(u,3) ~= i)
is all the matrix of the third dimension of u except for j = i
Finally,
sum(...,3)
is the sum of all the matrix by the thrid dimension.
Let me know if it does help!
:) I'm trying to code a Least Squares algorithm and I've come up with this:
function [y] = ex1_Least_Squares(xValues,yValues,x) % a + b*x + c*x^2 = y
points = size(xValues,1);
A = ones(points,3);
b = zeros(points,1);
for i=1:points
A(i,1) = 1;
A(i,2) = xValues(i);
A(i,3) = xValues(i)^2;
b(i) = yValues(i);
end
constants = (A'*A)\(A'*b);
y = constants(1) + constants(2)*x + constants(3)*x^2;
When I use this matlab script for linear functions, it works fine I think. However, when I'm passing 12 points of the sin(x) function I get really bad results.
These are the points I pass to the function:
xValues = [ -180; -144; -108; -72; -36; 0; 36; 72; 108; 144; 160; 180];
yValues = [sind(-180); sind(-144); sind(-108); sind(-72); sind(-36); sind(0); sind(36); sind(72); sind(108); sind(144); sind(160); sind(180) ];
And the result is sin(165°) = 0.559935259380508, when it should be sin(165°) = 0.258819
There is no reason why fitting a parabola to a full period of a sinusoid should give good results. These two curves are unrelated.
MATLAB already contains a least square polynomial fitting function, polyfit and a complementary function, polyval. Although you are probably supposed to write your own, trying out something like the following will be educational:
xValues = [ -180; -144; -108; -72; -36; 0; 36; 72; 108; 144; 160; 180];
% you may want to experiment with different ranges of xValues
yValues = sind(xValues);
% try this with different values of n, say 2, 3, and 4
p = polyfit(xValues,yValues,n);
x = -180:36:180;
y = polyval(p,x);
plot(xValues,yValues);
hold on
plot(x,y,'r');
Also, more generically, you should avoid using loops as you have in your code. This should be equivalent:
points = size(xValues,1);
A = ones(points,3);
A(:,2) = xValues;
A(:,3) = xValues.^2; % .^ and ^ are different
The part of the loop involving b is equivalent to doing b = yValues; either name the incoming variable b or just use the variable yValues, there's no need to make a copy of it.
After a few days of optimization this is my code for an enumeration process that consist in finding the best combination for every row of W. The algorithm separates the matrix W in one where the elements of W are grather of LimiteInferiore (called W_legali) and one that have only element below the limit (called W_nlegali).
Using some parameters like Media (aka Mean), rho_b_legali The algorithm minimizes the total cost function. In the last part, I find where is the combination with the lowest value of objective function and save it in W_ottimo
As you can see the algorithm is not so "clean" and with very large matrix (142506x3000) is damn slow...So, can somebody help me to speed it up a little bit?
for i=1:3000
W = PesoIncertezza * MatriceCombinazioni';
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
W_legali = W;
W_legali(W<LimiteInferiore) = nan;
if i==1
Media = W_legali;
rho_b_legale = ones(size (W_legali,1),size(MatriceCombinazioni,1));
else
Media = (repmat(sum(W_tot_migl,2),1,size(MatriceCombinazioni,1))+W_legali)/(size(W_tot_migl,2)+1);
rho_b_legale = repmat(((n_b+1)/i),1,size(MatriceCombinazioni,1));
end
[W_legali_migl,comb] = min(C_u .* Media .* (1./rho_b_legale) + (1./rho_b_legale) .* c_0 + (c_1./(i * rho_b_legale)),[],2);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
MatriceCombinazioni_2 = MatriceCombinazioni;
MatriceCombinazioni_2(sum(MatriceCombinazioni_2,2)<2,:)=[];
W_nlegali = PesoIncertezza * MatriceCombinazioni_2';
W_nlegali(W_nlegali>=LimiteInferiore) = nan;
if i==1
Media = W_nlegali;
rho_b_nlegale = zeros(size (W_nlegali,1),size(MatriceCombinazioni_2,1));
else
Media = (repmat(sum(W_tot_migl,2),1,size(MatriceCombinazioni_2,1))+W_nlegali)/(size(W_tot_migl,2)+1);
rho_b_nlegale = repmat(((n_b)/i),1,size(MatriceCombinazioni_2,1));
end
[W_nlegali_migliori,comb2] = min(C_u .* Media .* (1./rho_b_nlegale) + (1./rho_b_nlegale) .* c_0 + (c_1./(i * rho_b_nlegale)),[],2);
z = [W_legali_migl, W_nlegali_migliori];
[z_ott,comb3] = min(z,[],2);
%Increasing n_b
if i==1
n_b = zeros(size(W,1),1);
end
index = find(comb3==1);
increment = ones(size(index,1),1);
B = accumarray(index,increment);
nzIndex = (B ~= 0);
n_b(nzIndex) = n_b(nzIndex) + B(nzIndex);
%Using comb3 to find where is the best configuration, is in
%W_legali or in W_nLegali?
combinazione = comb.*logical(comb3==1) + comb2.*logical(comb3==2);
W_ottimo = W(sub2ind(size(W),[1:size(W,1)],combinazione'))';
W_tot_migl(:,i) = W_ottimo;
FunzObb(:,i) = z_ott;
[PesoCestelli] = Simulazione_GenerazioneNumeriCasuali (PianoSperimentale,NumeroCestelli,NumeroEsperimenti,Alfa);
[PesoIncertezza_2] = Simulazione_GenerazioneIncertezza (NumeroCestelli,NumeroEsperimenti,IncertezzaCella,PesoCestelli);
PesoIncertezza(MatriceCombinazioni(combinazione,:)~=0) = PesoIncertezza_2(MatriceCombinazioni(combinazione,:)~=0); %updating just the hoppers that has been discharged
end
When you see repmat you should think bsxfun. For example, replace:
Media = (repmat(sum(W_tot_migl,2),1,size(MatriceCombinazioni,1))+W_legali) / ...
(size(W_tot_migl,2)+1);
with
Media = bsxfun(#plus,sum(W_tot_migl,2),W_legali) / ...
(size(W_tot_migl,2)+1);
The purpose of bsxfun is to do a virtual "singleton expansion" like repmat, without actually replicating the array into a matrix of the same size as W_legali.
Also note that in the above code, sum(W_tot_migl,2) is computed twice. There are other small optimizations, but changing to bsxfun should give you a good improvement.
The values of 1./rho_b_legale are effectively computed three times. Store this quotient matrix.