Here is my codes for computing the average rank for each image from 1000 images. (We assume every 100 images are one catagory, e.g, 1-100, 101-200,....)
for z=1:1000
H{z}=imread(strcat(int2str(z-1),'.jpg'));
Im_red=H{z}(:,:,1);
Im_green= H{z}(:,:,2);
Im_blue= H{z}(:,:,3);
hist_im1=zeros(1,256);
[h,w]=size(Im_red);
for i=1:h
for j=1:w
value_pixel1=Im_red(i,j)+1;
hist_im1(value_pixel1)=hist_im1(value_pixel1)+1;
end
end
hist_im2=zeros(1,256);
[h,w]=size(Im_green);
for i=1:h
for j=1:w
value_pixel2=Im_green(i,j)+1;
hist_im2(value_pixel2)=hist_im2(value_pixel2)+1;
end
end
hist_im3=zeros(1,256);
[h,w]=size(Im_blue);
for i=1:h
for j=1:w
value_pixel3 = Im_blue(i,j) + 1;
hist_im3(value_pixel3) = hist_im3(value_pixel3)+1;
end
end
Q{z}=[hist_im1, hist_im2, hist_im3];
end
for r=1:1000
for i=1:1000
a(r,i)=matchfunction(Q{r},Q{i});
end
for j=1:1000
b(r,j)=j;
end
L=[a;b];
end
for r=1:1000
B=[L(r,:);L(r+1000,:)];
[d1,d2] = sort(B(1,:),'descend');
C=B(:,d2);
aaa=C(1,:);
bbb=C(2,:);
ccc=zeros(1,1000);
for g=1:1000
if ((bbb(g)>=fix((r-1)/100)*100+1) & (bbb(g)<=ceil(r/100)*100))
ccc(g)=g;
end
end
ddd=sum(ccc(g))/100;
s(r)=ddd
end
avgrank(1)=sum(s(1:100))/100
avgrank(2)=sum(s(101:200))/100
avgrank(3)=sum(s(201:300))/100
avgrank(4)=sum(s(301:400))/100
avgrank(5)=sum(s(401:500))/100
avgrank(6)=sum(s(501:600))/100
avgrank(7)=sum(s(601:700))/100
avgrank(8)=sum(s(701:800))/100
avgrank(9)=sum(s(801:900))/100
avgrank(10)=sum(s(901:1000))/100
xCoordinates = 1:10;
plot(xCoordinates,avgrank,'b:*');
The match function is a function computes the match value of two histograms of two images with two histograms as input. You can see that Q{z} is the histogram. I think my problem is within here:
for g=1:1000
if ((bbb(g)>=fix((r-1)/100)*100+1) & (bbb(g)<=ceil(r/100)*100))
ccc(g)=g;
end
end
This is how I calculate the rank. So I just give the rank to ccc(g)
since for g runs from 1 to 1000, it will just be the rank we nee if we have
(bbb(g)>=fix((r-1)/100)*100+1) & (bbb(g)<=ceil(r/100)*100)
for a g.
But why after I run this program I got the value of ccc is one thousand 0s? Why 0? Is there anything wrong with my way of getting the rank through ccc? And is there more errors of my code? I just get the average ranks and the ccc all 0 but cannot figure out why. Thanks in advance!!
Related
I have created this code to generate a 1 set of lottery numbers, but I am trying to make it so that the user can enter how many sets they want (input n), and it will print out as one long matrix of size nX6? I was messing around with a few options from online suggestions, but to no avail. I put the initial for i=1:1:n at the beginning, but I do not know how to store each run into a growing matrix. Right now it still generates just 1 set.
function lottery(n)
for i=1:1:n
xlow=1;
xhigh=69;
m=5;
i=1;
while (i<=m)
lottonum(i)=floor(xlow+rand*(xhigh-xlow+1));
flag=0;
for j=1:i-1
if (lottonum(i)==lottonum(j))
flag=1;
end
end
if flag==0
i=i+1;
end
end
ylow=1;
yhigh=26;
m=1;
lottonum1=floor(ylow+rand*(yhigh-ylow+1));
z = horzcat(lottonum, lottonum1);
end
disp('The lotto numbers picked are')
fprintf('%g ',z)
disp (' ')
The problem is that you are not storing or displaying the newly generated numbers, only the last set. To solve this, initialize z with NaNs or zeros, and later index z to store each set in a row of z, by using z(i,:) = lottonum.
However, you are using i as iterator in the while loop already, so you should use another variable, e.g. k.
You can also set z as an output of the function, so you can use this matrix in some other part of a program.
function z = lottery(n)
% init z
z = NaN(n,6);
for k = 1:n
xlow=1;
xhigh=69;
m=5;
i=1;
while (i<=m)
lottonum(i)=floor(xlow+rand*(xhigh-xlow+1));
flag=0;
for j=1:i-1
if (lottonum(i)==lottonum(j))
flag=1;
end
end
if flag==0
i=i+1;
end
end
ylow=1;
yhigh=26;
lottonum1 = floor(ylow+rand*(yhigh-ylow+1));
z(k,:) = horzcat(lottonum, lottonum1); % put the numbers in a row of z
end
disp('The lotto numbers picked are')
disp(z) % prettier display than fprintf in this case.
disp (' ')
end
The nice answer from rinkert corrected your basic mistakes (like trying to modify your loop iterator i from within the loop => does not work), and answered your question on how to store all your results.
This left you with a working code, however, I'd like to propose to you a different way to look at it.
The porposed architecture is to divide the tasks into separate functions:
One function draw_numbers which can draw N numbers randomly (and does only that)
One function draw_lottery which call the previous function as many times as it needs (your n), collect the results and display them.
draw_lottery
This architecture has the benefit to greatly simplify your main function. It can now be as simple as:
function Draws = draw_lottery(n)
% define your draw parameters
xmin = 1 ; % minimum number drawn
xmax = 69 ; % maximum number drawn
nballs = 5 ; % number of number to draw
% pre allocate results
Draws = zeros( n , nballs) ;
for iDraw=1:1:n
% draw "nballs" numbers
thisDraw = draw_numbers(xmin,xmax,nballs) ;
% add them to the result matrix
Draws(iDraw,:) = thisDraw ;
end
disp('The lotto numbers picked are:')
disp (Draws)
disp (' ')
end
draw_numbers
Instead of using a intricated set of if conditions and several iterators (i/m/k) to branch the program flow, I made the function recursive. It means the function may have to call itself a number of time until a condition is satisfied. In our case the condition is to have a set of nballs unique numbers.
The function:
(1) draws N integer numbers randomly, using randi.
(2) remove duplicate numbers (if any). Using unique.
(3) count how many unique numbers are left Nu
(4a) if Nu = N => exit function
(4b) if Nu < N => Call itself again, sending the existing Nu numbers and asking to draw an additional N-Nu numbers to add to the collection. Then back to step (2).
in code, it looks like that:
function draw = draw_numbers(xmin,xmax,nballs,drawn_set)
% check if we received a partial set
if nargin == 4
% if yes, adjust the number of balls to draw
n2draw = nballs - numel(drawn_set) ;
else
% if not, make a full draw
drawn_set = [] ;
n2draw = nballs ;
end
% draw "nballs" numbers between "xmin" and "xmax"
% and concatenate these new numbers with the partial set
d = [drawn_set , randi([xmin xmax],1,n2draw)] ;
% Remove duplicate
drawn_set = unique(d) ;
% check if we have some more balls to draw
if numel(drawn_set) < nballs
% draw some more balls
draw = draw_numbers(xmin,xmax,nballs,drawn_set) ;
else
% we're good to go, assign output and exit funtion
draw = drawn_set ;
end
end
You can have both functions into the same file if you want.
I encourage you to look at the documentation of a couple of Matlab built-in functions used:
randi
unique
l_0=1.5;
l_1=1.6;
Lambda_min=2*(1+1)*l_0;
Lambda_max=2*(1+1)*l_1;
n_0=linspace(2,2.11,10);
n_1=linspace(2.30,2.50,10);
for i=1:10
for j=1:10
for k=1:10
l(i) = Lambda_min * ( Lambda_max/Lambda_min)^(i/10)
sum=sum(l)
d_0(:,j)= l(i)/((n_0(i)/n_1(i)+1))
d_1(:,k)= (n_0(i)/n_1(i))*d_0(:,j)
end
end
end
First of all; I want to find values of l(i) which is a vector, then take the sum of that vector. second, for d_0(:,j) I want to create a matrix so I can plot it later, that takes different values from l(i),n_0,n_1 each time. If I take the values for n_0 and n_1 and put in the for loop I will get index error because it should be logic or integer number.
My matrix is overwritten and do not know how to avoid it. Note, I want in d_0 and d_1 n_0 and n_1 to take values from linspace. for example in the first iteration n_0= 2 n_1= 2.30 then second iteration take the next value in linspace.
I tried to see the value of n_0(i) and does it give me 10 iterations. It gives me more that that overwritten.
Try:
l_0=1.5;
l_1=1.6;
Lambda_min = 4*l_0;
Lambda_max = 4*l_1;
n_0 = linspace(2,2.11,10) % don't add semicolon so you can check this is giving 10 values
n_1 = linspace(2.30,2.50,10) %
for i=1:10
l(i) = Lambda_min * ( Lambda_max/Lambda_min)^(i/10) % should give you 10 values
end
d_0= l./((n_0./n_1+1)); % This will only give you a vector, not a matrix.
d_1= (n_0./n_1).*d_0;
Lsum = sum(l); % should give you one value
My code works in the following manner:
1.First, it obtains several images from the training set
2.After loading these images, we find the normalized faces,mean face and perform several calculation.
3.Next, we ask for the name of an image we want to recognize
4.We then project the input image into the eigenspace, and based on the difference from the eigenfaces we make a decision.
5.Depending on eigen weight vector for each input image we make clusters using kmeans command.
Source code i tried:
clear all
close all
clc
% number of images on your training set.
M=1200;
%Chosen std and mean.
%It can be any number that it is close to the std and mean of most of the images.
um=60;
ustd=32;
%read and show images(bmp);
S=[]; %img matrix
for i=1:M
str=strcat(int2str(i),'.jpg'); %concatenates two strings that form the name of the image
eval('img=imread(str);');
[irow icol d]=size(img); % get the number of rows (N1) and columns (N2)
temp=reshape(permute(img,[2,1,3]),[irow*icol,d]); %creates a (N1*N2)x1 matrix
S=[S temp]; %X is a N1*N2xM matrix after finishing the sequence
%this is our S
end
%Here we change the mean and std of all images. We normalize all images.
%This is done to reduce the error due to lighting conditions.
for i=1:size(S,2)
temp=double(S(:,i));
m=mean(temp);
st=std(temp);
S(:,i)=(temp-m)*ustd/st+um;
end
%show normalized images
for i=1:M
str=strcat(int2str(i),'.jpg');
img=reshape(S(:,i),icol,irow);
img=img';
end
%mean image;
m=mean(S,2); %obtains the mean of each row instead of each column
tmimg=uint8(m); %converts to unsigned 8-bit integer. Values range from 0 to 255
img=reshape(tmimg,icol,irow); %takes the N1*N2x1 vector and creates a N2xN1 matrix
img=img'; %creates a N1xN2 matrix by transposing the image.
% Change image for manipulation
dbx=[]; % A matrix
for i=1:M
temp=double(S(:,i));
dbx=[dbx temp];
end
%Covariance matrix C=A'A, L=AA'
A=dbx';
L=A*A';
% vv are the eigenvector for L
% dd are the eigenvalue for both L=dbx'*dbx and C=dbx*dbx';
[vv dd]=eig(L);
% Sort and eliminate those whose eigenvalue is zero
v=[];
d=[];
for i=1:size(vv,2)
if(dd(i,i)>1e-4)
v=[v vv(:,i)];
d=[d dd(i,i)];
end
end
%sort, will return an ascending sequence
[B index]=sort(d);
ind=zeros(size(index));
dtemp=zeros(size(index));
vtemp=zeros(size(v));
len=length(index);
for i=1:len
dtemp(i)=B(len+1-i);
ind(i)=len+1-index(i);
vtemp(:,ind(i))=v(:,i);
end
d=dtemp;
v=vtemp;
%Normalization of eigenvectors
for i=1:size(v,2) %access each column
kk=v(:,i);
temp=sqrt(sum(kk.^2));
v(:,i)=v(:,i)./temp;
end
%Eigenvectors of C matrix
u=[];
for i=1:size(v,2)
temp=sqrt(d(i));
u=[u (dbx*v(:,i))./temp];
end
%Normalization of eigenvectors
for i=1:size(u,2)
kk=u(:,i);
temp=sqrt(sum(kk.^2));
u(:,i)=u(:,i)./temp;
end
% show eigenfaces;
for i=1:size(u,2)
img=reshape(u(:,i),icol,irow);
img=img';
img=histeq(img,255);
end
% Find the weight of each face in the training set.
omega = [];
for h=1:size(dbx,2)
WW=[];
for i=1:size(u,2)
t = u(:,i)';
WeightOfImage = dot(t,dbx(:,h)');
WW = [WW; WeightOfImage];
end
omega = [omega WW];
end
% Acquire new image
% Note: the input image must have a bmp or jpg extension.
% It should have the same size as the ones in your training set.
% It should be placed on your desktop
ed_min=[];
srcFiles = dir('G:\newdatabase\*.jpg'); % the folder in which ur images exists
for b = 1 : length(srcFiles)
filename = strcat('G:\newdatabase\',srcFiles(b).name);
Imgdata = imread(filename);
InputImage=Imgdata;
InImage=reshape(permute((double(InputImage)),[2,1,3]),[irow*icol,1]);
temp=InImage;
me=mean(temp);
st=std(temp);
temp=(temp-me)*ustd/st+um;
NormImage = temp;
Difference = temp-m;
p = [];
aa=size(u,2);
for i = 1:aa
pare = dot(NormImage,u(:,i));
p = [p; pare];
end
InImWeight = [];
for i=1:size(u,2)
t = u(:,i)';
WeightOfInputImage = dot(t,Difference');
InImWeight = [InImWeight; WeightOfInputImage];
end
noe=numel(InImWeight);
% Find Euclidean distance
e=[];
for i=1:size(omega,2)
q = omega(:,i);
DiffWeight = InImWeight-q;
mag = norm(DiffWeight);
e = [e mag];
end
ed_min=[ed_min MinimumValue];
theta=6.0e+03;
%disp(e)
z(b,:)=InImWeight;
end
IDX = kmeans(z,5);
clustercount=accumarray(IDX, ones(size(IDX)));
disp(clustercount);
Running time for 100 images:Elapsed time is 103.947573 seconds.
QUESTIONS:
1.It is working fine for M=50(i.e Training set contains 50 images) but not for M=1200(i.e Training set contains 1200 images).It is not showing any error.There is no output.I waited for 10 min still there is no output.What is the problem?Where i was wrong?
To answer your second question, you can simply 'save' any generated variable as a .mat file in your working directory (Current folder) which can be accessed later. So in your code, if the 'training eigenfaces' is given by the variable 'u', you can use the following:
save('eigenface.mat','u')
This creates a .mat file with the name eigenface.mat which contains the variable 'u', the eigenfaces. Note that this variable is saved in your Current Folder.
In a later instance when you are trying out with your test data, you can simply 'load' this variable:
load('eigenface.mat')
This automatically loads 'u' into your workspace.
You can also save additional variables in the same .mat file if necessary
save('eigenface.mat','u','v'...)
The answer to the first question is that the code is simply not done running yet. Vectorizing the code (instead of using a for loop), as suggested in the comment section above can improve the speed significantly.
[EDIT]
Since the images are not very big, the code is not significantly slowed down by the first for loop. You can improve the performance of the rest of the code by vectorizing the code. Vectorized operations are faster than for-loops. This link might be useful in understanding vectorization:
http://www.mathworks.com/help/matlab/matlab_prog/vectorization.html
For example, the second for-loop can be replaced by the following vectorized form, as suggested in the comments
tempS = double(S);
meanS = mean(S,1);
stdS = std(S,0,1);
S = (tempS - meanS) ./ stdS;
Use MATLAB's timer functions tic and toc for finding how long the first for-loop alone is taking to execute. Add tic before the for loop and toc after it. If the time taken for 50 images is about 104 seconds, then it would be significantly more for 1200 images.
The problem is:
In a class of 26 students, a test with 10 questions is given. The students answer the question by tossing a coin. I have to find out how many students have two or less answers correct. This is the program that I wrote, but I'm not sure about it ... is it good?
correct=0;
students=0;
for i=1:26
for j=1:10
answ=ceil(rand);
if answ==1
correct=correct+1;
if correct==2
students=students+1;
end
end
end
end
disp(students)
It's neater, faster to run, and more readable if you do it vectorized:
answers = round(rand(10,26)); % 10x26 matrix of 0 / 1 values
correct = sum(answers); % sum along each column
students = sum(correct<=2) % how many values of correct are 2 or less
By the way, from your code it appears you want to know how many students have 2 or more correct answers (not 2 or less as you state). In that case change last line to
students = sum(correct>=2) % how many values of correct are 2 or more
Slight modification to your code:
correct=0;
students=0;
for i=1:26
for j=1:10
answ=round(rand); % Use round instead of ceil
if (answ==1)
correct=correct+1;
if correct==2
students=students+1;
break;
end
end
end
correct=0; % Was missing in original code
end
disp(students)
Additional Note:
Try testing code for correct ans 5 as this will have more chances of variation to result because of uniform distribution of random number. Most of the students will get at-least 2 correct ans assuming equal probability, which is not the practical case.
Try maintaining habit of proper indentation. It will be more readable and less prone to errors.
Just like this : (You have to reset correct counter for every student, and some end aren't at the right place.)
correct = 0;
students = 0;
for i = 1:26
for j = 1:10
answ = ceil(rand);
if(answ==1)
correct = correct + 1;
end
end
if(correct < 2)
students = students + 1;
end
correct= 0;
end
EdIT
Sorry, I didn't saw the less in your sentense, so I change the > for <
I am running a parfor loop in Matlab that takes a lot of time and I would like to know how many iterations are left. How can I get that info?
I don't believe you can get that information directly from MATLAB, short of printing something with each iteration and counting these lines by hand.
To see why, recall that each parfor iteration executes in its own workspace: while incrementing a counter within the loop is legal, accessing its "current" value is not (because this value does not really exist until completion of the loop). Furthermore, the parfor construct does not guarantee any particular execution order, so printing the iterator value isn't helpful.
cnt = 0;
parfor i=1:n
cnt = cnt + 1; % legal
disp(cnt); % illegal
disp(i); % legal ofc. but out of order
end
Maybe someone does have a clever workaround, but I think that the independent nature of the parfor iterations belies taking a reliable count. The restrictions mentioned above, plus those on using evalin, etc. support this conclusion.
As #Jonas suggested, you could obtain the iteration count via side effects occurring outside of MATLAB, e.g. creating empty files in a certain directory and counting them. This you can do in MATLAB of course:
fid = fopen(['countingDir/f' num2str(i)],'w');
fclose(fid);
length(dir('countingDir'));
Try this FEX file: http://www.mathworks.com/matlabcentral/fileexchange/32101-progress-monitor--progress-bar--that-works-with-parfor
You can easily modify it to return the iteration number instead of displaying a progress bar.
Something like a progress bar could be done similar to this...
Before the parfor loop :
fprintf('Progress:\n');
fprintf(['\n' repmat('.',1,m) '\n\n']);
And during the loop:
fprintf('\b|\n');
Here we have m is the total number of iterations, the . shows the total number of iterations and | shows the number of iterations completed. The \n makes sure the characters are printed in the parfor loop.
With Matlab 2017a or later you can use a data queue or a pollable data queue to achieve this. Here's the MathWorks documentation example of how to do a progress bar from the first link :
function a = parforWaitbar
D = parallel.pool.DataQueue;
h = waitbar(0, 'Please wait ...');
afterEach(D, #nUpdateWaitbar);
N = 200;
p = 1;
parfor i = 1:N
a(i) = max(abs(eig(rand(400))));
send(D, i);
end
function nUpdateWaitbar(~)
waitbar(p/N, h);
p = p + 1;
end
end
End result :
If you just want to know how much time is left approximately, you can run the program once record the max time and then do this
tStart = tic;
parfor i=1:n
tElapsed = toc(tStart;)
disp(['Time left in min ~ ', num2str( ( tMax - tElapsed ) / 60 ) ]);
...
end
I created a utility to do this:
http://www.mathworks.com/matlabcentral/fileexchange/48705-drdan14-parforprogress