I am trying to calculate the co-occurrence of some values in a vector in Matlab. I am using the following code to do so:
x = graph(:,1);
y = zeros(size(x));
for i = 1:length(x)
y(i) = sum(x==x(i));
end
The above code calculates the co-occurrence of every index inside the vector. I want to have the co-occurrence of the unique indexes. How can I do so?
I found the following implementation:
a = unique(x);
out = [a,histc(x(:),a)];
However, I want the indexes to be as it is, without sorting.
Let's see if this is what you need:
a=unique(x);
Coocurrence=zeros(length(a));
for ii=1:length(a)
Coocurrence(ii)=sum(x==a(ii));
end
or the vectorized solution
a=unique(x);
Coocurrence=sum(bsxfun(#eq,x,a'),2);
Related
The following is a function that takes two equal sized vectors X and Y, and is supposed to return a vector containing single correlation coefficients for image correspondence. The function is supposed to work similarly to the built in corr(X,Y) function in matlab if given two equal sized vectors. Right now my code is producing a vector containing multiple two-number vectors instead of a vector containing single numbers. How do I fix this?
function result = myCorr(X, Y)
meanX = mean(X);
meanY = mean(Y);
stdX = std(X);
stdY = std(Y);
for i = 1:1:length(X),
X(i) = (X(i) - meanX)/stdX;
Y(i) = (Y(i) - meanY)/stdY;
mult = X(i) * Y(i);
end
result = sum(mult)/(length(X)-1);
end
Edit: To clarify I want myCorr(X,Y) above to produce the same output at matlab's corr(X,Y) when given equal sized vectors of image intensity values.
Edit 2: Now the format of the output vector is correct, however the values are off by a lot.
I recommend you use r=corrcoef(X,Y) it will give you a normalized r value you are looking for in a 2x2 matrix and you can just return the r(2,1) entry as your answer. Doing this is equivalent to
r=(X-mean(X))*(Y-mean(Y))'/(sqrt(sum((X-mean(X)).^2))*sqrt(sum((Y-mean(Y)).^2)))
However, if you really want to do what you mentioned in the question you can also do
r=(X)*(Y)'/(sqrt(sum((X-mean(X)).^2))*sqrt(sum((Y-mean(Y)).^2)))
I'm newbie to Matlab. I have some questions.
How can I vectorise this loop:
epsilon = 0.45;
t = -3.033;
n = 100;
I = ones(n,n);
%// diagonal block 1
DB1 = gallery('tridiag',ones(1,n-1),ones(1,n),ones(1,n-1));
for k = 1:n
DB1(k,k) = epsilon;
end
for k = 1:n-1
DB1(k,k+1) = t*heaviside((-1)^(k+1));
end
for k = 2:n
DB1(k,k-1) = t*heaviside((-1)^k);
and this loop with LRG, R2, R1 are 3D arrays
for k = 2:N
LRG(:,:,k) = inv(R(:,:,k) - R2(:,:,k-1)*LRG(:,:,k-1)*R1(:,:,k-1));
end
Is there any way to handle the third dimension of an array (page) without writing (:,:,...) many times?
You don't need to call gallery: this will just initialize DB1 to a specific sparse tridiagonal matrix, which you manually overwrite afterwards. With n=100, I suspect that you don't actually want to work on sparse matrices. Here's the solution:
epsilon = 0.45; t = -3.033;
n = 100;
DB1 = zeros(n); %initialize to 0
%diagonal indices
inds=sub2ind([n n],1:n,1:n);
DB1(inds) = epsilon;
%superdiagonal
inds=sub2ind([n n],1:n-1,2:n);
DB1(inds) = t*heaviside((-1).^(2:n));
%subdiagonal
inds=sub2ind([n n],2:n,1:n-1);
DB1(inds) = t*heaviside((-1).^(2:n));
Here sub2ind lets you compute linear indices from your matrix indices, allowing you to access "subvectors" in your matrix. Watch out for the exponentialization in the heaviside: you need .^ instead of ^ to perform an element-wise operation using the vector 2:n.
If you insist on getting a sparse matrix, you could either convert this matrix to sparse by calling
DB1=sparse(DB1);
which is the easiest option if your matrix is small. If n was larger and you would really need sparse matrices, then you could set up the sparse matrix itself by
DB1new=sparse([1:n 1:n-1 2:n],[1:n 2:n 1:n-1],...
[epsilon*ones(1,n) t*heaviside((-1).^(2:n)) t*heaviside((-1).^(2:n))],n,n);
This call sets the elements of DB1new one-by-one according to its vector arguments: it takes the first index from the first input argument, the seconf index from the second, and the corresponding element from the third vector (see help sparse in MATLAB).
As for your second question: I'm not sure I understand correctly. Like in my example, sub2ind can be used to transform the multidimensional indices into linear ones, see help sub2ind. But personally, I think using the array notation is far more transparent and less prone to typos. And I think you can't vectorize that loop: you have to explicitly calculate the inverse matrix for each k.
I am working in matlab. I have five matrices in ,out, out_temp,ind_i , ind_j, all of identical dimensions say n x m. I want to implement the following loop in one line.
out = zeros(n,m)
out_temp = zeros(n,m)
for i = 1:n
for j = 1:m
out(ind_i(i,j),ind_j(i,j)) = in(ind_i(i,j),ind_j(i,j));
out_temp(ind_i(i,j),ind_j(i,j)) = some_scalar_value;
end
end
It is assured that the values in ind_i lies in range 1:n and values in ind_j lies in range 1:m.
I believe a way to implement line 3 would give the way to implement line 4 , but I wrote it to be clear about what I want.
Code
%// Calculate the linear indices in one go using all indices from ind_i and ind_j
%// keeping in mind that the sizes of both out and out_temp won't go beyond
%// the maximum of ind_i for the number of rows and maximum of ind_j for number
%// of columns
ind1 = sub2ind([n m],ind_i(:),ind_j(:))
%// Initialize out and out_temp
out = zeros(n,m)
out_temp = zeros(n,m)
%// Finally index into out and out_temp and assign them values
%// using indiced values from in and the scalar value respectively.
out(ind1) = in(ind1);
out_temp(ind1) = some_scalar_value;
I'm going to start off by stating that, yes, this is homework (my first homework question on stackoverflow!). But I don't want you to solve it for me, I just want some guidance!
The equation in question is this:
I'm told to take N = 50, phi1 = 300, phi2 = 400, 0<=x<=1, and 0<=y<=1, and to let x and y be vectors of 100 equally spaced points, including the end points.
So the first thing I did was set those variables, and used x = linspace(0,1) and y = linspace(0,1) to make the correct vectors.
The question is Write a MATLAB script file called potential.m which calculates phi(x,y) and makes a filled contour plot versus x and y using the built-in function contourf (see the help command in MATLAB for examples). Make sure the figure is labeled properly. (Hint: the top and bottom portions of your domain should be hotter at about 400 degrees versus the left and right sides which should be at 300 degrees).
However, previously, I've calculated phi using either x or y as a constant. How am I supposed to calculate it where both are variables? Do I hold x steady, while running through every number in the vector of y, assigning that to a matrix, incrementing x to the next number in its vector after running through every value of y again and again? And then doing the same process, but slowly incrementing y instead?
If so, I've been using a loop that increments to the next row every time it loops through all 100 values. If I did it that way, I would end up with a massive matrix that has 200 rows and 100 columns. How would I use that in the linspace function?
If that's correct, this is how I'm finding my matrix:
clear
clc
format compact
x = linspace(0,1);
y = linspace(0,1);
N = 50;
phi1 = 300;
phi2 = 400;
phi = 0;
sum = 0;
for j = 1:100
for i = 1:100
for n = 1:N
sum = sum + ((2/(n*pi))*(((phi2-phi1)*(cos(n*pi)-1))/((exp(n*pi))-(exp(-n*pi))))*((1-(exp(-n*pi)))*(exp(n*pi*y(i)))+((exp(n*pi))-1)*(exp(-n*pi*y(i))))*sin(n*pi*x(j)));
end
phi(j,i) = phi1 - sum;
end
end
for j = 1:100
for i = 1:100
for n = 1:N
sum = sum + ((2/(n*pi))*(((phi2-phi1)*(cos(n*pi)-1))/((exp(n*pi))-(exp(-n*pi))))*((1-(exp(-n*pi)))*(exp(n*pi*y(j)))+((exp(n*pi))-1)*(exp(-n*pi*y(j))))*sin(n*pi*x(i)));
end
phi(j+100,i) = phi1 - sum;
end
end
This is the definition of contourf. I think I have to use contourf(X,Y,Z):
contourf(X,Y,Z), contourf(X,Y,Z,n), and contourf(X,Y,Z,v) draw filled contour plots of Z using X and Y to determine the x- and y-axis limits. When X and Y are matrices, they must be the same size as Z and must be monotonically increasing.
Here is the new code:
N = 50;
phi1 = 300;
phi2 = 400;
[x, y, n] = meshgrid(linspace(0,1),linspace(0,1),1:N)
f = phi1-((2./(n.*pi)).*(((phi2-phi1).*(cos(n.*pi)-1))./((exp(n.*pi))-(exp(-n.*pi)))).*((1-(exp(-1.*n.*pi))).*(exp(n.*pi.*y))+((exp(n.*pi))-1).*(exp(-1.*n.*pi.*y))).*sin(n.*pi.*x));
g = sum(f,3);
[x1,y1] = meshgrid(linspace(0,1),linspace(0,1));
contourf(x1,y1,g)
Vectorize the code. For example you can write f(x,y,n) with:
[x y n] = meshgrid(-1:0.1:1,-1:0.1:1,1:10);
f=exp(x.^2-y.^2).*n ;
f is a 3D matrix now just sum over the right dimension...
g=sum(f,3);
in order to use contourf, we'll take only the 2D part of x,y:
[x1 y1] = meshgrid(-1:0.1:1,-1:0.1:1);
contourf(x1,y1,g)
The reason your code takes so long to calculate the phi matrix is that you didn't pre-allocate the array. The error about size happens because phi is not 100x100. But instead of fixing those things, there's an even better way...
MATLAB is a MATrix LABoratory so this type of equation is pretty easy to compute using matrix operations. Hints:
Instead of looping over the values, rows, or columns of x and y, construct matrices to represent all the possible input combinations. Check out meshgrid for this.
You're still going to need a loop to sum over n = 1:N. But for each value of n, you can evaluate your equation for all x's and y's at once (using the matrices from hint 1). The key to making this work is using element-by-element operators, such as .* and ./.
Using matrix operations like this is The Matlab Way. Learn it and love it. (And get frustrated when using most other languages that don't have them.)
Good luck with your homework!
I am working towards comparing multiple images. I have these image data as column vectors of a matrix called "images." I want to assess the similarity of images by first computing their Eucledian distance. I then want to create a matrix over which I can execute multiple random walks. Right now, my code is as follows:
% clear
% clc
% close all
%
% load tea.mat;
images = Input.X;
M = zeros(size(images, 2), size (images, 2));
for i = 1:size(images, 2)
for j = 1:size(images, 2)
normImageTemp = sqrt((sum((images(:, i) - images(:, j))./256).^2));
%Need to accurately select the value of gamma_i
gamma_i = 1/10;
M(i, j) = exp(-gamma_i.*normImageTemp);
end
end
My matrix M however, ends up having a value of 1 along its main diagonal and zeros elsewhere. I'm expecting "large" values for the first few elements of each row and "small" values for elements with column index > 4. Could someone please explain what is wrong? Any advice is appreciated.
Since you're trying to compute a Euclidean distance, it looks like you have an error in where your parentheses are placed when you compute normImageTemp. You have this:
normImageTemp = sqrt((sum((...)./256).^2));
%# ^--- Note that this parenthesis...
But you actually want to do this:
normImageTemp = sqrt(sum(((...)./256).^2));
%# ^--- ...should be here
In other words, you need to perform the element-wise squaring, then the summation, then the square root. What you are doing now is summing elements first, then squaring and taking the square root of the summation, which essentially cancel each other out (or are actually the equivalent of just taking the absolute value).
Incidentally, you can actually use the function NORM to perform this operation for you, like so:
normImageTemp = norm((images(:, i) - images(:, j))./256);
The results you're getting seem reasonable. Recall the behavior of the exp(-x). When x is zero, exp(-x) is 1. When x is large exp(-x) is zero.
Perhaps if you make M(i,j) = normImageTemp; you'd see what you expect to see.
Consider this solution:
I = Input.X;
D = squareform( pdist(I') ); %'# euclidean distance between columns of I
M = exp(-(1/10) * D); %# similarity matrix between columns of I
PDIST and SQUAREFORM are functions from the Statistics Toolbox.
Otherwise consider this equivalent vectorized code (using only built-in functions):
%# we know that: ||u-v||^2 = ||u||^2 + ||v||^2 - 2*u.v
X = sum(I.^2,1);
D = real( sqrt(bsxfun(#plus,X,X')-2*(I'*I)) );
M = exp(-(1/10) * D);
As was explained in the other answers, D is the distance matrix, while exp(-D) is the similarity matrix (which is why you get ones on the diagonal)
there is an already implemented function pdist, if you have a matrix A, you can directly do
Sim= squareform(pdist(A))