Convert to inHomogeneous cordinates Eigen::MatrixXf - eigen

I have a 3x1000 matrix Eigen::MatrixXf. I need to divide the matrix with the 3rd row.
A(0,:) = A(0,:) / A(2,:)
A(1,:) = A(1,:) / A(2,:)
A(2,:) = A(2,:) / A(2,:)
How can I perform this operation in Eigen. In particular I cannot find anything for componentwise division

Use as an array:
A.array() /= A.array().col(2);
For more information, see the documentation.

Related

Efficient way of computing multivariate gaussian varying the mean - Matlab

Is there a efficient way to do the computation of a multivariate gaussian (as below) that returns matrix p , that is, making use of some sort of vectorization? I am aware that matrix p is symmetric, but still for a matrix of size 40000x3, for example, this will take quite a long time.
Matlab code example:
DataMatrix = [3 1 4; 1 2 3; 1 5 7; 3 4 7; 5 5 1; 2 3 1; 4 4 4];
[rows, cols ] = size(DataMatrix);
I = eye(cols);
p = zeros(rows);
for k = 1:rows
p(k,:) = mvnpdf(DataMatrix(:,:),DataMatrix(k,:),I);
end
Stage 1: Hack into source code
Iteratively we are performing mvnpdf(DataMatrix(:,:),DataMatrix(k,:),I)
The syntax is : mvnpdf(X,Mu,Sigma).
Thus, the correspondence with our input becomes :
X = DataMatrix(:,:);
Mu = DataMatrix(k,:);
Sigma = I
For the sizes relevant to our situation, the source code mvnpdf.m reduces to -
%// Store size parameters of X
[n,d] = size(X);
%// Get vector mean, and use it to center data
X0 = bsxfun(#minus,X,Mu);
%// Make sure Sigma is a valid covariance matrix
[R,err] = cholcov(Sigma,0);
%// Create array of standardized data, and compute log(sqrt(det(Sigma)))
xRinv = X0 / R;
logSqrtDetSigma = sum(log(diag(R)));
%// Finally get the quadratic form and thus, the final output
quadform = sum(xRinv.^2, 2);
p_out = exp(-0.5*quadform - logSqrtDetSigma - d*log(2*pi)/2)
Now, if the Sigma is always an identity matrix, we would have R as an identity matrix too. Therefore, X0 / R would be same as X0, which is saved as xRinv. So, essentially quadform = sum(X0.^2, 2);
Thus, the original code -
for k = 1:rows
p(k,:) = mvnpdf(DataMatrix(:,:),DataMatrix(k,:),I);
end
reduces to -
[n,d] = size(DataMatrix);
[R,err] = cholcov(I,0);
p_out = zeros(rows);
K = sum(log(diag(R))) + d*log(2*pi)/2;
for k = 1:rows
X0 = bsxfun(#minus,DataMatrix,DataMatrix(k,:));
quadform = sum(X0.^2, 2);
p_out(k,:) = exp(-0.5*quadform - K);
end
Now, if the input matrix is of size 40000x3, you might want to stop here. But with system resources permitting, you can vectorize everything as discussed next.
Stage 2: Vectorize everything
Now that we see what's actually going on and that the computations look parallelizable, it's time to step-up to use bsxfun in 3D with his good friend permute for a vectorized solution, like so -
%// Get size params and R
[n,d] = size(DataMatrix);
[R,err] = cholcov(I,0);
%// Calculate constants : "logSqrtDetSigma" and "d*log(2*pi)/2`"
K1 = sum(log(diag(R)));
K2 = d*log(2*pi)/2;
%// Major thing happening here as we calclate "X0" for all iterations
%// in one go with permute and bsxfun
diffs = bsxfun(#minus,DataMatrix,permute(DataMatrix,[3 2 1]));
%// "Sigma" is an identity matrix, so it plays no in "/R" at "xRinv = X0 / R".
%// Perform elementwise squaring and summing rows to get vectorized "quadform"
quadform1 = squeeze(sum(diffs.^2,2))
%// Finally use "quadform1" and get vectorized output as a 2D array
p_out = exp(-0.5*quadform1 - K1 - K2)

Formula for calculating distance with decaying velocity

I have a moving graphic whose velocity decays geometrically every frame. I want to find the initial velocity that will make the graphic travel a desired distance in a given number of frames.
Using these variables:
v initial velocity
r rate
d distance
I can come up with d = v * (r0 + r1 + r2 + ...)
So if I want to find the v to travel 200 pixels in 3 frames with a decay rate of 90%, I would adapt to:
d = 200
r = .9
v = d / (r0 + r1 + r2)
That doesn't translate well to code, since I have to edit the expression if the number of frames changes. The only solution I can think of is this (in no specific language):
r = .9
numFrames = 3
d = 200
sum = 1
for (i = 1; i < numFrames; i++) {
sum = sum + power(r, i);
}
v = d / sum;
Is there a better way to do this without using a loop?
(I wouldn't be surprised if there is a mistake in there somewhere... today is just one of those days..)
What you have here is a geometric sequence. See the link:
http://www.mathsisfun.com/algebra/sequences-sums-geometric.html
To find the sum of a geometric sequence, you use this formula:
sum = a * ((1 - r^n) / (1 - r))
Since you are looking for a, the initial velocity, move the terms around:
a = sum * ((1-r) / (1 - r^n))
In Java:
int distanceInPixels = SOME_INTEGER;
int decayRate = SOME_DECIMAl;
int numberOfFrames = SOME_INTEGER;
int initialVelocity; //this is what we need to find
initialVelocity = distanceinPixel * ((1-decayRate) / (1-Math.pow(decayRate, NumberOfFrames)));
Using this formula you can get any one of the four variables if you know the values of the other three. Enjoy!
According to http://mikestoolbox.com/powersum.html, you should be able to reduce your for loop to:
F(x) = (x^n - 1)/(x-1)

Efficiently compute pairwise squared Euclidean distance in Matlab

Given two sets of d-dimensional points. How can I most efficiently compute the pairwise squared euclidean distance matrix in Matlab?
Notation:
Set one is given by a (numA,d)-matrix A and set two is given by a (numB,d)-matrix B. The resulting distance matrix shall be of the format (numA,numB).
Example points:
d = 4; % dimension
numA = 100; % number of set 1 points
numB = 200; % number of set 2 points
A = rand(numA,d); % set 1 given as matrix A
B = rand(numB,d); % set 2 given as matrix B
The usually given answer here is based on bsxfun (cf. e.g. [1]). My proposed approach is based on matrix multiplication and turns out to be much faster than any comparable algorithm I could find:
helpA = zeros(numA,3*d);
helpB = zeros(numB,3*d);
for idx = 1:d
helpA(:,3*idx-2:3*idx) = [ones(numA,1), -2*A(:,idx), A(:,idx).^2 ];
helpB(:,3*idx-2:3*idx) = [B(:,idx).^2 , B(:,idx), ones(numB,1)];
end
distMat = helpA * helpB';
Please note:
For constant d one can replace the for-loop by hardcoded implementations, e.g.
helpA(:,3*idx-2:3*idx) = [ones(numA,1), -2*A(:,1), A(:,1).^2, ... % d == 2
ones(numA,1), -2*A(:,2), A(:,2).^2 ]; % etc.
Evaluation:
%% create some points
d = 2; % dimension
numA = 20000;
numB = 20000;
A = rand(numA,d);
B = rand(numB,d);
%% pairwise distance matrix
% proposed method:
tic;
helpA = zeros(numA,3*d);
helpB = zeros(numB,3*d);
for idx = 1:d
helpA(:,3*idx-2:3*idx) = [ones(numA,1), -2*A(:,idx), A(:,idx).^2 ];
helpB(:,3*idx-2:3*idx) = [B(:,idx).^2 , B(:,idx), ones(numB,1)];
end
distMat = helpA * helpB';
toc;
% compare to pdist2:
tic;
pdist2(A,B).^2;
toc;
% compare to [1]:
tic;
bsxfun(#plus,dot(A,A,2),dot(B,B,2)')-2*(A*B');
toc;
% Another method: added 07/2014
% compare to ndgrid method (cf. Dan's comment)
tic;
[idxA,idxB] = ndgrid(1:numA,1:numB);
distMat = zeros(numA,numB);
distMat(:) = sum((A(idxA,:) - B(idxB,:)).^2,2);
toc;
Result:
Elapsed time is 1.796201 seconds.
Elapsed time is 5.653246 seconds.
Elapsed time is 3.551636 seconds.
Elapsed time is 22.461185 seconds.
For a more detailed evaluation w.r.t. dimension and number of data points follow the discussion below (#comments). It turns out that different algos should be preferred in different settings. In non time critical situations just use the pdist2 version.
Further development:
One can think of replacing the squared euclidean by any other metric based on the same principle:
help = zeros(numA,numB,d);
for idx = 1:d
help(:,:,idx) = [ones(numA,1), A(:,idx) ] * ...
[B(:,idx)' ; -ones(1,numB)];
end
distMat = sum(ANYFUNCTION(help),3);
Nevertheless, this is quite time consuming. It could be useful to replace for smaller d the 3-dimensional matrix help by d 2-dimensional matrices. Especially for d = 1 it provides a method to compute the pairwise difference by a simple matrix multiplication:
pairDiffs = [ones(numA,1), A ] * [B'; -ones(1,numB)];
Do you have any further ideas?
For squared Euclidean distance one can also use the following formula
||a-b||^2 = ||a||^2 + ||b||^2 - 2<a,b>
Where <a,b> is the dot product between a and b
nA = sum( A.^2, 2 ); %// norm of A's elements
nB = sum( B.^2, 2 ); %// norm of B's elements
distMat = bsxfun( #plus, nA, nB' ) - 2 * A * B' ;
Recently, I've been told that as of R2016b this method for computing square Euclidean distance is faster than accepted method.

Algorithm to express elements of a matrix as a vector

Statement of Problem:
I have an array M with m rows and n columns. The array M is filled with non-zero elements.
I also have a vector t with n elements, and a vector omega
with m elements.
The elements of t correspond to the columns of matrix M.
The elements of omega correspond to the rows of matrix M.
Goal of Algorithm:
Define chi as the multiplication of vector t and omega. I need to obtain a 1D vector a, where each element of a is a function of chi.
Each element of chi is unique (i.e. every element is different).
Using mathematics notation, this can be expressed as a(chi)
Each element of vector a corresponds to an element or elements of M.
Matlab code:
Here is a code snippet showing how the vectors t and omega are generated. The matrix M is pre-existing.
[m,n] = size(M);
t = linspace(0,5,n);
omega = linspace(0,628,m);
Conceptual Diagram:
This appears to be a type of integration (if this is the right word for it) along constant chi.
Reference:
Link to reference
The algorithm is not explicitly stated in the reference. I only wish that this algorithm was described in a manner reminiscent of computer science textbooks!
Looking at Figure 11.5, the matrix M is Figure 11.5(a). The goal is to find an algorithm to convert Figure 11.5(a) into 11.5(b).
It appears that the algorithm is a type of integration (averaging, perhaps?) along constant chi.
It appears to me that reshape is the matlab function you need to use. As noted in the link:
B = reshape(A,siz) returns an n-dimensional array with the same elements as A, but reshaped to siz, a vector representing the dimensions of the reshaped array.
That is, create a vector siz with the number m*n in it, and say A = reshape(P,siz), where P is the product of vectors t and ω; or perhaps say something like A = reshape(t*ω,[m*n]). (I don't have matlab here, or would run a test to see if I have the product the right way around.) Note, the link does not show an example with one number (instead of several) after the matrix parameter to reshape, but I would expect from the description that A = reshape(t*ω,m*n) might also work.
You should add a pseudocode or a link to the algorithm you want to implement. From what I could understood I have developed the following code anyway:
M = [1 2 3 4; 5 6 7 8; 9 10 11 12]' % easy test M matrix
a = reshape(M, prod(size(M)), 1) % convert M to vector 'a' with reshape command
[m,n] = size(M); % Your sample code
t = linspace(0,5,n); % Your sample code
omega = linspace(0,628,m); % Your sample code
for i=1:length(t)
for j=1:length(omega) % Acces a(chi) in the desired order
chi = length(omega)*(i-1)+j;
t(i) % related t value
omega(j) % related omega value
a(chi) % related a(chi) value
end
end
As you can see, I also think that the reshape() function is the solution to your problems. I hope that this code helps,
The basic idea is to use two separate loops. The outer loop is over the chi variable values, whereas the inner loop is over the i variable values. Referring to the above diagram in the original question, the i variable corresponds to the x-axis (time), and the j variable corresponds to the y-axis (frequency). Assuming that the chi, i, and j variables can take on any real number, bilinear interpolation is then used to find an amplitude corresponding to an element in matrix M. The integration is just an averaging over elements of M.
The following code snippet provides an overview of the basic algorithm to express elements of a matrix as a vector using the spectral collapsing from 2D to 1D. I can't find any reference for this, but it is a solution that works for me.
% Amp = amplitude vector corresponding to Figure 11.5(b) in book reference
% M = matrix corresponding to the absolute value of the complex Gabor transform
% matrix in Figure 11.5(a) in book reference
% Nchi = number of chi in chi vector
% prod = product of timestep and frequency step
% dt = time step
% domega = frequency step
% omega_max = maximum angular frequency
% i = time array element along x-axis
% j = frequency array element along y-axis
% current_i = current time array element in loop
% current_j = current frequency array element in loop
% Nchi = number of chi
% Nivar = number of i variables
% ivar = i variable vector
% calculate for chi = 0, which only occurs when
% t = 0 and omega = 0, at i = 1
av0 = mean( M(1,:) );
av1 = mean( M(2:end,1) );
av2 = mean( [av0 av1] );
Amp(1) = av2;
% av_val holds the sum of all values that have been averaged
av_val_sum = 0;
% loop for rest of chi
for ccnt = 2:Nchi % 2:Nchi
av_val_sum = 0; % reset av_val_sum
current_chi = chi( ccnt ); % current value of chi
% loop over i vector
for icnt = 1:Nivar % 1:Nivar
current_i = ivar( icnt );
current_j = (current_chi / (prod * (current_i - 1))) + 1;
current_t = dt * (current_i - 1);
current_omega = domega * (current_j - 1);
% values out of range
if(current_omega > omega_max)
continue;
end
% use bilinear interpolation to find an amplitude
% at current_t and current_omega from matrix M
% f_x_y is the bilinear interpolated amplitude
% Insert bilinear interpolation code here
% add to running sum
av_val_sum = av_val_sum + f_x_y;
end % icnt loop
% compute the average over all i
av = av_val_sum / Nivar;
% assign the average to Amp
Amp(ccnt) = av;
end % ccnt loop

MATLAB loop optimization

I have a matrix, matrix_logical(50000,100000), that is a sparse logical matrix (a lot of falses, some true). I have to produce a matrix, intersect(50000,50000), that, for each pair, i,j, of rows of matrix_logical(50000,100000), stores the number of columns for which rows i and j have both "true" as the value.
Here is the code I wrote:
% store in advance the nonzeros cols
for i=1:50000
nonzeros{i} = num2cell(find(matrix_logical(i,:)));
end
intersect = zeros(50000,50000);
for i=1:49999
a = cell2mat(nonzeros{i});
for j=(i+1):50000
b = cell2mat(nonzeros{j});
intersect(i,j) = numel(intersect(a,b));
end
end
Is it possible to further increase the performance? It takes too long to compute the matrix. I would like to avoid the double loop in the second part of the code.
matrix_logical is sparse, but it is not saved as sparse in MATLAB because otherwise the performance become the worst possible.
Since the [i,j] entry counts the number of non zero elements in the element-wise multiplication of rows i and j, you can do it by multiplying matrix_logical with its transpose (you should convert to numeric data type first, e.g matrix_logical = single(matrix_logical)):
inter = matrix_logical * matrix_logical';
And it works both for sparse or full representation.
EDIT
In order to calculate numel(intersect(a,b))/numel(union(a,b)); (as asked in your comment), you can use the fact that for two sets a and b, you have
length(union(a,b)) = length(a) + length(b) - length(intersect(a,b))
so, you can do the following:
unLen = sum(matrix_logical,2);
tmp = repmat(unLen, 1, length(unLen)) + repmat(unLen', length(unLen), 1);
inter = matrix_logical * matrix_logical';
inter = inter ./ (tmp-inter);
If I understood you correctly, you want a logical AND of the rows:
intersct = zeros(50000, 50000)
for ii = 1:49999
for jj = ii:50000
intersct(ii, jj) = sum(matrix_logical(ii, :) & matrix_logical(jj, :));
intersct(jj, ii) = intersct(ii, jj);
end
end
Doesn't avoid the double loop, but at least works without the first loop and the slow find command.
Elaborating on my comment, here is a distance function suitable for pdist()
function out = distfun(xi,xj)
out = zeros(size(xj,1),1);
for i=1:size(xj,1)
out(i) = sum(sum( xi & xj(i,:) )) / sum(sum( xi | xj(i,:) ));
end
In my experience, sum(sum()) is faster for logicals than nnz(), thus its appearance above.
You would also need to use squareform() to reshape the output of pdist() appropriately:
squareform(pdist(martrix_logical,#distfun));
Note that pdist() includes a 'jaccard' distance measure, but it is actually the Jaccard distance and not the Jaccard index or coefficient, which is the value you are apparently after.

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