I have a m X n X k matrix and I want to find the elements that have minimal absolute value along the third dimension for each unique 2D spatial coordinate. An additional constraint is that once I find these minimum values, the sign of these values (i.e. before I took the absolute value) must be maintained.
The code I wrote to accomplish this is shown below.
tmp = abs(dist); %the size(dist)=[m,n,k]
[v,ind] = min(tmp,[],3); %find the index of minimal absolute value in the 3rd dimension
ind = reshape(ind,m*n,1);
[col,row]=meshgrid(1:n,1:m); row = reshape(row,m*n,1); col = reshape(col,m*n,1);
ind2 = sub2ind(size(dist),row,col,ind); % row, col, ind are sub
dm = dist(ind2); %take the signed value from dist
dm = reshape(dm,m,n);
The resulting matrix dm which is m X n corresponds to the matrix that is subject to the constraints that I have mentioned previously. However, this code sounds a little bit inefficient since I have to generate linear indices. Is there any way to improve this?
If I'm interpreting your problem statement correctly, you wish to find the minimum absolute value along the third dimension for each unique 2D spatial coordinate in your 3D matrix. This is already being done by the first two lines of your code.
However, a small caveat is that once you find these minimum values, you must ensure that the original sign of these values (i.e. before taking the absolute value) are respected. That is the purpose of the rest of the code.
If you want to select the original values, you don't have a choice but to generate the correct linear indices and sample from the original matrix. However, a lot of code is rather superfluous. There is no need to perform any kind of reshaping.
We can simplify your method by using ndgrid to generate the correct spatial coordinates to sample from the 3D matrix then use ind from your code to reference the third dimension. After, use this to sample dist and complete your code:
%// From your code
[v,ind] = min(abs(dist),[],3);
%// New code
[row,col] = ndgrid(1:size(dist,1), 1:size(dist,2));
%// Output
dm = dist(sub2ind(size(dist), row, col, ind));
Related
The 6 faces method is a very cheap and fast way to calibrate and accelerometer like my MPU6050, here a great description of the method.
I made 6 tests to calibrate the accelerometer based on the g vector.
After that i build up a matrix and in each row is stored the mean of each axis expressed in m/s^2, thanks to this question i automatically calculated the mean for each column in each file.
The tests were randomly performed, i tested all the six positions, but i didn't follow any path.
So i sorted manually the final matrix, based on the sort of the Y matrix, my reference matrix.
The Y elements are fixed.
The matrix manually sorted is the following
Here how i manually sorted the matrix
meanmatrix=[ax ay az];
mean1=meanmatrix(1,:);
mean2=meanmatrix(2,:);
mean3=meanmatrix(3,:);
mean4=meanmatrix(4,:);
mean5=meanmatrix(5,:);
mean6=meanmatrix(6,:);
meanmatrix= [mean1; mean3; mean2; mean4;mean6;mean5];
Based on the Y matrix constrain how can sort my matrix without knwowing "a priori" wich is the test stored in the row?
Assuming that the bias on the accelerometer is not huge, you can look at the rows of your matrix and see with which of the rows in your Y matrix matches.
sorted_meanmatrix = zeros(size(meanmatrix));
for rows = 1:length(Y)
% Calculates the square of distance and see which row has a nearest distcance
[~,index] = min(sum((meanmatrix - Y(rows,:)).^2, 2));
sorted_meanmatrix(rows,:) = meanmatrix(index,:);
end
I want to know which elements from my data set went into the tallest bin in a bivariate histogram, and have not found information on how to do this online. I suspect this is possible since it is fairly useful.
I know I can do some other code that helps me find it but I was wondering if there is a succinct way of doing this. For example I could search through the dataset with a conditional that helps me extract the things falling into the bins but I'm not interested in that. Right now I have written
X = [Eavg,Estdev];
hist3(X,[15 15])
The result is a 15x15 bin bivariate histogram. I want to extract the elements in the tallest bin in a very terse manner.
I'm doing a statistical mechanics (Monte Carlo) simulation, if its worth mentioning...
The signature [N, CEN] = hist3(... returns bincounts and center of bins. Bin centers can be converted to bin edges. Then edges can be use to find which data elements fall into a specific bin.
X = randi([1 100],10,2);
[N, CEN] = hist3(X,[5 5]);
%find row and column of highest value of histogram
%since there may be multiple histogram values that
%are equal to maximum value then we select the first one
[r,c]= find(N==max(N(:)),1);
%convert cell of bin centers to vector
R = [CEN{1}];
C = [CEN{2}];
%convert bin centers to edges
%realmax used to include values that
%are beyond the first and the last computed edges
ER = [-realmax R(1:end-1)+diff(R)/2 realmax];
EC = [-realmax C(1:end-1)+diff(C)/2 realmax];
%logical indices of rows where data fall into specified bin
IDX = X(:,1)>= ER(r) & X(:,1)< ER(r+1) & X(:,2)>= EC(c) & X(:,2)< EC(c+1)
I am trying to find the point that is at a minimum distance from the candidate set. Z is a matrix where the rows are the dimension and columns indicate points. Computing the inter-point distances, and then recording the point with minimum distance and its distance as well. Below is the code snippet. The code works fine for a small dimension and small set of points. But, it takes a long time for large data set (N = 1 million data points and dimension is also high). Is there an efficient way?
I suggest that you use pdist to do the heavy lifting for you. This function will compute the pairwise distance between every two points in your array. The resulting vector has to be put into matrix form using squareform in order to find the minimal value for each pair:
N = 100;
Z = rand(2,N); % each column is a 2-dimensional point
% pdist assumes that the second index corresponds to dimensions
% so we need to transpose inside pdist()
distmatrix = squareform(pdist(Z.','euclidean')); % output is [N, N] in size
% set diagonal values to infinity to avoid getting 0 self-distance as minimum
distmatrix = distmatrix + diag(inf(1,size(distmatrix,1)));
mindists = min(distmatrix,[],2); % find the minimum for each row
sum_dist = sum(mindists); % sum of minimal distance between each pair of points
This computes every pair twice, but I think this is true for your original implementation.
The idea is that pdist computes the pairwise distance between the columns of its input. So we put the transpose of Z into pdist. Since the full output is always a square matrix with zero diagonal, pdist is implemented such that it only returns the values above the diagonal, in a vector. So a call to squareform is needed to get the proper distance matrix. Then, the row-wise minimum of this matrix have to be found, but first we have to exclude the zero in the diagonals. I was lazy so I put inf into the diagonals, to make sure that the minimum is elsewhere. In the end we just have to sum up the minimal distances.
I'm currently attempting to select a subset of 0's in a very large matrix (about 400x300 elements) and change their value to 1. I am able to do this, but it requires using a loop where each instance it selects the next value in a randperm vector. In other words, 50% of the 0's in the matrix are randomly selected, one-at-a-time, and changed to 1:
z=1;
for z=1:(.5*numberofzeroes)
A(zeroposition(rpnumberofzeroes(z),1),zeroposition(rpnumberofzeroes(z),2))=1;
z=z+1;
end
Where 'A' is the matrix, 'zeroposition' is a 2-column-wide matrix with the positions of the 0's in the matrix (the "coordinates" if you like), and 'rpnumberofzeros' is a randperm vector from 1 to the number of zeroes in the matrix.
So for example, for z=20, the code might be something like this:
A(3557,2684)=1;
...so that the 0 which appears in this location within A will now be a 1.
It performs this loop thousands of times, because .5*numberofzeroes is a very big number. This inevitably takes a long time, so my question is can this be done without using a loop? Or at least, in some way that takes less processing resources/time?
As I said, the only thing that needs to be done is an entirely random selection of 50% (or whatever proportion) of the 0's changed to 1.
Thanks in advance for the help, and let me know if I can clear anything up! I'm new here, so apologies in advance if I've made any faux pa's.
That's very easy. I'd like to introduce you to my friend sub2ind. sub2ind allows you to take row and column coordinates of a matrix and convert them into linear column-major indices so that you can access multiple values in a matrix simultaneously in a single call. As such, the equivalent code you want is:
%// First access the values in rpnumberofzeroes
vals = rpnumberofzeroes(1:0.5*numberofzeroes, :);
%// Now, use the columns of these to determine which rows and columns we want
%// to access A
rows = zeroposition(vals(:,1), 1);
cols = zeroposition(vals(:,2), 2);
%// Get linear indices via sub2ind
ind1 = sub2ind(size(A), rows, cols);
%// Now set these locations to 1
A(ind1) = 1;
The first statement gets the first half of your matrix of coordinates stored in rpnumberofzeroes. The first column is the row coordinates, the second column is the column coordinates. Notice that in your code, you wish to use the values in zeroposition to access the locations in A. As such, extract out the corresponding rows and columns from rpnumberofzeroes to figure out the right rows and columns from zeroposition. Once that's done, we wish to use these new rows and columns from zeroposition and index into A. sub2ind requires three inputs - the size of the matrix you are trying to access... so in our case, that's A, the row coordinates and the column coordinates. The output is a set of column major indices that are computed for each row and column pair.
The last piece of the puzzle is to use these to index into A and set the locations to 1.
This can be accomplished with linear indexing as well:
% find linear position of all zeros in matrix
ix=find(abs(A)<eps);
% set one half of those, selected at random, to one.
A(ix(randperm(round(numel(ix)*.5)))=1;
Please see the following issue:
P=rand(4,4);
for i=1:size(P,2)
for j=1:size(P,2)
[r,p]=corr(P(:,i),P(:,j))
end
end
Clearly, the loop will cause the number of correlations to be doubled (i.e., corr(P(:,1),P(:,4)) and corr(P(:,4),P(:,1)). Does anyone have a suggestion on how to avoid this? Perhaps not using a loop?
Thanks!
I have four suggestions for you, depending on what exactly you are doing to compute your matrices. I'm assuming the example you gave is a simplified version of what needs to be done.
First Method - Adjusting the inner loop index
One thing you can do is change your j loop index so that it only goes from 1 up to i. This way, you get a lower triangular matrix and just concentrate on the values within the lower triangular half of your matrix. The upper half would essentially be all set to zero. In other words:
for i = 1 : size(P,2)
for j = 1 : i
%// Your code here
end
end
Second Method - Leave it unchanged, but then use unique
You can go ahead and use the same matrix like you did before with the full two for loops, but you can then filter the duplicates by using unique. In other words, you can do this:
[Y,indices] = unique(P);
Y will give you a list of unique values within the matrix P and indices will give you the locations of where these occurred within P. Note that these are column major indices, and so if you wanted to find the row and column locations of where these locations occur, you can do:
[rows,cols] = ind2sub(size(P), indices);
Third Method - Use pdist and squareform
Since you're looking for a solution that requires no loops, take a look at the pdist function. Given a M x N matrix, pdist will find distances between each pair of rows in a matrix. squareform will then transform these distances into a matrix like what you have seen above. In other words, do this:
dists = pdist(P.', 'correlation');
distMatrix = squareform(dists);
Fourth Method - Use the corr method straight out of the box
You can just use corr in the following way:
[rho, pvals] = corr(P);
corr in this case will produce a m x m matrix that contains the correlation coefficient between each pair of columns an n x m matrix stored in P.
Hopefully one of these will work!
this works ?
for i=1:size(P,2)
for j=1:i
Since you are just correlating each column with the other, then why not just use (straight from the documentation)
[Rho,Pval] = corr(P);
I don't have the Statistics Toolbox, but according to http://www.mathworks.com/help/stats/corr.html,
corr(X) returns a p-by-p matrix containing the pairwise linear correlation coefficient between each pair of columns in the n-by-p matrix X.