I have an Nx2 matrix of data points where each row is a data point. I also have an NxK matrix of indices of the K nearest neighbours from the knnsearch function. I am trying to create a matrix that contains in each row the data point followed by the K neighbouring data points, i.e. for K = 2 we would have something like [data1, neighbour1, neighbour2] for each row.
I have been messing round with loops and attempting to index with matrices but to no avail, the fact that each datapoint is 1x2 is confusing me.
My ultimate aim is to calculate gradients to train an RBF network in a similar manner to:
D = (x_dist - y_dist)./(y_dist+(y_dist==0));
temp = y';
neg_gradient = -2.*sum(kron(D, ones(1,2)) .* ...
(repmat(y, 1, ndata) - repmat((temp(:))', ndata, 1)), 1);
neg_gradient = (reshape(neg_gradient, net.nout, ndata))';
You could use something along those lines:
K = 2;
nearest = knnsearch(data, data, 'K', K+1);%// Gets point itself and K nearest ones
mat = reshape(data(nearest.',:).',[],N).'; %// Extracts the coordinates
We generate data(nearest.',:) to get a 3*N-by-2 matrix, where every 3 consecutive rows are the points that correspond to each other. We transpose this to get the xy-coordinates into the same column. (MATLAB is column major, i.e. values in a column are stored consecutively). Then we reshape the data, so every column contains the xy-coordinates of the rows of nearest. So we only need to transpose once more in the end.
Related
I have a matrix that I generate from a CSV file as follows:
X = xlsread('filename.csv');
I am looping through the matrix based on the number of records and I need to find the Euclidean distance for each of the rows of this matrix :
for i = 1:length(X)
j = X(:, [2:5])
end
The resulting matrix is of 150 X 4. What would be the best way to calculate the Euclidean distance of each row (with 4 columns as the data points) with every row and getting an average of the same?
In order to find the Euclidean distance between any pair of rows, you could use the function pdist.
X = randn(6, 4);
D = pdist(X,'euclidean');
res=mean(D);
The average is stored in res.
I have 100 coordinates in a variable x in MATLAB . How can I make sure that distance between all combinations of two points is greater than 1?
You can do this in just one simple line, with the functions all and pdist:
if all(pdist(x)>1)
...
end
Best,
First you'll need to generate a matrix that gives you all possible pairs of coordinates. This post can serve as inspiration:
Generate a matrix containing all combinations of elements taken from n vectors
I'm going to assume that your coordinates are stored such that the columns denote the dimensionality and the rows denote how many points you have. As such, for 2D, you would have a 100 x 2 matrix, and in 3D you would have a 100 x 3 matrix and so on.
Once you generate all possible combinations, you simply compute the distance... which I will assume it to be Euclidean here... of all points and ensure that all of them are greater than 1.
As such:
%// Taken from the linked post
vectors = { 1:100, 1:100 }; %// input data: cell array of vectors
n = numel(vectors); %// number of vectors
combs = cell(1,n); %// pre-define to generate comma-separated list
[combs{end:-1:1}] = ndgrid(vectors{end:-1:1}); %// the reverse order in these two
%// comma-separated lists is needed to produce the rows of the result matrix in
%// lexicographical order
combs = cat(n+1, combs{:}); %// concat the n n-dim arrays along dimension n+1
combs = reshape(combs,[],n); %// reshape to obtain desired matrix
%// Index into your coordinates array
source_points = x(combs(:,1), :);
end_points = x(combs(:,2), :);
%// Checks to see if all distances are greater than 1
is_separated = all(sqrt(sum((source_points - end_points).^2, 2)) > 1);
is_separated will contain either 1 if all points are separated by a distance of 1 or greater and 0 otherwise. If we dissect the last line of code, it's a three step procedure:
sum((source_points - end_points).^2, 2) computes the pairwise differences between each component for each pair of points, squares the differences and then sums all of the values together.
sqrt(...(1)...) computes the square root which gives us the Euclidean distance.
all(...(2)... > 1) then checks to see if all of the distances computed in Step #2 were greater than 1 and our result thus follows.
Say we have a matrix m x n where the number of rows of the matrix is very big. If we assume each row is a vector, then how could one find the maximum/minimum distance between vectors in this matrix?
My suggestion would be to use pdist. This computes pairs of Euclidean distances between unique combinations of observations like #seb has suggested, but this is already built into MATLAB. Your matrix is already formatted nicely for pdist where each row is an observation and each column is a variable.
Once you do apply pdist, apply squareform so that you can display the distance between pairwise entries in a more pleasant matrix form. The (i,j) entry for each value in this matrix tells you the distance between the ith and jth row. Also note that this matrix will be symmetric and the distances along the diagonal will inevitably equal to 0, as any vector's distance to itself must be zero. If your minimum distance between two different vectors were zero, if we were to search this matrix, then it may possibly report a self-distance instead of the actual distance between two different vectors. As such, in this matrix, you should set the diagonals of this matrix to NaN to avoid outputting these.
As such, assuming your matrix is A, all you have to do is this:
distValues = pdist(A); %// Compute pairwise distances
minDist = min(distValues); %// Find minimum distance
maxDist = max(distValues); %// Find maximum distance
distMatrix = squareform(distValues); %// Prettify
distMatrix(logical(eye(size(distMatrix)))) = NaN; %// Ignore self-distances
[minI,minJ] = find(distMatrix == minDist, 1); %// Find the two vectors with min. distance
[maxI,maxJ] = find(distMatrix == maxDist, 1); %// Find the two vectors with max. distance
minI, minJ, maxI, maxJ will return the two rows of A that produced the smallest distance and the largest distance respectively. Note that with the find statement, I have made the second parameter 1 so that it only returns one pair of vectors that have this minimum / maximum distance between each other. However, if you omit this parameter, then it will return all possible pairs of rows that share this same distance, but you will get duplicate entries as the squareform is symmetric. If you want to escape the duplication, set either the upper triangular half, or lower triangular half of your squareform matrix to NaN to tell MATLAB to skip searching in these duplicated areas. You can use MATLAB's tril or triu commands to do that. Take note that either of these methods by default will include the diagonal of the matrix and so there won't be any extra work here. As such, try something like:
distValues = pdist(A); %// Compute pairwise distances
minDist = min(distValues); %// Find minimum distance
maxDist = max(distValues); %// Find maximum distance
distMatrix = squareform(distValues); %// Prettify
distMatrix(triu(true(size(distMatrix)))) = NaN; %// To avoid searching for duplicates
[minI,minJ] = find(distMatrix == minDist); %// Find pairs of vectors with min. distance
[maxI,maxJ] = find(distMatrix == maxDist); %// Find pairs of vectors with max. distance
Judging from your application, you just want to find one such occurrence only, so let's leave it at that, but I'll put that here for you in case you need it.
You mean the max/min distance between any 2 rows? If so, you can try that:
numRows = 6;
A = randn(numRows, 100); %// Example of input matrix
%// Compute distances between each combination of 2 rows
T = nchoosek(1:numRows,2); %// pairs of indexes for all combinations of 2 rows
for k=1:length(T)
d(k) = norm(A(T(k,1),:)-A(T(k,2),:));
end
%// Find min/max distance
[~, minIndex] = min(d);
[~, maxIndex] = max(d);
T(minIndex,:) %// Displays indexes of the 2 rows with minimum distance
T(maxIndex,:) %// Displays indexes of the 2 rows with maximum distance
I would like to have a program that makes the following actions:
Read several matrices having the same size (1126x1440 double)
Select the most occuring value in each cell (same i,j of the matrices)
write this value in an output matrix having the same size 1126x1440 in the corresponding i,j position, so that this output matrix will have in each cell the most occurent value from the same position of all the input matrices.
Building on #angainor 's answer, I think there is a simpler method using the mode function.
nmatrices - number of matrices
n, m - dimensions of a single matrix
maxval - maximum value of an entry (99)
First organize data into a 3-D matrix with dimensions [n X m X nmatrices]. As an example, we can just generate the following random data in a 3-D form:
CC = round(rand(n, m, nmatrices)*maxval);
and then the computation of the most frequent values is one line:
B = mode(CC,3); %compute the mode along the 3rd dimension
Here is the code you need. I have introduced a number of constants:
nmatrices - number of matrices
n, m - dimensions of a single matrix
maxval - maximum value of an entry (99)
I first generate example matrices with rand. Matrices are changed to vectors and concatenated in the CC matrix. Hence, the dimensions of CC are [m*n, nmatrices]. Every row of CC holds individual (i,j) values for all matrices - those you want to analyze.
CC = [];
% concatenate all matrices into CC
for i=1:nmatrices
% generate some example matrices
% A = round(rand(m, n)*maxval);
A = eval(['neurone' num2str(i)]);
% flatten matrix to a vector, concatenate vectors
CC = [CC A(:)];
end
Now we do the real work. I have to transpose CC, because matlab works on column-based matrices, so I want to analyze individual columns of CC, not rows. Next, using histc I find the most frequently occuring values in every column of CC, i.e. in (i,j) entries of all matrices. histc counts the values that fall into given bins (in your case - 1:maxval) in every column of CC.
% CC is of dimension [nmatrices, m*n]
% transpose it for better histc and sort performance
CC = CC';
% count values from 1 to maxval in every column of CC
counts = histc(CC, 1:maxval);
counts have dimensions [maxval, m*n] - for every (i,j) of your original matrices you know the number of times a given value from 1:maxval is represented. The last thing to do now is to sort the counts and find out, which is the most frequently occuring one. I do not need the sorted counts, I need the permutation that will tell me, which entry from counts has the highest value. That is exactly what you want to find out.
% sort the counts. Last row of the permutation will tell us,
% which entry is most frequently found in columns of CC
[~,perm] = sort(counts);
% the result is a reshaped last row of the permutation
B = reshape(perm(end,:)', m, n);
B is what you want.
I have a 128 x 100 matrix in matlab, where each column should be treated as a separate element. Lets call this matrix M.
I have another 128 x 2000 matrix(called V) composed of columns from matrix M.
How would I make a histogram that maps the frequency of each column being used in the second matrix?
hist(double(V),double(M)) gives the error:
Error using histc
Edge vector must be monotonically
non-decreasing.
what should I be doing?
Here is an example. We start with data that resembles what you described
%# a matrix of 100 columns
M = rand(128,100);
sz = size(M);
%# a matrix composed of randomly selected columns of M (with replacement)
V = M(:,randi([1 sz(2)],[1 2000]));
Then:
%# map the columns to indices starting at 1
[~,~,idx] = unique([M,V]', 'rows', 'stable');
idx = idx(sz(2)+1:end);
%# count how many times each column occurs
count = histc(idx, 1:sz(2));
%# plot histogram
bar(1:sz(2), count, 'histc')
xlabel('column index'), ylabel('frequency')
set(gca, 'XLim',[1 sz(2)])
[Lia,Locb] = ismember(A,B,'rows') also returns a vector, Locb,
containing the highest index in B for each row in A that is also a row
in B. The output vector, Locb, contains 0 wherever A is not a row of
B.
ismember with the rows argument can identify which row of one matrix the rows of another matrix come from. Since it works on rows, and you are looking for columns, just transpose both matrices.
[~,Locb]=ismember(V',M');
histc(Locb)