I am trying to do a density plot for a data containing two columns with different ranges. The RMSD column is [0-2] and Angle is [0-200] ranges.
My data in the file is like this:
0.0225370 37.088
0.1049553 35.309
0.0710002 33.993
0.0866880 34.708
0.0912664 33.011
0.0932054 33.191
0.1083590 37.276
0.1104145 34.882
0.1027977 34.341
0.0896688 35.991
0.1047578 36.457
0.1215936 38.914
0.1105484 35.051
0.0974138 35.533
0.1390955 33.601
0.1333878 32.133
0.0933365 35.714
0.1200465 33.038
0.1155794 33.694
0.1125247 34.522
0.1181806 37.890
0.1291700 38.871
I want both x and y axis to be binned 1/10th of the range
The 0 of both the axis to be starting in the same
Print the number of elements in each grid of the matrix like this and make a density plot based on these number of elements
0 0.1 0.2 (RMSD)
0 0 1 3
20 2 0 4
40 1 0 5
60 0 0 2
(Angle)
I can find ways to do 1-D binning but then I am stumped about how to make a density plot from those values and havent even dared to attempt2-D binning + plotting.
Thanks for the help
I think you want hist3. Assuming you want to specifty bin edges (not bin centers), use
result = hist3(data, 'Edges', {[0 .1 .2], [0 20 40 60]}).';
where data denotes your data.
From the linked documentation:
hist3(X,'Edges',edges), where edges is a two-element cell array of numeric vectors with monotonically non-decreasing values, uses a 2-D grid of bins with edges at edges{1} in the first dimension and at edges{2} in the second. The (i,j)th bin includes the value X(k,:) if
edges{1}(i) <= X(k,1) < edges{1}(i+1)
edges{2}(j) <= X(k,2) < edges{2}(j+1)
With your example data this gives
result =
0 0 0
8 14 0
0 0 0
0 0 0
For those who don't have Statistics and Machine Learning Toolbox to run bivariate histogram (hist3), it may be more practical using an alternative to solve 2-D hist problem. The following function generates the same output
function N = hist3_alt(x,y,edgesX,edgesY)
N = zeros(length(edgesY)-1,length(edgesX)-1);
[~,~,binX] = histcounts(x,edgesX);
for ii=1:numel(edgesX)-1
N(:,ii) = (histcounts(y(binX==ii),edgesY))';
end
It's simple and efficient. Then you could run the function like this:
N = hist3_alt(x,y,[0:0.1:2],[0:20:200])
Related
I have a matrix suppX in Matlab with size GxN and a matrix A with size MxN. I would like your help to construct a matrix Xresponse with size GxM with Xresponse(g,m)=1 if the row A(m,:) is equal to the row suppX(g,:) and zero otherwise.
Let me explain better with an example.
suppX=[1 2 3 4;
5 6 7 8;
9 10 11 12]; %GxN
A=[1 2 3 4;
1 2 3 4;
9 10 11 12;
1 2 3 4]; %MxN
Xresponse=[1 1 0 1;
0 0 0 0;
0 0 1 0]; %GxM
I have written a code that does what I want.
Xresponsemy=zeros(size(suppX,1), size(A,1));
for x=1:size(suppX,1)
Xresponsemy(x,:)=ismember(A, suppX(x,:), 'rows').';
end
My code uses a loop. I would like to avoid this because in my real case this piece of code is part of another big loop. Do you have suggestions without looping?
One way to do this would be to treat each matrix as vectors in N dimensional space and you can find the L2 norm (or the Euclidean distance) of each vector. After, check if the distance is 0. If it is, then you have a match. Specifically, you can create a matrix such that element (i,j) in this matrix calculates the distance between row i in one matrix to row j in the other matrix.
You can treat your problem by modifying the distance matrix that results from this problem such that 1 means the two vectors completely similar and 0 otherwise.
This post should be of interest: Efficiently compute pairwise squared Euclidean distance in Matlab.
I would specifically look at the answer by Shai Bagon that uses matrix multiplication and broadcasting. You would then modify it so that you find distances that would be equal to 0:
nA = sum(A.^2, 2); % norm of A's elements
nB = sum(suppX.^2, 2); % norm of B's elements
Xresponse = bsxfun(#plus, nB, nA.') - 2 * suppX * A.';
Xresponse = Xresponse == 0;
We get:
Xresponse =
3×4 logical array
1 1 0 1
0 0 0 0
0 0 1 0
Note on floating-point efficiency
Because you are using ismember in your implementation, it's implicit to me that you expect all values to be integer. In this case, you can very much compare directly with the zero distance without loss of accuracy. If you intend to move to floating-point, you should always compare with some small threshold instead of 0, like Xresponse = Xresponse <= 1e-10; or something to that effect. I don't believe that is needed for your scenario.
Here's an alternative to #rayryeng's answer: reduce each row of the two matrices to a unique identifier using the third output of unique with the 'rows' input flag, and then compare the identifiers with singleton expansion (broadcast) using bsxfun:
[~, ~, w] = unique([A; suppX], 'rows');
Xresponse = bsxfun(#eq, w(1:size(A,1)).', w(size(A,1)+1:end));
If I have a two column matrix A like below, I can plot the scatter plot using scatter/plot command. I would like to get the matrix corresponding to such outputs as in hist command. hist command gives the vector output too.
A=[7 1;3 2; 4 3]
For example out=scatter(A(:,1),A(:,2)) must give something like below:
[0 0 0;
0 0 0;
0 1 0;
0 0 1;
0 0 0;
0 0 0;
1 0 0]
Only the indices (7,1), (3,2) and (4,3) are only ones. Or Can someone give me a snippet code to realize this without using loops?
You can use a combination of sparse and full where you can specify the non-zero row and column locations, and the rest of the matrix would be zero:
A = [7 1; 3 2; 4 3];
B = full(sparse(A(:,1), A(:,2), 1, max(A(:,1)), max(A(:,2)))) == 1;
The sparse command takes in the row and column locations of what is non-zero for the first two inputs, the third input is what the non-zero location would be for each row and column location. We can specify a constant to mean that every non-zero location gets the same coefficient, which is 1. We can also specify the size of the matrix, where in this case the rows and columns of the output correspond to the largest number in the first and second columns respectively. Because this is a sparse matrix, you will want to convert this to a full matrix and because you want it to be logical, you will want to compare all elements with the number 1.
We thus get for the output, which is B:
B =
7×3 logical array
0 0 0
0 0 0
0 1 0
0 0 1
0 0 0
0 0 0
1 0 0
Alternatively, we can use sub2ind to create linear indices to index into a pre-allocated matrix of logical false and set only those non-zero row locations to true:
A = [7 1; 3 2; 4 3];
B = false(max(A(:,1)), max(A(:,2)));
ind = sub2ind(size(B), A(:,1), A(:,2));
B(ind) = true;
We first allocate the matrix, then calculate the linear indices to index into the matrix, then finally set the right locations to true. The output here would be the same as the sparse approach.
Just to add: rayryeng's solution is fine if you really want your result to be logical in the sense that it is equal to one if there is anything at the coordinate and zero otherwise. Still, since you added a note on hist, I was wondering if you actually want to count the number of times a specific coordinate is hit. In this case, consider using
S = histcounts2(A(:,2),A(:,1));
if you have access to R2015b+. If not, there is a hist2 function on fileexchange you can use for the purpose.
Here is my solution. Matlab provides a command called accumarray.
S = logical(accumarray(A, 1) )
will give the result too.
I need to assign weights to edges of a graph, from the following papers:
"Fast linear iterations for distributed averaging" by L. Xiao and S. Boyd
"Convex Optimization of Graph Laplacian Eigenvalues" by S. Boyd
I have the adjacency matrix for my graph (a 50 by 50 matrix), with 512 non-zero values.
I also have a 256 by 1 vector with the optimal weights.
For the software I'm using, I need a 50 by 50 matrix with the weight of edge (i,j) in the relevant position of the adjacency matrix (and with the opposite sign for edge (j,i)).
My attempt is below, but I can't get it working.
function weights = construct_weight_mtx(weight_list, Adj)
weights = zeros(size(Adj));
positions = find(Adj);
for i=1:length(positions)/2
if Adj(i) == 1
weights(i) = weight_list(i);
end
end
weights = weights - weights';
find(Adj) == find(weights);
end
You're finding the nonzero positions in the original adjacency matrix, but you're finding all of them. To get around this, you then take only the first half of those positions.
for i=1:length(positions)/2 ...
Unfortunately, this takes the indices from complete columns rather than just the positions below the diagonal. So if your matrix was all 1's, you'd be taking:
1 1 1 0 0 ...
1 1 1 0 0 ...
1 1 1 0 0 ...
...
instead of:
1 0 0 0 0 ...
1 1 0 0 0 ...
1 1 1 0 0 ...
...
To take the correct values, we just take the lower triangular portion of Adj and then find the nonzero positions of that:
positions = find(tril(Adj));
Now we have only the 256 positions below the diagonal and we can loop over all of the positions. Next, we need to fix the assignment in the loop:
for i=1:length(positions)
if Adj(i) == 1 %// we already know Adj(i) == 1 for all indices in positions
weights(i) = weight_list(i); %// we need to update weights(positions(i))
end
end
So this becomes:
for i=1:length(positions)
weights(positions(i)) = weight_list(i);
end
But if all we're doing is assigning 256 values to 256 positions, we can do that without a for loop:
weights(position) = weight_list;
Note that the elements of weight_list must be in the proper order with the nonzero elements of the lower-triangular portion ordered by columns.
Completed code:
function weights = construct_weight_mtx(weight_list, Adj)
weights = zeros(size(Adj));
positions = find(tril(Adj));
weights(positions) = weight_list;
weights = weights - weights.'; %// ' is complex conjugate; not a big deal here, but something to know
find(Adj) == find(weights); %// Not sure what this is meant to do; maybe an assert?
end
I'm implementing a finite difference scheme for a 2D PDE problem. I wish to avoid using a loop to generate the finite differences. For instance to generate a 2nd order central difference of u(x,y)_xx, I can multiply u(x,y) by the following:
Is there a nice matrix representation for u_xy = (u_{i+1,j+1} + u_{i-1,j-1} - u_{i-1,j+1} - u_{i+1,j-1})/(4dxdy)? It's a harder problem to code as it's in 2D - I'd like to multiply some matrix by u(x,y) to avoid looping. Many thanks!
If your points are stored in a N-by-N matrix then, as you said, left multiplying by your finite difference matrix gives an approximation to the second derivative with respect to u_{xx}. Right-multiplying by the transpose of the finite difference matrix is equivalent to an approximation u_{yy}. You can get an approximation to the mixed derivative u_{xy} by left-multiplying and right-multiplying by e.g. a central difference matrix
delta_2x =
0 1 0 0 0
-1 0 1 0 0
0 -1 0 1 0
0 0 -1 0 1
0 0 0 -1 0
(then divide by the factor 4*Dx*Dy), so something like
U_xy = 1/(4*Dx*Dy) * delta_2x * U_matrix * delta_2x';
If you cast a matrix as a N^2 vector
U_vec = U_matrix(:);
then these operators can be expressed using a Kronecker product, implemented in MATLAB as kron: We have
A*X*B = kron(B',A)*X(:);
so for your finite difference matrices
U_xy_vec = 1/(4*Dx*Dy)*(kron(delta_2x,delta_2x)*U_vec);
If instead you have an N-by-M matrix U_mat, then left matrix multiplication is equivalent to kron(eye(M),delta_2x_N) and right multiplication to kron(delta_2y_M,eye(N)), where delta_2y_M (delta_2x_N) is the M-by-M (N-by-N) central difference matrix, so the operation is
U_xy_vec = 1/(4*Dx*Dy) * kron(delta_2y_M,delta_2y_N)*U_vec;
Here is an MATLAB code example:
N = 20;
M = 30;
Dx = 1/N;
Dy = 1/M;
[Y,X] = meshgrid((1:(M))./(M+1),(1:(N))/(N+1));
% Example solution and mixed derivative (chosen for 0 BCs)
U_mat = sin(2*pi*X).*(sin(2*pi*Y.^2));
U_xy = 8*pi^2*Y.*cos(2*pi*X).*cos(2*pi*Y.^2);
% Centred finite difference matrices
delta_x_N = 1/(2*Dx)*(diag(ones(N-1,1),1) - diag(ones(N-1,1),-1));
delta_y_M = 1/(2*Dy)*(diag(ones(M-1,1),1) - diag(ones(M-1,1),-1));
% Cast U as a vector
U_vec = U_mat(:);
% Mixed derivative operator
A = kron(delta_y_M,delta_x_N);
U_xy_num = A*U_vec;
U_xy_matrix = reshape(U_xy_num,N,M);
subplot(1,2,1)
contourf(X,Y,U_xy_matrix)
colorbar
title 'Numeric U_{xy}'
subplot(1,2,2)
contourf(X,Y,U_xy)
colorbar
title 'Analytic U_{xy}'
You can obviously create the matrix yourself, but in Matlab there is tridiag for this purpose.
For example
>> full(gallery('tridiag',5,-1,2,-1))
ans =
2 -1 0 0 0
-1 2 -1 0 0
0 -1 2 -1 0
0 0 -1 2 -1
0 0 0 -1 2
Using sparse functionality available in MATLAB to generate finite difference approximation matrix is a good option.. It saves lot (indeed very much) of memory...
Given a matrix A with dimension m x n and the entries in the matrix lies [0,1]
For example
A = [0.5 0 0 0.5 0
0 0.5 0 0 0.5
1 0 0 0 0]
I would like to calculate sum(sum(a_ij log(a_ij))), where a_ij is the i th row and j th col entry in the matrix A. Since there exist an 0 entry in the matrix, i always get NAN as a result.
How do i consider only non-zero entries to calculate sum(sum(a_ij log(a_ij))) [entropy of the matrix].
To consider only specific elements of a matrix you can use logical indexing. For example if you only want to select non-zero entries of A you can use A(A~=0). So for your problem the solution can be written:
sum(A(A~=0).*log(A(A~=0)));
EDIT: wow that is some kind of coincidence, I've just seen your comment after posting this. Well, glad you've worked it out yourself.
If it is a very large array:
sum(A.*log(A+eps))
which should be faster than indexing.
Another possibility:
x = A(:);
E = x' * log(x + (x==0))