If I have data in a 1-D vector (Y) in matlab, I can access the last value by calling 'end', such as answer = Y(end). However, in my code, I am using a cell array to store a list of vectors of various lengths. Each vector is one cell in a 1D cell array. Is there some way in matlab to reference the last value of the vector stored in a specific cell array?
You can do this by writing y{k}(end) to grab the end of the kth item. Like this:
% Make some random vectors:
y1 = [ 1 2 3];
y2 = [ 1 5 9 12];
y3 = [9 48 2 1];
% create a cell array of them:
x = {y1, y2, y3};
% grab the end of the second vector:
k = 2;
x{k}(end)
Related
I have a matrix X of size n x m. I resized X to a vector a of length n x m.
How can I know "automatically" that the ith element in vector a corresponds to what element position (coordinates) in X?
I have written the following MATLAB code but I do not know how to continue.
X = rand(10,10);
[n,m] = size(X);
a = reshape(X, [n*m, 1]);
t = zeros(length(a),1);
for i = 1 : length(a)
t(i) = % I want to perform here the sum over the x and y coordinate values of the element in X
% that corresponds to the ith element in vector a.
end
Any help will be very appreciated.
That's what ind2sub does:
[row, col] = ind2sub([m n], i);
However, you may prefer to do it manually:
row = mod(i-1,m)+1;
col = floor((i-1)/m)+1;
This works because Matlab used column-major order for storing array elements. For example, in a 3×4 matrix the order in which the elements are stored in memory is as follows:
1 4 7 10
2 5 8 11
3 6 9 12
So the entry in the 2nd row, 3rd column is the 8th element in (column-major) linear order. When this matrix is reshaped into a vector (or into any other shape) this linear order is preserved. Therefore you can retrieve the original coordinates by divisions and modulus operations. Note also that, since Matlab's indexing is 1-based (as opposed to 0-based), the modulus operations need to be shifted by 1.
I am writing a script that operates on matrices, and I have run into the problem of needing to add the sum of the diagonals of a previous matrix to the diagonal elements of a new matrix. The code I have so far for this particular function (described in more detail below) is:
t = 1;
for k = (m-1):-1:-(m-1)
C = bsxfun(#plus, diag(B, k), d);
g(t) = sum(diag(B, k));
t = t + 1;
end
where d is a 1x3 array, and C is supposed to be a 3x3 array; however, C is being output as a 1x3 array in such a way that the first diagonal is being summed and added to d, then the main diagonal is being summed and added to d, and the final diagonal is being summed and added to d.
Is there a way I can get the values of C to be such that the first diagonal is the sum of it's individual elements added to the last element of d, the main diagonal's individual elements added to the middle element of d, and the bottom diagonal's elements added to the first element of d? (while still working for any array size?)
Here is a picture that describes what I'm trying to achieve:
Thanks!
You can use toeplitz to generate a matrix containing the values that need to be added to your original matrix:
M = [5 5 5; 7 7 7; 9 9 9]; %// data matrix
v = [1 11 4 3 2]; %// data vector
S = toeplitz(v);
S = S(1:(numel(v)+1)/2, (numel(v)+1)/2:end);
result = M+S;
Or, as noted by #thewaywewalk, you can do this more directly as follows:
M = [5 5 5; 7 7 7; 9 9 9]; %// data matrix
v = [1 11 4 3 2]; %// data vector
result = M + toeplitz(v(size(M,1):-1:1), v(size(M,2):end));
Assuming B to be a square shaped matrix, listed in this post would be one bsxfun based vectorized approach. Here's the implementation -
N = size(B,1) %// Store size of B for later usage
%// Find a 2D grid of all indices with kth column representing kth diagonal of B
idx = bsxfun(#plus,[N-numel(B)+1:N+1:N]',[0:2*N-2]*N) %//'
%// Mask of all valid indices as we would see many from the 2D grid
%// going out of bounds of 2D array, B
mask = idx>numel(B) | idx<1
%// Set all out-of-bounds indices to one, so that in next step
%// we could index into B in a vectorized manner and sum those up with d
idx(mask)=1
sum1 = bsxfun(#plus,B(idx),d(:).') %//'
%// Store the summations at proper places in B with masking again
B(idx(~mask)) = sum1(~mask)
Sample run -
B =
1 9 0
7 9 4
6 8 7
d =
4 9 5 8 2
B =
6 17 2
16 14 12
10 17 12
Code:
The following code adds the sums of the diagonals of A to the corresponding diagonals in the matrix B. The code works for matrices A, B of equal size, not necessarily square.
A = magic(4);
B = magic(4);
D = bsxfun(#minus, size(A,2)+(1:size(A,1)).', 1:size(A,2)); %'
sumsDiagsA = accumarray(D(:), A(:)); %// Compute sums of diagonals (your 'd')
B = B + sumsDiagsA(D); %// Add them to the matrix
Explanation:
First we build a matrix that numbers all diagonals beginning from the rightmost diagonal:
>> D = bsxfun(#minus, size(A,2)+(1:size(A,1)).', 1:size(A,2))
D =
4 3 2 1
5 4 3 2
6 5 4 3
7 6 5 4
Then we compute sumsDiagsA as the sum of the diagonals via accumarray:
sumsDiagsA = accumarray(D(:), A(:));
The variable sumsDiagsA is what you refer to as d in your code.
Now we use indexing to the vector containing the sums and add them to the matrix B:
C = B + sumsDiagsA(D);
Assuming you have already computed your vector d, you don't need the accumarray-step and all you need to do is:
D = bsxfun(#minus, size(B,2)+(1:size(B,1)).', 1:size(B,2)); %'
C = B + d(D);
I have a three dimensional domain in MATLAB. For each point in the domain I have defined three arrays of size (NX,NY,NZ) at each point of the domain:
A1; % size(A1) = [NX NY NZ]
A2; % size(A2) = [NX NY NZ]
A3; % size(A3) = [NX NY NZ]
For each element, I am trying to construct an array which holds the value of A1, A2, and A3. Would the following be a good candidate for having a 1×3 vector at each point?
B = [A1(:) A2(:) A3(:)];
B = reshape(B, [size(A1) 1 3]);
If the 1×3 array is named C, I am trying to find C'*C at each point.
C = [A1(i,j,k) A2(i,j,k) A3(i,j,k)]; % size(C) = [1 3]
D = C'*C; % size(D) = [3 3]
My ultimate goal is to find the array D with size 3×3 for all the points in the domain in a vectorize fashion? In fact, the output which consists of array D for each point will have the size [NX NY NZ 3 3]. Could someone help me?
Basically we concatenate A1, A2 and A3 along the 4th and 5th dimensions separately that leaves singleton dimensions in the 5th and 4th dimensions respectively, which are then used by bsxfun [Apply element-by-element binary operation to two arrays with singleton expansion enable] to expand as 3x3 matrices along the 4th-5th dimensions for matrix multiplication result from each triplet of [A1(i,j,k),A2(i,j,k),A3(i,j,k)].
D = bsxfun(#times,cat(4,A1,A2,A3),cat(5,A1,A2,A3));
I have a 3X3 cell array and each element store a (x,y) point.
The point are generate by random number from [0,1].
What I want to do is sort the cell array so that it looks like following
ex: 9 points
each circle is one 2D point
index:(1,1) in the left top corner and (3,3) to the right bottom corner as the usual array index
that is to ensure the topological order.
How do I do it?
Thank in advance.
for the example
pairs = [4 9 2 6 5 1 7 8 3; 9 6 2 1 3 8 7 4 5] (row 1 = x-values, row 2 = y-values))
what I want to do is put them in the cell array so that they can be connected by read lines like the image's topology.
The number of permutations is factorial(9), which is not terribly large. So a brute-froce approach is feasible: test all permutations for your desired conditions, and pick the first that is valid.
In the following I'm using a 2x3x3 array, instead of a 3x3 cell array containing length-2 vectors, because it's much easier that way.
N = 3;
data = rand(2,N,N);
permutations = perms(1:N^2); %// generate all permutations
for k = 1:numel(permutations)
dx = reshape(data(1,permutations(k,:)),N,N); %// permuted x data
dy = reshape(data(2,permutations(k,:)),N,N); %// permuted y data
if all(all(diff(dy,[],1)<0)) && all(all(diff(dx,[],2)>0)) %// solution found
disp(dx) %// display solution: x values
disp(dy) %// y values
break %// we only want one solution
end
end
Note that for some choices of data there may not be a solution.
In MATLAB, is there a way to rotate the elements of a array to another dimension, like:
y=[-1,0,1] --> y=[-1; 0; 1] (like transpose)
y=[-1,0,1] --> y(:,:,1)=-1, y(:,:,2)=0, y(:,:,3)=1
y=[-1,0,1] --> y(:,:,1,1)=-1, y(:,:,1,2)=0, y(:,:,1,3)=1
I would like to avoid for loops.
You can do these sorts of matrix operations using transposition, the function RESHAPE, or the function PERMUTE. For example:
y = [-1 0 1]; %# Your 1-by-3 sample array
y2 = y.'; %'# Transposing y gives you a 3-by-1 array
y2 = reshape(y,[3 1]); %# This also gives you a 3-by-1 array
y2 = permute(y,[2 1]); %# This also gives you a 3-by-1 array
y3 = reshape(y,[1 1 3]); %# This gives you a 1-by-1-by-3 array
y3 = permute(y,[3 1 2]); %# This also gives you a 1-by-1-by-3 array
y4 = reshape(y,[1 1 1 3]); %# This gives you a 1-by-1-by-1-by-3 array
y4 = permute(y,[4 1 2 3]); %# This also gives you a 1-by-1-by-1-by-3 array
While reshape and permute are more powerful tools, you can solve the given example quite easy with:
y = [-1 0 1];
y2(:,1)=y;
y3(1,1,:)=y;
y4(1,1,1,:)=y;