I have a question to the mapping of a matrix with another matrix which contains only 1 and 0.
Here an example of my problem: A is the matrix with doubles
A = [ 1 4 3;
2 3 4;
4 3 1;
4 5 5;
1 2 1];
B is a matrix with ones and zeros:
B = [ 0 0 0;
0 0 0;
1 1 1;
1 1 1;
0 0 0];
I want to achieve a matrix C which is the result of A mapped by B, just like that:
C = [ 0 0 0;
0 0 0;
4 3 1;
4 5 5;
0 0 0];
I tried B as a logical array and as a matrix. Both lead to the same error:
"Subscript indices must either be real positive integers or logicals."
Just multiply A and B element-wise:
C = A.*B
I like Dan's solution, but this would be another way:
C = zeros(size(A));
C(B==1) = A(B==1);
Related
I have a matrix B and I would like to obtain a new matrix C from B by adding its last w*a rows to the first w*a rows (w and a will be defined afterwards).
My matrix B is generally defined by :
I would like to obtain matrix C defined in a general way by:
The characteristics of matrices B and C are:
L and w are defined real values;
B0,B1,...,Bw are of dimension: a by b;
B is of dimension: [(L+w)×a] by (L×b);
C is of dimension: (L×a) by (L×b).
Example: For L = 4 and w = 2 I obtain the following matrix B:
The w*a = 2*1 = 2 last rows of B are:
The w*a = 2*1 = 2 first rows of B are:
By adding the two matrices we have:
The matrix C thus obtained is then:
For B0 = [1 0], B1 = [0 1] and B2 = [1 1]. We obtain :
B0, B1 and B2 are of dimension a by b i.e. 1 by 2;
B is of dimension: [(L+w )×(a)] by (L×b) i.e. [(4+2)×1] by (4×2) i.e. 6 by 8;
C is of dimension: (L×a) by (L×b) i.e. (4×1) by (4×2) i.e. 4 by 8.
The matrices B and C that I get are as follows:
B =
1 0 0 0 0 0 0 0
0 1 1 0 0 0 0 0
1 1 0 1 1 0 0 0
0 0 1 1 0 1 1 0
0 0 0 0 1 1 0 1
0 0 0 0 0 0 1 1
C =
1 0 0 0 1 1 0 1
0 1 1 0 0 0 1 1
1 1 0 1 1 0 0 0
0 0 1 1 0 1 1 0
I would like to have some suggestions on how to program this construction so that from a given matrix B I can deduce the matrix C.
Matlab's range indexing should help you do this in a few steps. The key things to remember are that ranges are inclusive, i.e. A[1:3] is a three 3x1 matrix, and that you can use the keyword end to automatically index the end of the matrix row or column.
%% Variables from OP example
w = 2;
L = 4;
B0 = [1 0];
B1 = [0 1];
B2 = [1 1];
[a, b] = size(B0);
% Construct B
BX = [B0;B1;B2]
B = zeros((L+w)*a, L*b);
for ii = 0:L-1
B(ii+1:ii+w+1, ii*b+1:ii*b+b) = BX;
end
%% Construct C <- THIS PART IS THE ANSWER TO THE QUESTION
% Grab first rows of B
B_first = B(1:end-w*a, :) % Indexing starts at first row, continues to w*a rows before the end, and gets all columns
% Grab last rows of B
B_last = B(end-w*a+1:end, :); % Indexing starts at w*a rows before the end, continues to end. Plus one is needed to avoid off by one error.
% Initialize C to be the same as B_first
C = B_first;
% Add B_last to the first rows of C
C(1:w*a, :) = C(1:w*a, :) + B_last;
I get the output
C =
1 0 0 0 0 0 1 1 0 1
0 1 1 0 0 0 0 0 1 1
1 1 0 1 1 0 0 0 0 0
0 0 1 1 0 1 1 0 0 0
0 0 0 0 1 1 0 1 1 0
I've been trying to make a 2-dimensional array that has the largest number in the center, and numbers around it decrement by one like this:
[0 0 0 0 0 0 0;
0 1 1 1 1 1 0;
0 1 2 2 2 1 0;
0 1 2 3 2 1 0;
0 1 2 2 2 1 0;
0 1 1 1 1 1 0;
0 0 0 0 0 0 0]
Any help?
This is easy using implicit expansion:
M = 7; % desired size. Assumed to be odd
t = [0:(M-1)/2 (M-3)/2:-1:0].';
result = min(t, t.');
Alternatively, you can use the gallery function with the 'minij' option to produce one quadrant of the result, and then extend symmetrically:
M = 7; % desired size. Assumed to be odd
result = gallery('minij',(M+1)/2)-1;
result = [result result(:,end-1:-1:1)];
result = [result; result(end-1:-1:1,:)];
Another approach, using padarray from the Image Processing toolbox:
result = 0;
for k = 1:(M-1)/2;
result = padarray(result+1, [1 1]);
end
I would like to create a square matrix of size n x n where the diagonal elements as well as the left-diagonal are all equal to 1. The rest of the elements are equal to 0.
For example, this would be the expected result if the matrix was 5 x 5:
1 0 0 0 0
1 1 0 0 0
0 1 1 0 0
0 0 1 1 0
0 0 0 1 1
How could I do this in MATLAB?
Trivial using the tril function:
tril(ones(n),0) - tril(ones(n),-2)
And if you wanted a thicker line of 1s just adjust that -2:
n = 10;
m = 4;
tril(ones(n),0) - tril(ones(n),-m)
If you prefer to use diag like excaza suggested then try
diag(ones(n,1)) + diag(ones(n-1,1),-1)
but you can't control the 'thickness' of the stripe this way. However, for a thickness of 2, it might perform better. You'd have to test it though.
You can also use spdiags too to create that matrix:
n = 5;
v = ones(n,1);
d = full(spdiags([v v], [-1 0], n, n));
We get:
>> d
d =
1 0 0 0 0
1 1 0 0 0
0 1 1 0 0
0 0 1 1 0
0 0 0 1 1
The first two lines define the desired size of the matrix, assuming a square n x n as well as a vector of all ones that is of length n x 1. We then call spdiags to define where along the diagonal of this matrix this vector will be populating. We want to define the main diagonal to have all ones as well as the diagonal to the left of the main diagonal, or -1 away from the main diagonal. spdiags will adjust the total number of elements for the diagonal away from the main to compensate.
We also ensure that the output is of size n x n, but this matrix is actually sparse . We need to convert the matrix to full to complete the result.,
With a bit of indices juggling, you can also do this:
N = 5;
ind = repelem(1:N, 2); % [1 1 2 2 3 3 ... N N]
M = full(sparse(ind(2:end), ind(1:end-1), 1))
Simple approach using linear indexing:
n = 5;
M = eye(n);
M(2:n+1:end) = 1;
This can also be done with bsxfun:
n = 5; %// matrix size
d = [0 -1]; %// diagonals you want set to 1
M = double(ismember(bsxfun(#minus, 1:n, (1:n).'), d));
For example, to obtain a 5x5 matrix with the main diagonal and the two diagonals below set to 1, define n=5 and d = [0 -1 -2], which gives
M =
1 0 0 0 0
1 1 0 0 0
1 1 1 0 0
0 1 1 1 0
0 0 1 1 1
I want to draw a simple cubic lattice using MATLAB.
I have read How to plot 3D grid (cube) in Matlab, however, I want to color every small cube.
I have a three-dimensional array in MATLAB, such as,
cube(:,:,1) = [1 0 1
0 1 1
1 1 0]
cube(:,:,2) = [0 0 1
1 1 1
0 1 0]
cube(:,:,3) = [1 1 1
0 1 1
1 0 1]
How can I draw a simple cubic lattice using this array, in which cube(:,:,1) denotes the first floor of the cubic lattice, cube(:,:,2) denotes the second floor, and cube(:,:,3) the third floor.
A 0 denotes a small white cube, whilst a 1 denotes a small black cube.
The desired result is something like this: http://www.instructables.com/id/Puzzle-Cube/
I couldn't find anything simpler, so this is what it is!
C = randi(2,[3 3 3])-1;
colorC = char(repmat('k',[3 3 3]));
colorC(C == 0) = 'y';
figure(1);
for x = 0 : 2
for y = 0 : 2
for z = 0 : 2
vert = [1 1 0;
0 1 0;
0 1 1;
1 1 1;
0 0 1;
1 0 1;
1 0 0;
0 0 0];
vert(:,1) = vert(:,1) + x;
vert(:,2) = vert(:,2) + y;
vert(:,3) = vert(:,3) + z;
fac = [1 2 3 4;
4 3 5 6;
6 7 8 5;
1 2 8 7;
6 7 1 4;
2 3 5 8];
patch('Faces',fac,'Vertices',vert,'FaceColor',colorC(x + 1, y + 1, z + 1));
axis([0, 3, 0, 3, 0, 3]);
alpha('color');
alphamap('rampdown');
axis equal
hold on
end
end
end
Gives you this,
If you delete alpha('color'); and alphamap('rampdown');and use axis off, you get,
How is it possible to expand a quadratic - let's say NxN - matrix to a bigger on like a (N+k)x(N+k) matrix?
It's really all about resizing the matrix and filling the missing rows/columns with zeros such that not dimension mismatch occurs.
No need to do it more difficult than it is. MATLAB automatically pads with zeros if you assign something to an element outside of the original size:
n = 4;
A = [1 2; 3 4];
A(n,n) = 0;
A =
1 2 0 0
3 4 0 0
0 0 0 0
0 0 0 0
you can add zeros to a matrix using padarray... For example:
A = [1 2; 3 4];
B = padarray(A,[2 2],'post')
B =
1 2 0 0
3 4 0 0
0 0 0 0
0 0 0 0
Or, if you don't have the image processing toolbox, you can use matrix indexing:
B = zeros(size(A)+k, class(A));
B(k:end-k+1,k:end-k+1) = A;