I have a matrix A in Matlab of dimension Nx(N-1), e.g.
N=5;
A=[1 2 3 4;
5 6 7 8;
9 10 11 12;
13 14 15 16;
17 18 19 20];
I want to rearrange the elements of A in a certain way. Specifically I want to create a matrix B of dimension (N-1)xN such that:
for i=1,...,N,
B(:,i) collects
1) the first i-1 elements of the i-1th column of A and
2) the last N-i elements of the ith column of A.
Notice that for i=1 the i-1th column of A does not exist and therefore 1) is skipped; similarly, for i=N theith column of A does not exist and therefore 2) is skipped.
In the example above
B=[5 1 2 3 4
9 10 6 7 8
13 14 15 11 12
17 18 19 20 16];
This code does what I want. I am asking your help to vectorise it in an efficient way.
B=zeros(N-1,N);
for i=1:N
if i>1 && i<N
step1=A(1:i-1,i-1);
step2=A(i+1:N,i);
B(:,i)=[step1;step2];
elseif i==1
B(:,i)=A(i+1:N,i);
elseif i==N
B(:,i)=A(1:i-1,i-1);
end
end
Extract the lower and upper triangular matrices of A. Then reassemble them with a "diagonal shift":
u = triu(A);
l = tril(A,-1);
B = padarray(u(1:end-1,:),[0 1],'pre') + padarray(l(2:end,:),[0 1],'post');
Another valid approach using logical indexing combined with tril and triu:
B = zeros(size(A'));
B(tril(true(size(B)))) = A(tril(true(size(A)), -1));
B(triu(true(size(B)), 1)) = A(triu(true(size(A))));
Result:
>> B
B =
5 1 2 3 4
9 10 6 7 8
13 14 15 11 12
17 18 19 20 16
Related
I want to create matrix B from matrix A, with the following rules:
Non-diagonal elements A are like non-diagonal elements of B.
The main diagonal of A is the antidiagonal of B
The antidiagonal of A is the main diagonal of B.
For example:
A = [ 1 2 3 4;
7 8 9 10;
13 14 15 16;
19 20 21 22 ];
B = [ 4 2 3 1;
7 9 8 10;
13 15 14 16;
22 20 21 19 ];
How can I create B given A?
You can create all of the indices, then it's a single assignment.
% Get size of square matrix A
n = size(A,1);
% Indicies are 1:n^2 by default
idx = 1:n^2;
% Swap diagonal and antidiagonal indices
idx( [1:(n+1):n^2, n^2-n+1:1-n:n] ) = [n^2-n+1:1-n:n, 1:(n+1):n^2];
% Use the indexing array to create B from A, reshape to be n*n
B = reshape( A( idx ), n, n );
Output for your example A:
B =
4 2 3 1
7 9 8 10
13 15 14 16
22 20 21 19
There are so many ways to reach that results, it is just an indexing exercise. Here is one (of the many) way to reach that result for any square matrix of size n:
%% input
A=[ 1 2 3 4 ;
7 8 9 10 ;
13 14 15 16 ;
19 20 21 22 ];
%% Calculate linear indices for the diagonal and antidiagonal
n=size(A,1) ;
idxdiag = 1:(n+1):n^2 ; % => idxdiag = [1 6 11 16]
idxantidiag = n:(n-1):n^2-1 ; % => idxantidiag = [4 7 10 13]
%% Generate B
B = A ; % start with a simple copy (for the non-diagonal elements)
% Method 1: direct indice assignment
B(idxdiag) = diag(fliplr(A)) ; % Assign diagonal elements of B
B(idxantidiag) = flipud(diag(A)) ; % Assign antidiagonal elements of B
% Method 2: summation
B([idxdiag idxantidiag]) = 0 ;
B = B + diag(diag(fliplr(A))) + fliplr(diag(diag(A))) ;
B =
4 2 3 1
7 9 8 10
13 15 14 16
22 20 21 19
Both methods return exactly the same matrix B.
I suggest you familiarise yourself with the MATLAB function used to understand what is going on behind the scene:
fliplr
flipud
diag
and may be have a read at:
Matrix Indexing in MATLAB
I thought a little differently and came to a conclusion
A=[1 2 3 4;7 8 9 10;13 14 15 16; 19 20 21 22];;
n=size(A,1) ;
B=zeros(n,n) ;
for i=1:n
for j=1:n
if i==j
B(i,j)=A(i,n-i+1);
elseif j==n-i+1
B(i,j)=A(i,i);
else
B(i,j)=A(i,j);
end
end
end
B
Here's a variant using eye, find, and flip to generate linear indices:
ind1 = find(eye(size(A)));
ind2 = flip(find(flip(eye(size(A)))));
B = A;
B([ind1 ind2]) = B([ind2 ind1]);
B =
4 2 3 1
7 9 8 10
13 15 14 16
22 20 21 19
And here's a variant of the above that uses just eye and flip to generate logical indices:
ind1 = eye(size(A), 'logical');
ind2 = flip(ind1);
B = A;
B(ind1) = flip(A(ind2));
B(ind2) = flip(A(ind1));
B =
4 2 3 1
7 9 8 10
13 15 14 16
22 20 21 19
Consider a row vector A and row vector B. For example:
A = [1 2 3 7 8 10 12];
B = [1 1 2 2 2 3 5 6 6 7 7 7 8 8 10 10 10 11 12 12 12 13 15 16 18 19];
A has previously been checked to be a subset of B. By subset, I specifically mean that all elements in A can be found in B. I know that elements in A will not ever repeat. However, the elements in B are free to repeat as many or as few times as they like. I checked this condition using:
is_subset = all(ismember(A,B));
With all that out of the way, I need to know the indices of the elements of A within B including the times when these elements repeat within B. For the example A and B above, the output would be:
C = [1 2 3 4 5 6 10 11 12 13 14 15 16 17 19 20 21];
Use ismember to find the relevant logical indices. Then convert them to linear indices using find.
C = find(ismember(B,A));
You can find the difference of each element of A with B, and get the indices you want. Something like below:
A = [1 2 3 7 8 10 12];
B = [1 1 2 2 2 3 5 6 6 7 7 7 8 8 10 10 10 11 12 12 12 13 15 16 18 19];
C = [1 2 3 4 5 6 10 11 12 13 14 15 16 17 19 20 21];
tol = 10^-3 ;
N = length(A) ;
iwant = cell(N,1) ;
for i = 1:N
idx = abs(A(i)-B)<=tol ;
iwant{i} = find(idx) ;
end
iwant = [iwant{:}] ;
If I have this matrix:
A:
X Y Z
1 1 2
0 3 4
0 5 6
2 7 8
7 9 10
8 11 12
3 13 14
12 14 16
15 17 18
How could I create new matrix B, C, D and E which contains:
B:
0 3 4
0 5 6
C:
X Y Z
1 1 2
2 7 8
3 13 14
D:
7 9 10
8 11 12
E:
12 14 16
15 17 18
The idea is to construct a loop asking if 0<A<1 else 1<A<5 else 6<A<10 else 11<A<15. and create new matrix from that condition. Any idea about how to store the results of the loop?
I suggest you an approach that uses the discretize function in order to group the matrix rows into different categories based on their range. Here is the full implementation:
A = [
1 1 2;
0 3 4;
0 5 6;
2 7 8;
7 9 10;
8 11 12;
3 13 14;
12 14 16;
15 17 18
];
A_range = [0 1 5 10 15];
bin_idx = discretize(A(:,1),A_range);
A_split = arrayfun(#(bin) A(bin_idx == bin,:),1:(numel(A_range) - 1),'UniformOutput',false);
celldisp(A_split);
Since you want to consider 5 different ranges based on the first column values, the arguments passed to discretize must be the first matrix column and a vector containing the group limits (first number inclusive left, second number exclusive right, second number inclusive left, third number exclusive right, and so on...). Since your ranges are a little bit messed up, feel free to adjust them to respect the correct output. The latter is returned in the form of a cell array of double matrices in which every element contains the rows belonging to a distinct group:
A_split{1} =
0 3 4
0 5 6
A_split{2} =
1 1 2
2 7 8
3 13 14
A_split{3} =
7 9 10
8 11 12
A_split{4} =
12 14 16
15 17 18
Instead of using a loop, use logical indexing to achieve what you want. Use the first column of A and check for the ranges that you want to look for, then use this to subset into the final matrix A to get what you want.
For example, to create the matrix C, find all locations in the first column of A that are between 1 and 5, then subset the matrix along the rows using these locations:
m = A(:,1) >= 1 & A(:,1) <= 5;
C = A(m,:);
You can repeat this in a similar way for the rest of the matrices you want to create.
Assuming we have 3 matrices, A, B and C, all are of the same size 256x256. It is known that last 20% of columns of Matrix A is identical to first 20% of Matrix B and last 10% of Matrix B is identical to first 10% of Matrix C. So in these cases since we know the overlapping amount, I do not need to compare the 3 matrices, but i want to join them at the overlap.
Taking a smaller Matrix as an example here are the 3 Matrices
A = [1 2 3 4 ; 5 6 7 8; 9 10 11 12];
B = [3 4 13 14; 7 8 15 16; 11 12 17 18];
C = [14 19 20 21; 16 22 23 24; 18 25 26 27];
So I would like my output to be
D = [1 2 3 4 13 14 19 20 21
5 6 7 8 15 16 22 23 24
9 10 11 12 17 18 25 26 27
I hope this might explain it better. I am extremely new to matlab. I tried using matrix shift, but we have only circular shift available.
Concatenation does not work because it just joins the 3 matrices. What would be the best way to overlay these 3 matrices together ?
Make proper use of matrix indexing and concatenation
For your example
D = [A B(:,3) C];
For a 256x256 Matrix and your concatenation conditions:
D = [A B(:, 0.2*256+1 : 0.9*256) C]
Since 256/10 is no integer you may adjust the index values
The input is an N-by-1 matrix. I need to reshape it to L-by-M matrix. The following is an example.
Input:
b =
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Set length = 18, Output:
X =
1 2 3
2 3 4
3 4 5
4 5 6
5 6 7
6 7 8
7 8 9
8 9 10
9 10 11
10 11 12
11 12 13
12 13 14
13 14 15
14 15 16
15 16 17
16 17 18
17 18 19
18 19 20
Because I have a very big matrix, using a loop to reshape is very inefficient. How can I improve the reshape speed?
Your example output matrix X is the perfect matrix to index a vector of length N to get what you want. It's also very easy to create using bsxfun:
N = 20;
b = rand(N,1);
M = 3; %// number of columns
L = N-M; %// Note that N-M is an upper limit for L!
idx = bsxfun(#plus, (0:L)', 1:M)
X = b(idx)
That's exactly what im2col (from the Image Processing Toolbox) does:
b = (1:20).'; %'// example data
L = 18; % // desired length of sliding blocks
x = im2col(b, [L 1]); % // result
I'd use horzcat. For example:
function X = reshaper(b,len)
diff = length(b) - len + 1;
X = b(1:len);
for i=2:diff
X = horzcat(X,b(i:len+(i-1)));
end
You could probably remove the for loop with some further thought.