Loopless submatrix assignment in Matlab - matlab

I have a matrix F of size D-by-N and a vector A of length N of random integers in the range [1,a]. I want to create a matrix M of size D * a such that each colum M(:,i) has the vector F(:,i) starting from the index (A(i)-1)*D+1 to (A(i)-1)*D+D.
Example:
F = [1 2 3 10
4 5 6 22]
A = [3 2 1 2]
a = 4
M = [0 0 3 0
0 0 6 0
0 2 0 10
0 5 0 22
1 0 0 0
4 0 0 0
0 0 0 0
0 0 0 0]
I can do it with a simple loop
for i = 1 : N
M((A(i)-1)*D+1:(A(i)-1)*D+D,i) = F(:,i);
end
but for large N this might take a while. I am looking for a way to do it without loop.


You can use bsxfun for a linear-indexing based approach -
[D,N] = size(F); %// Get size of F
start_idx = (A-1)*D+1 + [0:N-1]*D*a; %// column start linear indices
all_idx = bsxfun(#plus,start_idx,[0:D-1]'); %//'# all linear indices
out = zeros(D*a,N); %// Initialize output array with zeros
out(all_idx) = F; %// Insert values from F into output array
Sample run -
F =
1 2 3 10
4 5 6 22
A =
3 2 1 2
a =
4
out =
0 0 3 0
0 0 6 0
0 2 0 10
0 5 0 22
1 0 0 0
4 0 0 0
0 0 0 0
0 0 0 0

Related

Generating unitriangular matrices in Octave|MATLAB [duplicate]

Suppose I have an NxN matrix A, an index vector V consisting of a subset of the numbers 1:N, and a value K, and I want to do this:
for i = V
A(i,i) = K
end
Is there a way to do this in one statement w/ vectorization?
e.g. A(something) = K
The statement A(V,V) = K will not work, it assigns off-diagonal elements, and this is not what I want. e.g.:
>> A = zeros(5);
>> V = [1 3 4];
>> A(V,V) = 1
A =
1 0 1 1 0
0 0 0 0 0
1 0 1 1 0
1 0 1 1 0
0 0 0 0 0

I usually use EYE for that:
A = magic(4)
A(logical(eye(size(A)))) = 99
A =
99 2 3 13
5 99 10 8
9 7 99 12
4 14 15 99
Alternatively, you can just create the list of linear indices, since from one diagonal element to the next, it takes nRows+1 steps:
[nRows,nCols] = size(A);
A(1:(nRows+1):nRows*nCols) = 101
A =
101 2 3 13
5 101 10 8
9 7 101 12
4 14 15 101
If you only want to access a subset of diagonal elements, you need to create a list of diagonal indices:
subsetIdx = [1 3];
diagonalIdx = (subsetIdx-1) * (nRows + 1) + 1;
A(diagonalIdx) = 203
A =
203 2 3 13
5 101 10 8
9 7 203 12
4 14 15 101
Alternatively, you can create a logical index array using diag (works only for square arrays)
diagonalIdx = false(nRows,1);
diagonalIdx(subsetIdx) = true;
A(diag(diagonalIdx)) = -1
A =
-1 2 3 13
5 101 10 8
9 7 -1 12
4 14 15 101

>> tt = zeros(5,5)
tt =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
>> tt(1:6:end) = 3
tt =
3 0 0 0 0
0 3 0 0 0
0 0 3 0 0
0 0 0 3 0
0 0 0 0 3
and more general:
>> V=[1 2 5]; N=5;
>> tt = zeros(N,N);
>> tt((N+1)*(V-1)+1) = 3
tt =
3 0 0 0 0
0 3 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 3
This is based on the fact that matrices can be accessed as one-dimensional arrays (vectors), where the 2 indices (m,n) are replaced by a linear mapping m*N+n.

>> B=[0,4,4;4,0,4;4,4,0]
B =
0 4 4
4 0 4
4 4 0
>> v=[1,2,3]
v =
1 2 3
>> B(eye(size(B))==1)=v
%insert values from v to eye positions in B
B =
1 4 4
4 2 4
4 4 3

A = zeros(7,6);
V = [1 3 5];
[n m] = size(A);
diagIdx = 1:n+1:n*m;
A( diagIdx(V) ) = 1
A =
1 0 0 0 0 0
0 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 1 0
0 0 0 0 0 0
0 0 0 0 0 0

I'd use sub2ind and pass the diagonal indices as both x and y parameters:
A = zeros(4)
V=[2 4]
idx = sub2ind(size(A), V,V)
% idx = [6, 16]
A(idx) = 1
% A =
% 0 0 0 0
% 0 1 0 0
% 0 0 0 0
% 0 0 0 1

Suppose K is the value. The command
A=A-diag(K-diag(A))
may be a bit faster
>> A=randn(10000,10000);
>> tic;A(logical(eye(size(A))))=12;toc
Elapsed time is 0.517575 seconds.
>> tic;A=A+diag((99-diag(A)));toc
Elapsed time is 0.353408 seconds.
But it consumes more memory.

I use this small inline function in finite difference code.
A=zeros(6,3);
range=#(A,i)[1-min(i,0):size(A,1)-max(i+size(A,1)-size(A,2),0 ) ];
Diag=#(A,i) sub2ind(size(A), range(A,i),range(A,i)+i );
A(Diag(A, 0))= 10; %set diagonal
A(Diag(A, 1))= 20; %equivelent to diag(A,1)=20;
A(Diag(A,-1))=-20; %equivelent to diag(A,-1)=-20;
It can be easily modified to work on a sub-range of the diagonal by changing the function range.

insert a row and a line in a matrix

I create a matrix b from a matrix a in the following way:
a=[1 2 ; 3 4];
b= [a zeros(2); zeros(2) a]
b =
1 2 0 0
3 4 0 0
0 0 1 2
0 0 3 4
Successively, I want to insert a line and a column of zeros at a certain point of the matrix. Let's say at middle way:
idx=2;
c=[b(1:idx,:); zeros(1,4); b(idx+1:end,:)]
c =
1 2 0 0
3 4 0 0
0 0 0 0
0 0 1 2
0 0 3 4
c=[c(:,1:idx) zeros(5,1) c(:,idx+1:end)]
c =
1 2 0 0 0
3 4 0 0 0
0 0 0 0 0
0 0 0 1 2
0 0 0 3 4
Is there a more intelligent way of doing this?

Here is another way(I don't know if it is a more intelligent way).
Assuming that you have the row index as row and the column index as col:
sc = size(b) + 1;
c = zeros(sc);
ROW = true(sc(1), 1);
ROW(row) = false;
COL = true(1, sc(2));
COL(col) = false;
Then in MATLAB r2016b /Octave you can write
c(ROW & COL)=b;
In pre 2016b you can use bsxfun
c(bsxfun(#and, ROW , COL))=b;

How to create this matrix in MATLAB

I have a vector such as
A=[4;3;1;6]
and I want to create a matrix with the elements below from A
B=[6 5 4 3 2 1;4 3 2 1 0 0;3 2 1 0 0 0;1 0 0 0 0 0];
How can I do this in MATLAB ? the number of columns equal to the max of A.

Here are two ways to do this: one vectorized, and one in a loop.
A=[4;3;1;6];
B = max(bsxfun(#minus, sort(A, 'descend'), 0:(max(A)-1)), 0);
or
S = sort(A, 'descend');
m = numel(A); n = S(1);
C = zeros(m,n);
for k = 1:m
C(k,1:S(k)) = S(k):-1:1;
end
Results:
B =
6 5 4 3 2 1
4 3 2 1 0 0
3 2 1 0 0 0
1 0 0 0 0 0

How to vectorize matrix subsetting such that find() returns a matrix?

Given a matrix A, I need to find the indices corresponding to the values 1 and 2. I could do this sequentially as follows:
>> B
B =
1 2 3
4 1 6
7 8 9
4 5 1
>> find(B==1)
ans =
1
6
12
>> find(B==2)
ans =
5
But if I do this kind of operation in a loop, Matlab will only use one core of my processor. How can I vectorise it, so that I obtain a matrix from find? I want this result:
>> my_find( B, [1 2] )
ans =
1 5
6 0
12 0
(or some other padding)
How can I obtain this?

Just don't use find
B==1
ans =
1 0 0
0 1 0
0 0 0
0 0 1
B==2
ans =
0 1 0
0 0 0
0 0 0
0 0 0
And then add or logical OR those together.
i.e.
(B==1) + (B==2)
ans =
1 1 0
0 1 0
0 0 0
0 0 1
or
(B==1) | (B==2)
ans =
1 1 0
0 1 0
0 0 0
0 0 1

[i, j] = ind2sub(size(B), find(logical(sum(bsxfun(#eq, B(:), [1 2]), 2))))

How to assign values to a MATLAB matrix on the diagonal?

Suppose I have an NxN matrix A, an index vector V consisting of a subset of the numbers 1:N, and a value K, and I want to do this:
for i = V
A(i,i) = K
end
Is there a way to do this in one statement w/ vectorization?
e.g. A(something) = K
The statement A(V,V) = K will not work, it assigns off-diagonal elements, and this is not what I want. e.g.:
>> A = zeros(5);
>> V = [1 3 4];
>> A(V,V) = 1
A =
1 0 1 1 0
0 0 0 0 0
1 0 1 1 0
1 0 1 1 0
0 0 0 0 0

I usually use EYE for that:
A = magic(4)
A(logical(eye(size(A)))) = 99
A =
99 2 3 13
5 99 10 8
9 7 99 12
4 14 15 99
Alternatively, you can just create the list of linear indices, since from one diagonal element to the next, it takes nRows+1 steps:
[nRows,nCols] = size(A);
A(1:(nRows+1):nRows*nCols) = 101
A =
101 2 3 13
5 101 10 8
9 7 101 12
4 14 15 101
If you only want to access a subset of diagonal elements, you need to create a list of diagonal indices:
subsetIdx = [1 3];
diagonalIdx = (subsetIdx-1) * (nRows + 1) + 1;
A(diagonalIdx) = 203
A =
203 2 3 13
5 101 10 8
9 7 203 12
4 14 15 101
Alternatively, you can create a logical index array using diag (works only for square arrays)
diagonalIdx = false(nRows,1);
diagonalIdx(subsetIdx) = true;
A(diag(diagonalIdx)) = -1
A =
-1 2 3 13
5 101 10 8
9 7 -1 12
4 14 15 101

>> tt = zeros(5,5)
tt =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
>> tt(1:6:end) = 3
tt =
3 0 0 0 0
0 3 0 0 0
0 0 3 0 0
0 0 0 3 0
0 0 0 0 3
and more general:
>> V=[1 2 5]; N=5;
>> tt = zeros(N,N);
>> tt((N+1)*(V-1)+1) = 3
tt =
3 0 0 0 0
0 3 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 3
This is based on the fact that matrices can be accessed as one-dimensional arrays (vectors), where the 2 indices (m,n) are replaced by a linear mapping m*N+n.

>> B=[0,4,4;4,0,4;4,4,0]
B =
0 4 4
4 0 4
4 4 0
>> v=[1,2,3]
v =
1 2 3
>> B(eye(size(B))==1)=v
%insert values from v to eye positions in B
B =
1 4 4
4 2 4
4 4 3

A = zeros(7,6);
V = [1 3 5];
[n m] = size(A);
diagIdx = 1:n+1:n*m;
A( diagIdx(V) ) = 1
A =
1 0 0 0 0 0
0 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 1 0
0 0 0 0 0 0
0 0 0 0 0 0

I'd use sub2ind and pass the diagonal indices as both x and y parameters:
A = zeros(4)
V=[2 4]
idx = sub2ind(size(A), V,V)
% idx = [6, 16]
A(idx) = 1
% A =
% 0 0 0 0
% 0 1 0 0
% 0 0 0 0
% 0 0 0 1

Suppose K is the value. The command
A=A-diag(K-diag(A))
may be a bit faster
>> A=randn(10000,10000);
>> tic;A(logical(eye(size(A))))=12;toc
Elapsed time is 0.517575 seconds.
>> tic;A=A+diag((99-diag(A)));toc
Elapsed time is 0.353408 seconds.
But it consumes more memory.

I use this small inline function in finite difference code.
A=zeros(6,3);
range=#(A,i)[1-min(i,0):size(A,1)-max(i+size(A,1)-size(A,2),0 ) ];
Diag=#(A,i) sub2ind(size(A), range(A,i),range(A,i)+i );
A(Diag(A, 0))= 10; %set diagonal
A(Diag(A, 1))= 20; %equivelent to diag(A,1)=20;
A(Diag(A,-1))=-20; %equivelent to diag(A,-1)=-20;
It can be easily modified to work on a sub-range of the diagonal by changing the function range.