Let us say that I have 7 values: a, b, c, d, e, f and g
I would like to construct an m by n matrix in this way:
[ a b c d 0 0 0 0 . . . .
[ b e f g 0 0 0 0 . . . .
[ c f a b c d 0 0 . . . .
[ d g b e f g 0 0 . . . .
[ 0 0 c f a b c d 0 0 . .
[ 0 0 d g b e f g 0 0 . .
[ . . 0 0 c f a b c d . .
[ . . 0 0 d g b e f g . .
And so forth...
Therefore, the desired matrix is symmetrical. Values a and e alternate on the main diagonal; values b and f alternate on the 1st upper diagonal; values c and g alternate on the 2nd upper diagonal; values d and 0 alternate on the 3rd upper diagonal. I would like to be able to specify the matrix size with m by n parameters.
I used to do this easily with the SparseArray and Band functions in Mathematica, but I cannot find equivalent functions in Matlab. Would there be an efficient way to construct this kind of matrix in Matlab?
You can use spdiags to specify the upper diagonals in a sparse matrix and then add the transpose of the strictly upper triangular part for exact symmetry:
>> n = 6;
>> a = 1;b = 2;c = 3;d = 4;e = 5;f = 6;g = 7;
>> n = 6;
>> A = spdiags(repmat([[a;e] , [f;b] , [c;g] , [0;d]],n/2,1),0:3,n,n);
>> A = A + triu(A,1).';
>> issymmetric(A)
ans =
1
>> full(A)
ans =
1 2 3 4 0 0
2 5 6 7 0 0
3 6 1 2 3 4
4 7 2 5 6 7
0 0 3 6 1 2
0 0 4 7 2 5
You might notice I flipped b/f and 0/d to adjust for the filling behavior; there may be a better way to do this.
For non-square matrices with potentially odd dimension number, I would build the minimum-sized square sparse matrix that has the actual one as a sub-matrix and mask out the unneeded portion at the end:
>> m = 13;
>> n = 6;
>> p = max([m,n]);
>> p = p + mod(p,2); % added to make p even.
>> A = spdiags(repmat([[a;e] , [f;b] , [c;g] , [0;d]],p/2,1),0:3,p,p);
>> A = A + triu(A,1).';
>> A = A(1:m,1:n);
>> full(A)
ans =
1 2 3 4 0 0
2 5 6 7 0 0
3 6 1 2 3 4
4 7 2 5 6 7
0 0 3 6 1 2
0 0 4 7 2 5
0 0 0 0 3 6
0 0 0 0 4 7
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 0 0 0 0 0
It's a tad inefficient to build, but easy and straight-forward.
You can use the spdiags function
Using numbers instead of letters (sparse does not support symbols) like this:
m = 10;
n = 10;
B = repmat([1:4;[5:7 0]],n/2,1);
B=[B(:,end:-1:2) B];
d=-3:3;
A = spdiags(B,d,m,n);
full(A)
which outputs:
1 6 3 0 0 0 0 0 0 0
2 5 2 7 4 0 0 0 0 0
3 6 1 6 3 0 0 0 0 0
4 7 2 5 2 7 4 0 0 0
0 0 3 6 1 6 3 0 0 0
0 0 4 7 2 5 2 7 4 0
0 0 0 0 3 6 1 6 3 0
...
Explanation:
The B matrix contains the values needed on the diagonals in its columns. It is first constructed using B = repmat([1:4;[5:7 0]],n/2,1); which makes the first two rows repeat n/2 times. At that point it looks like:
1 2 3 4
5 6 7 0
1 2 3 4
5 6 7 0
1 2 3 4
5 6 7 0
1 2 3 4
...
Then it is mirrored to give it the diagonals that are in the lower diagonal part using B=[B(:,end:-1:2) B]; which makes it look like:
4 3 2 1 2 3 4
0 7 6 5 6 7 0
4 3 2 1 2 3 4
0 7 6 5 6 7 0
4 3 2 1 2 3 4
0 7 6 5 6 7 0
4 3 2 1 2 3 4
...
These are the diagonals passed to the spdiags function.
The end is a MATLAB keyword which refers to the last column or row. That indexing line says, in words: "Take all the rows of B, and the columns starting from the end, going back 1 step at a time until the second column, and place that submatrix before B." That is how the mirroring is accomplished.
Related
I have a matrix in MATLAB with zeroes and I would like to get another matrix with the first N non-zero elements in each row. Let's say for example N = 3, and the matrix is
A = [ 0 0 2 0 6 7 9;
3 2 4 7 0 0 6;
0 1 0 3 4 8 6;
1 2 0 0 0 1 3]
I'd like the result to be:
B = [2 6 7;
3 2 4;
1 3 4;
1 2 1]
I have a huge matrix so I would like to do it without a loop, could you please help me? Thanks a lot!
Since MATLAB stores a matrix according to column-major order, I first transpose A, bubble up the non-zeros, and pick the first N lines, and transpose back:
N = 3;
A = [ 0 0 2 0 6 7 9;
3 2 4 7 0 0 6;
0 1 0 3 4 8 6;
1 2 0 0 0 1 3];
Transpose and preallocate output B
At = A';
B = zeros(size(At));
At =
0 3 0 1
0 2 1 2
2 4 0 0
0 7 3 0
6 0 4 0
7 0 8 1
9 6 6 3
Index zeros
idx = At == 0;
idx =
1 0 1 0
1 0 0 0
0 0 1 1
1 0 0 1
0 1 0 1
0 1 0 0
0 0 0 0
Bubble up the non-zeros
B(~sort(idx)) = At(~idx);
B =
2 3 1 1
6 2 3 2
7 4 4 1
9 7 8 3
0 6 6 0
0 0 0 0
0 0 0 0
Select first N rows and transpose back
B(1:N,:)'
You can do the bubbling in row-major order, but you would need to retrieve the row and column subscripts with find, and do some sorting and picking there. It becomes more tedious and less readable.
Using accumarray with no loops:
N = 3;
[ii,jj] = find(A); [ii,inds]=sort(ii); jj = jj(inds);
lininds = ii+size(A,1)*(jj-1);
C = accumarray(ii,lininds,[],#(x) {A(x(1:N)')}); %' cell array output
B = vertcat(C{:})
B =
2 6 7
3 2 4
1 3 4
1 2 1
Usually I don't go with a for loop solution, but this is fairly intuitive:
N = 3;
[ii,jj] = find(A);
B = zeros(size(A,1),N);
for iRow = 1:size(A,1),
nzcols = jj(ii==iRow);
B(iRow,:) = A(iRow,nzcols(1:N));
end
Since you are guaranteed to have more than N nonzeros per row of A, that should get the job done.
One-liner solution:
B = cell2mat(cellfun(#(c) c(1:N), arrayfun(#(k) nonzeros(A(k,:)), 1:size(A,1), 'uni', false), 'uni', false)).'
Not terribly elegant or efficient, but so much fun!
N = 3;
for ii=1:size(A,1);
B(ii,:) = A( ii,find(A(ii,:),N) );
end
Actually , you can do it like the code blow:
N=3
for n=1:size(A,1)
[a b]=find(A(n,:)>0,N);
B(n,:)=A(n,transpose(b));
end
Then I think this B matrix will be what you want.
I am having trouble creating this matrix in matlab, basically I need to create a matrix that has -1 going across the center diagonal followed be 4s on the diagonal outside of that (example below). All the other values can be zero.
A5 = [-1 4 0 0 0;
4 -1 4 0 0;
0 4 -1 4 0;
0 0 4 -1 4;
0 0 0 4 -1];
I have tried using a command v = [4]; D = diag(v)
but that only works for the center diagonal.
This can also be done using a toeplitz matrix:
function out = tridiag(a,b,c,N)
% TRIDIAG generates a tri-diagonal matrix of size NxN.
% lower diagonal is a
% main diagonal is b
% upper diagonal is c
out = toeplitz([b,a,zeros(1,N-2)],[b,c,zeros(1,N-2)]);
>> tridiag(4,-1,4,5)
ans =
-1 4 0 0 0
4 -1 4 0 0
0 4 -1 4 0
0 0 4 -1 4
0 0 0 4 -1
Note #1: When your desired output is symmetric, you can omit the 2nd input to toeplitz.
Note #2: As the size of the matrix increases, there comes a point where it makes more sense to store it as sparse, as this saves memory and improves performance (assuming your matrix is indeed sparse, i.e. comprised mostly of zeros, as it happens with a tridiagonal matrix). Some useful functions are spdiags, sptoeplitzFEX and blktridiagFEX.
A little hackish, but here it goes:
N = 7; % matrix size
v = [11 22 33]; % row vector containing the diagonal values
w = [0 v(end:-1:1)];
result = w(max(numel(v)+1-abs(bsxfun(#minus, 1:N, (1:N).')),1))
This gives
result =
11 22 33 0 0 0 0
22 11 22 33 0 0 0
33 22 11 22 33 0 0
0 33 22 11 22 33 0
0 0 33 22 11 22 33
0 0 0 33 22 11 22
0 0 0 0 33 22 11
To understand how it works, see some intermediate steps:
>> abs(bsxfun(#minus, 1:N, (1:N).'))
ans =
0 1 2 3 4 5 6
1 0 1 2 3 4 5
2 1 0 1 2 3 4
3 2 1 0 1 2 3
4 3 2 1 0 1 2
5 4 3 2 1 0 1
6 5 4 3 2 1 0
>> max(numel(v)+1-abs(bsxfun(#minus, 1:N, (1:N).')),1)
ans =
4 3 2 1 1 1 1
3 4 3 2 1 1 1
2 3 4 3 2 1 1
1 2 3 4 3 2 1
1 1 2 3 4 3 2
1 1 1 2 3 4 3
1 1 1 1 2 3 4
Use D = diag(u,k) to shift u in k levels above the main diagonal, and D = diag(u,-k) for the opposite direction. Keep in mind that you need u to be in the right length of the k diagonal you want, so if the final matrix is n*n, the k's diagonal will have only n-abs(k) elements.
For you case:
n = 5; % the size of the matrix
v = ones(n,1)-2; % make the vector for the main diagonal
u = ones(n-1,1)*4; % make the vector for +1 and -1 diagonal
A5 = diag(v)+diag(u,1)+diag(u,-1) % combine everything together
Which gives:
A5 =
-1 4 0 0 0
4 -1 4 0 0
0 4 -1 4 0
0 0 4 -1 4
0 0 0 4 -1
I need to transform a neural network output matrix with size 2 X N in zeros and ones, where 0 will represent the minimum value of the column and 1 contrariwise. This will be necessary in order to calculate the confusion matrix.
For example, consider this matrix 2 X 8:
2 33 4 5 6 7 8 9
1 44 5 4 7 5 2 1
I need to get this result:
1 0 0 1 0 1 1 1
0 1 1 0 1 0 0 0
How can I do this in MATLAB without for loops? Thanks in advance.
>> d = [ 2 33 4 5 6 7 8 9;
1 44 5 4 7 5 2 1];
>> bsxfun(#rdivide, bsxfun(#minus, d, min(d)), max(d) - min(d))
ans =
1 0 0 1 0 1 1 1
0 1 1 0 1 0 0 0
The bsxfun function is necessary to broadcast the minus and division operations to matrices of different dimensions (min and max have only 1 row each).
Other solution is the following (works only for 2 rows):
>> [d(1,:) > d(2,:); d(1,:) < d(2,:)]
ans =
1 0 0 1 0 1 1 1
0 1 1 0 1 0 0 0
If it's just 2xN, then this will work:
floor(A./[max(A); max(A)])
In general:
floor(A./repmat(max(A),size(A,1),1))
I have a matrix in MATLAB with zeroes and I would like to get another matrix with the first N non-zero elements in each row. Let's say for example N = 3, and the matrix is
A = [ 0 0 2 0 6 7 9;
3 2 4 7 0 0 6;
0 1 0 3 4 8 6;
1 2 0 0 0 1 3]
I'd like the result to be:
B = [2 6 7;
3 2 4;
1 3 4;
1 2 1]
I have a huge matrix so I would like to do it without a loop, could you please help me? Thanks a lot!
Since MATLAB stores a matrix according to column-major order, I first transpose A, bubble up the non-zeros, and pick the first N lines, and transpose back:
N = 3;
A = [ 0 0 2 0 6 7 9;
3 2 4 7 0 0 6;
0 1 0 3 4 8 6;
1 2 0 0 0 1 3];
Transpose and preallocate output B
At = A';
B = zeros(size(At));
At =
0 3 0 1
0 2 1 2
2 4 0 0
0 7 3 0
6 0 4 0
7 0 8 1
9 6 6 3
Index zeros
idx = At == 0;
idx =
1 0 1 0
1 0 0 0
0 0 1 1
1 0 0 1
0 1 0 1
0 1 0 0
0 0 0 0
Bubble up the non-zeros
B(~sort(idx)) = At(~idx);
B =
2 3 1 1
6 2 3 2
7 4 4 1
9 7 8 3
0 6 6 0
0 0 0 0
0 0 0 0
Select first N rows and transpose back
B(1:N,:)'
You can do the bubbling in row-major order, but you would need to retrieve the row and column subscripts with find, and do some sorting and picking there. It becomes more tedious and less readable.
Using accumarray with no loops:
N = 3;
[ii,jj] = find(A); [ii,inds]=sort(ii); jj = jj(inds);
lininds = ii+size(A,1)*(jj-1);
C = accumarray(ii,lininds,[],#(x) {A(x(1:N)')}); %' cell array output
B = vertcat(C{:})
B =
2 6 7
3 2 4
1 3 4
1 2 1
Usually I don't go with a for loop solution, but this is fairly intuitive:
N = 3;
[ii,jj] = find(A);
B = zeros(size(A,1),N);
for iRow = 1:size(A,1),
nzcols = jj(ii==iRow);
B(iRow,:) = A(iRow,nzcols(1:N));
end
Since you are guaranteed to have more than N nonzeros per row of A, that should get the job done.
One-liner solution:
B = cell2mat(cellfun(#(c) c(1:N), arrayfun(#(k) nonzeros(A(k,:)), 1:size(A,1), 'uni', false), 'uni', false)).'
Not terribly elegant or efficient, but so much fun!
N = 3;
for ii=1:size(A,1);
B(ii,:) = A( ii,find(A(ii,:),N) );
end
Actually , you can do it like the code blow:
N=3
for n=1:size(A,1)
[a b]=find(A(n,:)>0,N);
B(n,:)=A(n,transpose(b));
end
Then I think this B matrix will be what you want.
I want to create a matrix like
A = [0 0 0 0 1;
0 0 0 1 1;
0 0 0 1 1;
0 0 0 1 1;
0 0 1 1 1;
0 1 1 1 1]
based on a vector indicating how many '0's should precede '1's on each row:
B = [4 3 3 3 2 1]
Is there a loopless way to do this ?
You don't mention in your question how the horizontal size of the array should be defined (the number of ones).
For predefined width you can use this code:
width = 5;
A = cell2mat(arrayfun(#(x) [ zeros(1,x), ones(1,width-x) ], B, 'UniformOutput', false)');
If you want that A has minimal width, but still at least one 1 in every row:
A = cell2mat(arrayfun(#(x) [ zeros(1,x), ones(1,max(B)+1-x) ], B, 'UniformOutput', false)');
A shorter “old-school” way to achieve this without a loop would be as follows:
A = repmat(B',1,max(B)+1)<repmat([1:max(B)+1],size(B,2),1)
If you want to have a minimum number of ones
min_ones=1; %or whatever
A = repmat(B',1,max(B)+min_ones)<repmat([1:max(B)+min_ones],size(B,2),1)
I don’t know how this compares speedwise to #nrz’s approach (I’ve only got Octave to hand right now), but to me it's more intuitive as it’s simply comparing a max(B) + min_ones * column tiling of B:
4 4 4 4 4
3 3 3 3 3
3 3 3 3 3
3 3 3 3 3
2 2 2 2 2
1 1 1 1 1
with a row tiling of [1 : max(B) + min_ones]
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
To generate:
A =
0 0 0 0 1
0 0 0 1 1
0 0 0 1 1
0 0 0 1 1
0 0 1 1 1
0 1 1 1 1
This requires only one line, and seems to be faster than previous solutions based on repmat or arrayfun:
%// Example data
ncols = 5;
B = [4 3 3 3 2 1];
%// Generate A
A = bsxfun(#gt, 1:ncols, B(:));