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i have a quite complicated question here which i am working on. It's extremely difficult to describe in words, so i will try to explain it with an example.
Assume i have a matrix of values:
A =
[31 85 36 71 51]
[12 33 74 39 12]
[67 11 13 14 18]
[35 36 84 33 57]
Now, i want to first find a maximum vector in the first dimension, which is easy:
[max_vector,~] = max(A,[],1);
max_vector=[67,85, 84, 71,57]
Now i want to get a "slimmed" matrix with values around the maxima (periodical indices):
Desired_Matrix =
[12 36 36 33 18]
[67 85 84 71 57]
[35 33 13 39 51]
This is the matrix with the vectors around the maximum values of matrix A. Can someone tell me how to do this without using a double for loop?
Thank you!
% Input.
A = [31 85 36 71 51; 12 33 74 39 12; 67 11 13 14 18; 35 36 84 33 57]
% Dimensions needed.
nRows = size(A, 1);
nCols = size(A, 2);
% Get maxima and corresponding indices in input.
[max_vector, ind] = max(A);
% Get neighbouring indices.
ind = [ind - 1; ind; ind + 1];
% Modulo indices to prevent dimension overflow.
ind = mod(ind, nRows);
% Correct zero indices.
ind(ind == 0) = nRows;
% Calculate correct indices in A.
temp = repmat(0:nRows:nRows*(nCols-1), 3, 1);
ind = ind + temp;
% Output.
B = A(ind)
Since we have max indices per column, but later want to access these elements in the original array A, we need proper linear indices for A. Here, the trick is to add the number of rows multiplied by the column index (starting by 0). The easiest way to understand might be to remove the semicolons, and inspect the intermediate values of ind.
#HansHirse's answer is more efficient, as it does not create an intermediate matrix.
Try this:
[~, ind_max] = max(A,[],1);
A_ext = A([end 1:end 1],:);
ind_lin = bsxfun(#plus, bsxfun(#plus, ind_max, (0:2).'), (0:size(A_ext,2)-1)*size(A_ext,1));
result = reshape(A_ext(ind_lin), 3, []);
For Matlab R2016b or newer, you can simplify the third line:
[~, ind_max] = max(A,[],1);
A_ext = A([end 1:end 1],:);
ind_lin = ind_max + (0:2).' + (0:size(A_ext,2)-1)*size(A_ext,1);
result = reshape(A_ext(ind_lin), 3, []);
Here is another solution. This is similar to HansHirse's answer, with two improvements:
Slightly more elegantly handles the modular indexing
Is more flexible for specifying which neighbours your want
Code:
% Input
A = [31 85 36 71 51;
12 33 74 39 12;
67 11 13 14 18;
35 36 84 33 57];
% Relative rows of neighbours, i.e. this is [-1, 0, 1] for +/- one row
p = -1:1;
% Get A row and column counts for ease
[nr, nc] = size(A);
% Get max indices
[~,idx] = max( A, [], 1 );
% Handle overflowing indices to wrap around rows
% You don't have to redefine "idx", could use this directly in the indexing line
idx = mod( idx + p.' - 1, nr ) + 1;
% Output B. The "+ ... " is to convert to linear indices, as "idx"
% currently just refers to the row number.
B = A(idx + (0:nr:nr*nc-1));
You can use the Image Processing Toolbox to generate the result, though is less efficient than other solutions.
[~,idx] = max(A, [], 1);
d = imdilate( idx == (1:size(A,1) ).', [1;1;1], 'full');
p = padarray(A, 1, 'circular');
Desired_Matrix = reshape(p(d), 3, []);
just for your information, here is the generalized form for the 3D-Case:
A = zeros(3,5,5);
for id = 1: 20
A(:,:,id) = id;
if id == 10
A(:,:,id) = 100;
end
end
% Relative rows of neighbours, i.e. this is [-1, 0, 1] for +/- one row
p = -1:1;
% Get A row and column counts for ease
[nr, nc, nz] = size(A);
% Get max indices
[~,idx] = max( A, [], 3 );
% Handle overflowing indices to wrap around rows
% You don't have to redefine "idx", could use this directly in the indexing line
idx = mod( idx + reshape(p,1,1,3) - 1, nz ) + 1;
% Output B. The "+ ... " is to convert to linear indices, as "idx"
% currently just refers to the row number.
INDICES = ((idx-1) * (nr*nc)+1 )+ reshape(0:1:nc*nr-1,nr,nc);
B = A(INDICES);
Say I have a cell array in the following format:
A = {4;[22 16 4]; 23; 51; [16 22]; 32; [4 50]};
I want to output the unique columns gained by any permutations of the vectors in the rows.
For example, for the example in the above, the only column vectors that would satisfy this would be
[4; 22; 23; 51; 16; 32; 50] and [4; 16; 23; 51; 22; 32; 50].
I can't choose 4 from the second or last rows since 4 is the only option in the first row. Moreover, I can't choose 22 in both the second and fifth rows since this would make the column non-unique. Although empty choices in some rows are not allowed, if there are no unique columns, then I would need to output an empty column.
Does anyone have a smart way of doing this (fairly quickly) in Matlab? Any help would be much appreciated.
Many thanks!
Here's a brute-force approach:
Generate all possible columns obtained by choosing one element from each vector (Cartesian product). This can be done with this approach.
Keep only those columns for which all elements are different. This is easily done with sort, diff, all, and logical indexing.
Code:
A = {4; [22 16 4]; 23; 51; [16 22]; 32; [4 50]}; % Data
n = numel(A); % Step 1
c = cell(1,n);
[c{end:-1:1}] = ndgrid(A{:});
c = cat(n+1, c{end:-1:1});
c = reshape(c,[],n).'; % Step 1
result = c(:,all(diff(sort(c,1),[],1),1)); % Step 2
How to select a submatrix (not in any pattern) in Matlab? For example, for a matrix of size 10 by 10, how to select the submatrix consisting of intersection of the 1st 2nd and 9th rows and the 4th and 6th columns?
Thanks for any helpful answers!
TLDR: Short Answer
As for your question, suppose you have an arbitrary 10-by-10 matrix A. The simplest way to extract the desired sub-matrix would be with an index vector:
B = A([1 2 9], [4 6]);
Indexing in MATLAB
There's an interesting article in the official documentation that comprehensively explains indexing in MATLAB.
Basically, there are several ways to extract a subset of values, I'll summarize them for you:
1. Indexing Vectors
Indexing vectors indicate the indices of the element to be extracted. They can either contain a single index or several, like so:
A = [10 20 30 40 50 60 70 80 90]
%# Extracts the third and the ninth element
B = A([3 9]) %# B = [30 90]
Indexing vectors can be specified for each dimension separately, for instance:
A = [10 20 30; 40 50 60; 70 80 90];
%# Extract the first and third rows, and the first and second columns
B = A([1 3], [1 2]) %# B = [10 30; 40 60]
There are also two special subscripts: end and the colon (:):
end simply indicates the last index in that dimension.
The colon is just a short-hand notation for "1:end".
For example, instead of writing A([1 2 3], [2 3]), you can write A(:, 2:end). This is especially useful for large matrices.
2. Linear Indexing
Linear indexing treats any matrix as if it were a column vector by concatenating the columns into one column vector and assigning indices to the elements respectively. For instance, we have:
A = [10 20 30; 40 50 60; 70 80 90];
and we want to compute b = A(2). The equivalent column vector is:
A = [10;
40;
70;
20;
50;
80;
30;
60;
90]
and thus b equals 40.
The special colon and end subscripts are also allowed, of course. For that reason, A(:) converts any matrix A into a column vector.
Linear indexing with matrix subscripts:
It is also possible to use another matrix for linear indexing. The subscript matrix is simply converted into a column vector, and used for linear indexing. The resulting matrix is, however always of the same dimensions as the subscript matrix.
For instance, if I = [1 3; 1 2], then A(I) is the same as writing reshape(A(I(:)), size(I)).
Converting from matrix subscripts to linear indices and vice versa:
For that you have sub2ind and ind2sub, respectively. For example, if you want to convert the subscripts [1, 3] in matrix A (corresponding to element 30) into a linear index, you can write sub2ind(size(A), 1, 3) (the result in this case should be 7, of course).
3. Logical Indexing
In logical indexing the subscripts are binary, where a logical 1 indicates that the corresponding element is selected, and 0 means it is not. The subscript vector must be either of the same dimensions as the original matrix or a vector with the same number of elements. For instance, if we have:
A = [10 20 30; 40 50 60; 70 80 90];
and we want to extract A([1 3], [1 2]) using logical indexing, we can do either this:
Ir = logical([1 1 0]);
Ic = logical([1 0 1]);
B = A(Ir, Ic)
or this:
I = logical([1 0 1; 1 0 1; 0 0 0]);
B = A(I)
or this:
I = logical([1 1 0 0 0 0 1 1 0]);
B = A(I)
Note that in the latter two cases is a one-dimensional vector, and should be reshaped back into a matrix if necessary (for example, using reshape).
Let me explain with an example:
Let's define a 6x6 matrix
A = magic(6)
A =
35 1 6 26 19 24
3 32 7 21 23 25
31 9 2 22 27 20
8 28 33 17 10 15
30 5 34 12 14 16
4 36 29 13 18 11
From this matrix you want the elements in rows 1, 2 and 5, and in the columns 4 and 6
B = A([1 2 5],[4 6])
B =
26 24
21 25
12 16
Hope this helps.
function f = sub(A,i,j)
[m,n] = size(A);
row = 1:m;
col = 1:n;
x = row;
x(i) = [];
y=col;
y(j) = [];
f= A(x,y);
Returns the matrix A, with the ith row and jth column removed.
How to select a submatrix (not in any pattern) in Matlab? For example, for a matrix of size 10 by 10, how to select the submatrix consisting of intersection of the 1st 2nd and 9th rows and the 4th and 6th columns?
Thanks for any helpful answers!
TLDR: Short Answer
As for your question, suppose you have an arbitrary 10-by-10 matrix A. The simplest way to extract the desired sub-matrix would be with an index vector:
B = A([1 2 9], [4 6]);
Indexing in MATLAB
There's an interesting article in the official documentation that comprehensively explains indexing in MATLAB.
Basically, there are several ways to extract a subset of values, I'll summarize them for you:
1. Indexing Vectors
Indexing vectors indicate the indices of the element to be extracted. They can either contain a single index or several, like so:
A = [10 20 30 40 50 60 70 80 90]
%# Extracts the third and the ninth element
B = A([3 9]) %# B = [30 90]
Indexing vectors can be specified for each dimension separately, for instance:
A = [10 20 30; 40 50 60; 70 80 90];
%# Extract the first and third rows, and the first and second columns
B = A([1 3], [1 2]) %# B = [10 30; 40 60]
There are also two special subscripts: end and the colon (:):
end simply indicates the last index in that dimension.
The colon is just a short-hand notation for "1:end".
For example, instead of writing A([1 2 3], [2 3]), you can write A(:, 2:end). This is especially useful for large matrices.
2. Linear Indexing
Linear indexing treats any matrix as if it were a column vector by concatenating the columns into one column vector and assigning indices to the elements respectively. For instance, we have:
A = [10 20 30; 40 50 60; 70 80 90];
and we want to compute b = A(2). The equivalent column vector is:
A = [10;
40;
70;
20;
50;
80;
30;
60;
90]
and thus b equals 40.
The special colon and end subscripts are also allowed, of course. For that reason, A(:) converts any matrix A into a column vector.
Linear indexing with matrix subscripts:
It is also possible to use another matrix for linear indexing. The subscript matrix is simply converted into a column vector, and used for linear indexing. The resulting matrix is, however always of the same dimensions as the subscript matrix.
For instance, if I = [1 3; 1 2], then A(I) is the same as writing reshape(A(I(:)), size(I)).
Converting from matrix subscripts to linear indices and vice versa:
For that you have sub2ind and ind2sub, respectively. For example, if you want to convert the subscripts [1, 3] in matrix A (corresponding to element 30) into a linear index, you can write sub2ind(size(A), 1, 3) (the result in this case should be 7, of course).
3. Logical Indexing
In logical indexing the subscripts are binary, where a logical 1 indicates that the corresponding element is selected, and 0 means it is not. The subscript vector must be either of the same dimensions as the original matrix or a vector with the same number of elements. For instance, if we have:
A = [10 20 30; 40 50 60; 70 80 90];
and we want to extract A([1 3], [1 2]) using logical indexing, we can do either this:
Ir = logical([1 1 0]);
Ic = logical([1 0 1]);
B = A(Ir, Ic)
or this:
I = logical([1 0 1; 1 0 1; 0 0 0]);
B = A(I)
or this:
I = logical([1 1 0 0 0 0 1 1 0]);
B = A(I)
Note that in the latter two cases is a one-dimensional vector, and should be reshaped back into a matrix if necessary (for example, using reshape).
Let me explain with an example:
Let's define a 6x6 matrix
A = magic(6)
A =
35 1 6 26 19 24
3 32 7 21 23 25
31 9 2 22 27 20
8 28 33 17 10 15
30 5 34 12 14 16
4 36 29 13 18 11
From this matrix you want the elements in rows 1, 2 and 5, and in the columns 4 and 6
B = A([1 2 5],[4 6])
B =
26 24
21 25
12 16
Hope this helps.
function f = sub(A,i,j)
[m,n] = size(A);
row = 1:m;
col = 1:n;
x = row;
x(i) = [];
y=col;
y(j) = [];
f= A(x,y);
Returns the matrix A, with the ith row and jth column removed.
I have an NxM matrix in MATLAB that I would like to reorder in similar fashion to the way JPEG reorders its subblock pixels:
(image from Wikipedia)
I would like the algorithm to be generic such that I can pass in a 2D matrix with any dimensions. I am a C++ programmer by trade and am very tempted to write an old school loop to accomplish this, but I suspect there is a better way to do it in MATLAB.
I'd be rather want an algorithm that worked on an NxN matrix and go from there.
Example:
1 2 3
4 5 6 --> 1 2 4 7 5 3 6 8 9
7 8 9
Consider the code:
M = randi(100, [3 4]); %# input matrix
ind = reshape(1:numel(M), size(M)); %# indices of elements
ind = fliplr( spdiags( fliplr(ind) ) ); %# get the anti-diagonals
ind(:,1:2:end) = flipud( ind(:,1:2:end) ); %# reverse order of odd columns
ind(ind==0) = []; %# keep non-zero indices
M(ind) %# get elements in zigzag order
An example with a 4x4 matrix:
» M
M =
17 35 26 96
12 59 51 55
50 23 70 14
96 76 90 15
» M(ind)
ans =
17 35 12 50 59 26 96 51 23 96 76 70 55 14 90 15
and an example with a non-square matrix:
M =
69 9 16 100
75 23 83 8
46 92 54 45
ans =
69 9 75 46 23 16 100 83 92 54 8 45
This approach is pretty fast:
X = randn(500,2000); %// example input matrix
[r, c] = size(X);
M = bsxfun(#plus, (1:r).', 0:c-1);
M = M + bsxfun(#times, (1:r).'/(r+c), (-1).^M);
[~, ind] = sort(M(:));
y = X(ind).'; %'// output row vector
Benchmarking
The following code compares running time with that of Amro's excellent answer, using timeit. It tests different combinations of matrix size (number of entries) and matrix shape (number of rows to number of columns ratio).
%// Amro's approach
function y = zigzag_Amro(M)
ind = reshape(1:numel(M), size(M));
ind = fliplr( spdiags( fliplr(ind) ) );
ind(:,1:2:end) = flipud( ind(:,1:2:end) );
ind(ind==0) = [];
y = M(ind);
%// Luis' approach
function y = zigzag_Luis(X)
[r, c] = size(X);
M = bsxfun(#plus, (1:r).', 0:c-1);
M = M + bsxfun(#times, (1:r).'/(r+c), (-1).^M);
[~, ind] = sort(M(:));
y = X(ind).';
%// Benchmarking code:
S = [10 30 100 300 1000 3000]; %// reference to generate matrix size
f = [1 1]; %// number of cols is S*f(1); number of rows is S*f(2)
%// f = [0.5 2]; %// plotted with '--'
%// f = [2 0.5]; %// plotted with ':'
t_Amro = NaN(size(S));
t_Luis = NaN(size(S));
for n = 1:numel(S)
X = rand(f(1)*S(n), f(2)*S(n));
f_Amro = #() zigzag_Amro(X);
f_Luis = #() zigzag_Luis(X);
t_Amro(n) = timeit(f_Amro);
t_Luis(n) = timeit(f_Luis);
end
loglog(S.^2*prod(f), t_Amro, '.b-');
hold on
loglog(S.^2*prod(f), t_Luis, '.r-');
xlabel('number of matrix entries')
ylabel('time')
The figure below has been obtained with Matlab R2014b on Windows 7 64 bits. Results in R2010b are very similar. It is seen that the new approach reduces running time by a factor between 2.5 (for small matrices) and 1.4 (for large matrices). Results are seen to be almost insensitive to matrix shape, given a total number of entries.
Here's a non-loop solution zig_zag.m. It looks ugly but it works!:
function [M,index] = zig_zag(M)
[r,c] = size(M);
checker = rem(hankel(1:r,r-1+(1:c)),2);
[rEven,cEven] = find(checker);
[cOdd,rOdd] = find(~checker.'); %'#
rTotal = [rEven; rOdd];
cTotal = [cEven; cOdd];
[junk,sortIndex] = sort(rTotal+cTotal);
rSort = rTotal(sortIndex);
cSort = cTotal(sortIndex);
index = sub2ind([r c],rSort,cSort);
M = M(index);
end
And a test matrix:
>> M = [magic(4) zeros(4,1)];
M =
16 2 3 13 0
5 11 10 8 0
9 7 6 12 0
4 14 15 1 0
>> newM = zig_zag(M) %# Zig-zag sampled elements
newM =
16
2
5
9
11
3
13
10
7
4
14
6
8
0
0
12
15
1
0
0
Here's a way how to do this. Basically, your array is a hankel matrix plus vectors of 1:m, where m is the number of elements in each diagonal. Maybe someone else has a neat idea on how to create the diagonal arrays that have to be added to the flipped hankel array without a loop.
I think this should be generalizeable to a non-square array.
% for a 3x3 array
n=3;
numElementsPerDiagonal = [1:n,n-1:-1:1];
hadaRC = cumsum([0,numElementsPerDiagonal(1:end-1)]);
array2add = fliplr(hankel(hadaRC(1:n),hadaRC(end-n+1:n)));
% loop through the hankel array and add numbers counting either up or down
% if they are even or odd
for d = 1:(2*n-1)
if floor(d/2)==d/2
% even, count down
array2add = array2add + diag(1:numElementsPerDiagonal(d),d-n);
else
% odd, count up
array2add = array2add + diag(numElementsPerDiagonal(d):-1:1,d-n);
end
end
% now flip to get the result
indexMatrix = fliplr(array2add)
result =
1 2 6
3 5 7
4 8 9
Afterward, you just call reshape(image(indexMatrix),[],1) to get the vector of reordered elements.
EDIT
Ok, from your comment it looks like you need to use sort like Marc suggested.
indexMatrixT = indexMatrix'; % ' SO formatting
[dummy,sortedIdx] = sort(indexMatrixT(:));
sortedIdx =
1 2 4 7 5 3 6 8 9
Note that you'd need to transpose your input matrix first before you index, because Matlab counts first down, then right.
Assuming X to be the input 2D matrix and that is square or landscape-shaped, this seems to be pretty efficient -
[m,n] = size(X);
nlim = m*n;
n = n+mod(n-m,2);
mask = bsxfun(#le,[1:m]',[n:-1:1]);
start_vec = m:m-1:m*(m-1)+1;
a = bsxfun(#plus,start_vec',[0:n-1]*m);
offset_startcol = 2- mod(m+1,2);
[~,idx] = min(mask,[],1);
idx = idx - 1;
idx(idx==0) = m;
end_ind = a([0:n-1]*m + idx);
offsets = a(1,offset_startcol:2:end) + end_ind(offset_startcol:2:end);
a(:,offset_startcol:2:end) = bsxfun(#minus,offsets,a(:,offset_startcol:2:end));
out = a(mask);
out2 = m*n+1 - out(end:-1:1+m*(n-m+1));
result = X([out2 ; out(out<=nlim)]);
Quick runtime tests against Luis's approach -
Datasize: 500 x 2000
------------------------------------- With Proposed Approach
Elapsed time is 0.037145 seconds.
------------------------------------- With Luis Approach
Elapsed time is 0.045900 seconds.
Datasize: 5000 x 20000
------------------------------------- With Proposed Approach
Elapsed time is 3.947325 seconds.
------------------------------------- With Luis Approach
Elapsed time is 6.370463 seconds.
Let's assume for a moment that you have a 2-D matrix that's the same size as your image specifying the correct index. Call this array idx; then the matlab commands to reorder your image would be
[~,I] = sort (idx(:)); %sort the 1D indices of the image into ascending order according to idx
reorderedim = im(I);
I don't see an obvious solution to generate idx without using for loops or recursion, but I'll think some more.