I am trying to match RGB values in an image.
% R G B
RGBset = [ 3 9 12;
4 8 13;
11 13 13;
8 3 2]
img(:,:,1) = [1 2 3
6 5 4
7 9 8
10 11 12];
img(:,:,2) = [3 4 8;
6 7 8;
11 10 9;
12 13 14];
img(:,:,3)= [3 7 2;
4 9 10;
5 11 12;
6 13 14]
In this image, only one RGB value matches from the RGBset which is [11,13,13], so the expected output is:
[0 0 0;
0 0 0;
0 0 0;
0 1 0]; % reshape(img(4,2,:),1,3) = [11, 13 13] is available in RGBset
% no other RGB value is present in the image
I have made this code but it is very slow for larger images.
matched= zeros(img(:,:,1));
for r=1:size(img(:,:,1),1)
for c=1:size(img(:,:,2),2)
matched(r,c)=ismember(reshape(img(r,c,:),1,3),RGBset,'rows');
end
end
What will be the solution with more speed?
We can reduce each RGB triplet into a scalar each and we will do this for both RGBset and img. This would reduce them to 2D and 1D matrices respectively. We are calling this as dimensionality-reduction. With this reduced data, we are achieving memory efficiency and that hopefully should lead to performance boost.
Thus, a solution with those bases covered would look something like this -
% Scaling array for dim reduction
s = [256^2, 256, 1].';
% Reduce dims for RGBset and img
RGBset1D = RGBset*s;
img1D = reshape(img,[],3)*s;
% Finally use find membership and reshape to 2D
out = reshape(ismember(img1D, RGBset1D), size(img,1), []);
Benchmarking for the vectorized solutions
Benchmarking code -
% R G B
RGBset = [ 3 9 12;
4 8 13;
11 13 13;
8 3 2]
% Setup inputs
img = randi(255, 2000, 2000, 3);
img(3,2,:) = RGBset(4,:);
% Luis's soln
disp('--------------------- Reshape + Permute ------------------')
tic
img2 = reshape(permute(img, [3 1 2]), 3, []).';
matched = ismember(img2, RGBset, 'rows');
matched = reshape(matched, size(img,1), []);
toc
% Proposed in this post
disp('--------------------- Dim reduction ------------------')
tic
s = [256^2, 256, 1].';
RGBset1D = RGBset*s;
img1D = reshape(img,[],3)*s;
out = reshape(ismember(img1D, RGBset1D), size(img,1), []);
toc
Benchmarking output -
--------------------- Reshape + Permute ------------------
Elapsed time is 3.101870 seconds.
--------------------- Dim reduction ------------------
Elapsed time is 0.031589 seconds.
You can replace the loops by permute and reshape:
img2 = reshape(permute(img, [3 1 2]), 3, []).';
matched = ismember(img2, RGBset, 'rows');
matched = reshape(matched, size(img,1), []);
This creates a 3-column matrix img2 in which each row corresponds to one pixel from img. That way ismember(..., 'rows') can be applied in a vectorized manner. The obtained result is then reshaped as needed.
You can just loop over the colours, which is much quicker than looping over each pixel.
% Set up your colours into the 3rd dimension, so they match along the same axis
RGB3D = reshape(RGBset,[],1,3);
% Loop over them
for ii = 1:size(RGB3D, 1)
% See if all 3 members of the colour match any pixel
matched = all(ismember(img, RGB3D(ii,:,:)),3)
if any(matched)
disp(matched)
disp(['matched color: ' num2str(ii)]);
% do something else with the matched pixels
end
end
Related
For example I have this matrix(3x3) and I want to make a new 2x2 matrix with the max values from all the submatrixes :
๐ด = [
5 4 6
3 2 3
8 8 9
first submatrix:
[5 4
3 2 ]
max value=5
second submatrix:
[4 6
2 3]
max value=6
third submatrix:
[3 2
8 8 ]
max value=8
fourth submatrix:
[2 3
8 9]
max value=9
and I want to get this matrix(2x2) that has all the max values as elements of the previous submatrixes:
NewA=[5 6
8 9]
One last thing the only thing you can use are basic things like for loops if statements....(simple solutions,begginer solutions)
Also you can't use the max function,you have to write the code for that,and the solution should work for every square matrix
Using the Image Processing Toolbox
A = [5 4 6; 3 2 3; 8 8 9];
block_size = [2 2];
NewA = reshape(max(im2col(A, block_size, 'sliding'), [], 1), size(A)-block_size+1);
How it works:
im2col(A, block_size, 'sliding') arranges each sliding submatrix of size block_size as a column;
max(..., [], 1) takes the maximum of each column;
reshape(..., size(A)-block_size+1) reshapes the result into a matrix of the appropriate size.
Note that steps 1 and 3 both use column-major order, so the maxima in the result are arranged consistently with the input data.
Without the toolbox
Using linear indexing and implicit expansion, im2col's behaviour can be emulated as follows:
A = [5 4 6; 3 2 3; 8 8 9];
block_size = [2 2];
ind_base = (1:block_size(1)).' + (0:block_size(2)-1)*size(A,1);
ind_corner = (1:size(A,1)-block_size(1)+1).'+ (0:size(A,2)-block_size(2))*size(A,2);
ind_cols = ind_base(:) + ind_corner(:).' - 1;
NewA = reshape(max(A(ind_cols), [], 1) , size(A)-block_size+1);
The three variables ind_base, ind_corner and ind_cols have the following interpretation:
ind_base defines the linear indices of the first (uppermost, leftmost) submatrix;
ind_corner defines the linear indices of the upper-left corner of each submatrix;
ind_cols contains the linear indices of each submatrix arranged as columns.
Since the sub-matrices are 2x2 you can manually compute the max:
B = max(max(max(A(1:end-1, 1:end-1), A(1:end-1, 2:end)), A(2:end, 1:end-1)), A(2:end, 2:end));
or
B = max(cat(3, A(1:end-1, 1:end-1), A(1:end-1, 2:end), A(2:end, 1:end-1), A(2:end, 2:end)), [], 3);
If A is a square matrix both methods can be written more compact as:
n = size(A, 1);
x = 1:n-1;
y = 2:n;
B = max(max(max(A(x,x), A(x,y)), A(y,x)), A(y,y));
B = max(cat(3, A(x,x), A(x,y), A(y,x), A(y,y)), [], 3);
Using For-Loops (sliding window method)
Uses a set of nested for-loops to grab a neighbourhood of 4 (2 by 2). The position of the neighbourhood it relative to the variables Row_Index and Column_Index which are incremented/index the for-loop.
A = [5 4 6;
3 2 3;
8 8 9];
[Number_Of_Rows,Number_Of_Columns] = size(A);
Result = zeros(Number_Of_Rows-1,Number_Of_Columns-1);
for Row_Index = 1: Number_Of_Rows - 1
for Column_Index = 1: Number_Of_Columns - 1
Window = A(Row_Index:Row_Index+1,Column_Index:Column_Index+1);
Maximum = Window(1);
for Index = 2: 4
if Window(Index) > Maximum
Maximum = Window(Index);
end
end
Result(Row_Index,Column_Index) = Maximum;
end
end
Result
Ran using MATLAB R2019b
I have a 3x3 Matrix and want to save the indices and values into a new 9x3 matrix. For example A = [1 2 3 ; 4 5 6 ; 7 8 9] so that I will get a matrix x = [1 1 1; 1 2 2; 1 3 3; 2 1 4; 2 2 5; ...] With my code I only be able to store the last values x = [3 3 9].
A = [1 2 3 ; 4 5 6 ; 7 8 9];
x=[];
for i = 1:size(A)
for j = 1:size(A)
x =[i j A(i,j)]
end
end
Thanks for your help
Vectorized approach
Here's one way to do it that avoids loops:
A = [1 2 3 ; 4 5 6 ; 7 8 9];
[ii, jj] = ndgrid(1:size(A,1), 1:size(A,2)); % row and column indices
vv = A.'; % values. Transpose because column changes first in the result, then row
x = [jj(:) ii(:) vv(:)]; % result
Using your code
You're only missing concatenation with previous x:
A = [1 2 3 ; 4 5 6 ; 7 8 9];
x = [];
for i = 1:size(A)
for j = 1:size(A)
x = [x; i j A(i,j)]; % concatenate new row to previous x
end
end
Two additional suggestions:
Don't use i and j as variable names, because that shadows the imaginary unit.
Preallocate x instead of having it grow in each iteration, to increase speed.
The modified code is:
A = [1 2 3 ; 4 5 6 ; 7 8 9];
x = NaN(numel(A),3); % preallocate
n = 0;
for ii = 1:size(A)
for jj = 1:size(A)
n = n + 1; % update row counter
x(n,:) = [ii jj A(ii,jj)]; % fill row n
end
end
I developed a solution that works much faster. Here is the code:
% Generate subscripts from linear index
[i, j] = ind2sub(size(A),1:numel(A));
% Just concatenate subscripts and values
x = [i' j' A(:)];
Try it out and let me know ;)
This problem is a succession of my previous problem:
1) Extract submatrices, 2) vectorize and then 3) put back
Now, I have two patients, named Ann and Ben.
Indeed the matrices A and B are data for Ann and the matrix C is data for Ben:
Now, I need to design a matrix M such that y = M*x where
y = [a11, a21, a12, a22, b11, b21, b12, b22]' which is a vector, resulting from concatenation of the top-left sub-matrices, Ann and Ben;
x = [2, 5, 4, 6, 7, 9, 6, 2, 9, 3, 4, 2]' which is a vector, resulting from concatenation of sub-matrices A, B and C.
Here, the M is a 8 by 12 matrix that
a11 = 2 + 7, a21 = 5 + 9, .., a22 = 6 + 2 and b11 = 9, ..b22 = 2.
I design the M manually by:
M=zeros(8,12)
M(1,1)=1; M(1,5)=1; % compute a11
M(2,2)=1; M(2,6)=1; % compute a21
M(3,3)=1; M(3,7)=1; % compute a12
M(4,4)=1; M(4,8)=1; % compute a22
M(5,9)=1; % for setting b11 = 9, C(1,1)
M(6,10)=1; % for setting b21 = 3, C(2,1)
M(7,11)=1; % for setting b12 = 4, C(1,2)
M(8,12)=1 % for setting b22 = 2, C(2,2)
Obviously, in general for M(i,j), i means the 8 linear-index position of vector y and j means linear-index position of vector x.
However, I largely simplified my original problem that I want to construct this M automatically.
Thanks in advance for giving me a hand.
Here you have my solution. I have essentially build the matrix M automatically (from the proper indexes) as you suggested.
A = [2 4 8;
5 6 3;
10 3 6];
B = [7 6 3;
9 2 9;
10 2 3];
C = [9 4 7;
3 2 5;
10 3 4];
% All matrices in the same array
concat = cat(3, A, B, C);
concat_sub = concat(1:2,1:2,:);
x = concat_sub(:);
n = numel(x)/3; %Number of elements in each subset
M2 = zeros(12,8); %Transpose of the M matrix (it could be implemented directly over M but that was my first approach)
% The indexes you need
idx1 = 1:13:12*n; % Indeces for A
idx2 = 5:13:12*2*n; % Indices for B and C
M2([idx1 idx2]) = 1;
M = M2';
y = M*x
I have taken advantage of the shape that the matrix M shold take:
You can index into things and extract what you want without multiplication. For your example:
A = [2 4 8; 5 6 3; 10 3 6];
B = [7 6 3; 9 2 9; 10 2 3];
C = [9 4 7;3 2 5; 10 3 4];
idx = logical([1 1 0;1 1 0; 0 0 0]);
Ai = A(idx);
Bi = B(idx);
Ci = C(idx);
output = [Ai; Bi; Ci];
y = [Ai + Bi; Ci]; % desired y vector
This shows each step individually but they can be done in 2 lines. Define the index and then apply it.
idx = logical([1 1 0;1 1 0;0 0 0]);
output = [A(idx); B(idx); C(idx)];
y = [Ai + Bi; Ci]; % desired y vector
Also you can use linear indexing with idx = [1 2 4 5]' This will produce the same subvector for each of A B C. Either way works.
idx = [1 2 4 5]';
or alternatively
idx = [1;2;4;5];
output = [A(idx); B(idx); C(idx)];
y = [Ai + Bi; Ci]; % desired y vector
Either way works. Check out doc sub2ind for some examples of indexing from MathWorks.
I'm currently bringing some GAUSS code over to Matlab and I'm stuck trying to use the GAUSS "rotater" function.
The command reference entry for rotater says:
Purpose Rotates the rows of a matrix
Format y = rotater(x,r)
Input x: N x K matrix to be rotated. r: N x 1 or 1 x 1 matrix specifying the amount of rotation.
Output y: N x K rotated matrix.
Remarks The rotation is performed horizontally within each row of the matrix. A positive rotation value will cause the elements to move
to the right. A negative rotation will cause the elements to move to
the left. In either case, the elements that are pushed off the end of
the row will wrap around to the opposite end of the same row. If the rotation value is greater than or equal to the number of columns in x, then the rotation value will be calculated using (r % cols(x)).
Example 1
(I'm following Matlab's notation here, with straight brackets for matrices and a semicolon for a new ro)
If x = [1 2 3; 4 5 6], and r = [1; -1],then y = [3 1 2; 5 6 4]
Example 1
If x = [1 2 3; 4 5 6; 7 8 9; 10, 11, 12], and r = [0; 1; 2; 3], then y = [1 2 3; 6 4 5; 8 9 7; 10 11 12]
Maybe someone has found a function like that somewhere or can give me advice how to write it?
This can be done using bsxfun twice:
Compute rotated row indices by subtracting r with bsxfun and using mod. As usual, mod needs indices starting at 0, not 1. The rotated row indices are left as 0-based, because that's more convenient for step 2.
Get a linear index from columns and rotated rows, again using bsxfun. This linear index applied to x gives y:
Code:
[s1 s2] = size(x);
rows = mod(bsxfun(#plus, 1:s2, -r(:))-1, s2); % // step 1
y = x(bsxfun(#plus, rows*s1, (1:s1).')); %'// step 2
circshift is pretty close to what you're looking for except that 1) it works on columns rather than rows, and 2) it shifts the entire matrix by the same offset.
The first one is easy to fix, we just transpose. For the second one I haven't been able to find a vectorized approach, but in the meantime, here's a version with a for loop:
x = [1 2 3; 4 5 6; 7 8 9; 10 11 12]
r = [0 1 2 3]
B = x'
C = zeros(size(B));
for ii = 1:size(B,2)
C(:,ii) = circshift(B(:,ii),r(ii));
end
y = C'
The output is:
x =
1 2 3
4 5 6
7 8 9
10 11 12
r =
0 1 2 3
B =
1 4 7 10
2 5 8 11
3 6 9 12
y =
1 2 3
6 4 5
8 9 7
10 11 12
This can be done using a simple for loop to iterate over each row, and a function called 'circshift' from matlab.
I created a function the goes through each row and applies the appropriate shift to it. There may be more efficient ways to implement this, but this way works with your examples. I created a function
function rotated_arr = GaussRotate(input_array, rotation_vector)
[N,K] = size(input_array)
%creates array for return values
rotated_arr = zeros(N,K);
%if the rotation vector is a scalar
if (length(rotation_vector) == 1)
%replicate the value once for each row
rotation_vector = repmat(rotation_vector, [1,N]);
end
%if the rotation vector doesn't have as many entries as there are rows
%in the input array
if (length(rotation_vector) ~= N)
disp('ERROR GaussRotate: rotation_vector is the wrong size')
disp('if input_Array is NxK, rotation_vector must be Nx1 or 1x1')
return
end
%for each row
for idx=1:size(input_array,1)
%shift the row by the appropriate number of columns
%we use [0,shift] because we want to shift the columns, the row
%stays where it is (even though this is a 1xN at this point we
%still specify rows vs columns)
rotated_arr(idx,:) = circshift(input_array(idx,:),[0,rotation_vector(idx)]);
end
end
then simply called it with your examples
x = [1 2 3; 4 5 6];
r = [1; -1];
y = GaussRotate(x,r)
%produces [3 1 2; 5 6 4]
%I also made it support the 1x1 case
r = [-1]
%this will shift all elements one column to the left
y = GaussRotate(x,r)
%produces [2 3 1; 5 6 4]
x = [1 2 3; 4 5 6; 7 8 9; 10, 11, 12]
r = [0; 1; 2; 3]
y = GaussRotate(x,r)
%produces [1 2 3; 6 4 5; 8 9 7; 10 11 12]
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:
ยป M
M =
17 35 26 96
12 59 51 55
50 23 70 14
96 76 90 15
ยป 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.