Very Simple Question that i couldn't find easily on the web so i thought i would ask here:
You can make a 1D linear array like this:
1:10 = 1 2 3 4 5 6 7 8 9 10
1:2:10 = 1 3 5 7 9
How can you easily initialise a 2D array ie.
1 2 3 4
1 2 3 4
1 2 3 4
1 2 3 4
and also the same thing but for columns:
1 1 1 1
2 2 2 2
3 3 3 3
4 4 4 4
Should be a command to do it in one line.
v = 1:5;
A = repmat(v, 4, 1);
B = repmat(v', 1, 4);
A and B will have what you need.
Another option is to use MATLAB indexing as follows:
v = 1:5;
A = v(ones(4, 1), :);
v = [1:5]';
B = v(:, ones(1, 4));
Alternatively i have learned that you can use meshgrid:
meshgrid(1:4, 1:4) =>
1 2 3 4
1 2 3 4
1 2 3 4
1 2 3 4
Thanks for the help and upvotes.
Related
Suppose I have a list of length 2k, say {1,2,...,2k}. The number of possible ways of grouping the 2k numbers into k (unordered) pairs is n(k) = 1*3* ... *(2k-1). So for k=2, we have the following three different ways of forming 2 pairs
(1 2)(3 4)
(1 3)(2 4)
(1 4)(2 3)
How can I use Matlab to create the above list, i.e., create a matrix of n(k)*(2k) such that each row contains a different way of grouping the list of 2k numbers into k pairs.
clear
k = 3;
set = 1: 2*k;
p = perms(set); % get all possible permutations
% sort each two column
[~, col] = size(p);
for i = 1: 2: col
p(:, i:i+1) = sort(p(:,i:i+1), 2);
end
p = unique(p, 'rows'); % remove the same row
% sort each row
[row, col] = size(p);
for i = 1: row
temp = reshape(p(i,:), 2, col/2)';
temp = sortrows(temp, 1);
p(i,:) = reshape(temp', 1, col);
end
pairs = unique(p, 'rows'); % remove the same row
pairs =
1 2 3 4 5 6
1 2 3 5 4 6
1 2 3 6 4 5
1 3 2 4 5 6
1 3 2 5 4 6
1 3 2 6 4 5
1 4 2 3 5 6
1 4 2 5 3 6
1 4 2 6 3 5
1 5 2 3 4 6
1 5 2 4 3 6
1 5 2 6 3 4
1 6 2 3 4 5
1 6 2 4 3 5
1 6 2 5 3 4
As someone think my former answer is not useful, i post this.
I have the following brute force way of enumerating the pairs. Not particularly efficient. It can also cause memory problem when k>9. In that case, I can just enumerate but not create Z and store the result in it.
function Z = pair2(k)
count = [2*k-1:-2:3];
tcount = prod(count);
Z = zeros(tcount,2*k);
x = [ones(1,k-2) 0];
z = zeros(1,2*k);
for i=1:tcount
for j=k-1:-1:1
if x(j)<count(j)
x(j) = x(j)+1;
break
end
x(j) = 1;
end
y = [1:2*k];
for j=1:k-1
z(2*j-1) = y(1);
z(2*j) = y(x(j)+1);
y([1 x(j)+1]) = [];
end
z(2*k-1:2*k) = y;
Z(i,:) = z;
end
k = 3;
set = 1: 2*k;
combos = combntns(set, k);
[len, ~] = size(combos);
pairs = [combos(1:len/2,:) flip(combos(len/2+1:end,:))];
pairs =
1 2 3 4 5 6
1 2 4 3 5 6
1 2 5 3 4 6
1 2 6 3 4 5
1 3 4 2 5 6
1 3 5 2 4 6
1 3 6 2 4 5
1 4 5 2 3 6
1 4 6 2 3 5
1 5 6 2 3 4
You can also use nchoosek instead of combntns. See more at combntns or nchoosek
Ok so I am clustering data into clusters which are then indexed using a column. The data is in the form of motion vectors and so my data will look like this after being clustered:
[index x y x' y']
for example:
[1 3 5 4 6;
1 4 6 5 7;
2 3 5 4 6;
2 8 9 9 3;
3 2 3 2 4]
in above array there are 3 clusters, with clusters 1 and 2 each containing 2 vectors.
My problem is that I sometimes have to delete clusters based on certain criteria, and may be left with:
[2 3 5 4 6;
2 8 9 9 3;
3 2 3 2 4]
I want to be able to correct the index after deletion, so that it starts at 1 and ends with the number of clusters. So in this case replace the 2s with 1s and 3s with 2s.
Im sure there must be a simple way using a for loop but Ive been trying for a while and can't get ti right?
Assuming your matrix is called data, try this:
>> data = [2 3 5 4 6;
2 8 9 9 3;
3 2 3 2 4]
data =
2 3 5 4 6
2 8 9 9 3
3 2 3 2 4
>> data(:,1) = cumsum(diff(data([1 1:end], 1)) ~= 0) + 1
data =
1 3 5 4 6
1 8 9 9 3
2 2 3 2 4
A simple call to unique will help you do that. You can use the third output of it to assign each unique and new ID using the first column of the new data matrix (index vector) to replace its first column. Also, make sure you use the 'stable' flag so that it assigns IDs in order of occurrence from top to bottom:
%// Data setup
A = [1 3 5 4 6;
1 4 6 5 7;
2 3 5 4 6;
2 8 9 9 3;
3 2 3 2 4];
%-----
B = A(3:end,:); %// Remove first two rows
%// Go through the other IDs and reassign to unique IDs from 1 up to whatever
%// is left
[~,~,id] = unique(B(:,1), 'stable');
%// Replace the first column of the new matrix with the new IDs
B(:,1) = id; %// Replace first column with new IDs
We get:
>> B
B =
1 3 5 4 6
1 8 9 9 3
2 2 3 2 4
Given any number. Lets say for example 5, I need to generate a matrix similar to this:
1 2 3 4 5
2 2 3 4 5
3 3 3 4 5
4 4 4 4 5
5 5 5 5 5
How to generate a matrix similar to this using Matlab?
I'd use bsxfun:
n = 5;
matrix = bsxfun(#max, 1:n, (1:n).');
An alternative (probably slower) is to use ndgrid:
n = 5;
[ii, jj] = ndgrid(1:n);
matrix = max(ii, jj);
Nothing will ever beat bsxfun as used by Luis Mendo., but for the sake of reminding people of the existence of Matlab's gallery function, here another approach:
n = 5;
A = gallery('minij',n)
B = n + 1 - A(end:-1:1,end:-1:1)
A =
1 1 1 1 1
1 2 2 2 2
1 2 3 3 3
1 2 3 4 4
1 2 3 4 5
B =
1 2 3 4 5
2 2 3 4 5
3 3 3 4 5
4 4 4 4 5
5 5 5 5 5
Given the matrix I = [1,2;3,4], I would like to duplicate the elements to create a matrix I2 such that:
I2 = [1 1 1 2 2 2
1 1 1 2 2 2
1 1 1 2 2 2
3 3 3 4 4 4
3 3 3 4 4 4
3 3 3 4 4 4]
Other than using repmat, what other methods or functions are available?
Use kron:
>> N = 3 %// Number of times to replicate a number in each dimension
>> I = [1,2;3,4];
>> kron(I, ones(N))
ans =
1 1 1 2 2 2
1 1 1 2 2 2
1 1 1 2 2 2
3 3 3 4 4 4
3 3 3 4 4 4
3 3 3 4 4 4
This probably deserves some explanation in case you're not aware of what kron does. kron stands for the Kronecker Tensor Product. kron between two matrices A of size m x n and B of size p x q creates an output matrix of size mp x nq such that:
Therefore, for each coefficient in A, we take this value, multiply it with every value in the matrix B and we position these matrices in the same order as we see in A. As such, if we let A = I, and B be the 3 x 3 matrix full of ones, you thus get the above result.
Using indexing:
I = [1, 2; 3, 4]; %// original matrix
n = 3; %// repetition factor
I2 = I(ceil(1/n:1/n:size(I,1)), ceil(1/n:1/n:size(I,2))); %// result
One-liner with bsxfun -
R = 3; %// Number of replications
I2 = reshape(bsxfun(#plus,permute(I,[3 1 4 2]),zeros(R,1,R)),R*size(I,1),[])
Sample run -
I =
3 2 5
9 8 9
I2 =
3 3 3 2 2 2 5 5 5
3 3 3 2 2 2 5 5 5
3 3 3 2 2 2 5 5 5
9 9 9 8 8 8 9 9 9
9 9 9 8 8 8 9 9 9
9 9 9 8 8 8 9 9 9
My matrix is this:
0 3 0
0 1 2
4 4 1
I use im2col on it like this:
im2col(A, [2 2], 'sliding')
which correctly yields this:
0 0 3 1
0 4 1 4
3 1 0 2
1 4 2 1
I call this matrix K. Now I use col2im to go back to my original matrix. From the Matlab documentation I use this:
col2im(K, [2 2], [5 5],'sliding')
But this doesn't gives me my original matrix A. Reason being [5 5] should be [4 4] to get a 3*3 matrix for starters. But when I do that I get
??? Error using ==> reshape
To RESHAPE the number of elements must not change.
Why is that? And how can I get my original matrix back?
Fromthe docs:
A = col2im(B,[m n],[mm nn],'sliding') rearranges the row vector B into
a matrix of size (mm-m+1)-by-(nn-n+1). B must be a vector of size
1-by-(mm-m+1)*(nn-n+1). B is usually the result of processing the
output of im2col(...,'sliding') using a column compression function
(such as sum).
So that says to me you should be trying something like:
col2im(sum(K), [2 2], [4 4],'sliding')
however that would require K to have 9 columns. I don't have the image processing toolbox handy to test this right now
Your col2im doesn't work because it uses reshape and for that the number of elements of the matrix you wish to reshape (K) and the new one, need to be the same. This is not the case anymore, as through your transformation of A with im2col you obviously changed that. A has 9 and K 16 elements.
So you basically need to get back to a 3*3 matrix again by getting rid of the redundand doubled elements (due to the overlapping 2*2 blocks used in im2col) in K.
For that you could just make a new matrix (C) with the elements that you need:
C = [K([1,3,11;2,4,12;6,8,16])]
As long as you first went from a 3*3 to a 4*4 matrix using the same order of blocks this should work.
Maybe you could tell us more about what you really want to achieve, because I don't see any reason for this in the first place. It may also be possible that you might be better off using other functions instead, but I can only see that if I know what the reasoning behind your question is.
clear
clc
img = double(imread('tire.tif'));
[r c] = size(img);
w = 8;
imgBlock = im2col(img,[w w],'sliding'); imgBlock = imgBlock(:);
[x y] = meshgrid(1:c,1:r);
xx = im2col(x,[w w], 'sliding'); xx = xx(:);
yy = im2col(y,[w w], 'sliding'); yy = yy(:);
img2 = accumarray([yy xx], imgBlock, [], #mean);
figure,imshow(img, []);
figure,imshow(img2,[]);
% random matrix as image
img = randi(10,4)
img =
6 2 2 7
5 8 7 8
1 4 3 5
4 6 7 1
% matrix size
[r c] = size(img)
% patch size
w = 2;
% image to patch
imgBlock = im2col(img,[w w],'sliding')
% image patchs matrix to a vector
imgBlock = imgBlock(:);
r =
4
c =
4
imgBlock =
6 5 1 2 8 4 2 7 3
5 1 4 8 4 6 7 3 7
2 8 4 2 7 3 7 8 5
8 4 6 7 3 7 8 5 1
% index matrix size equal image size
[x y] = meshgrid(1:c,1:r)
% index matric to patchs;to vector
xx = im2col(x,[w w], 'sliding'); xx = xx(:);
yy = im2col(y,[w w], 'sliding'); yy = yy(:);
x =
1 2 3 4
1 2 3 4
1 2 3 4
1 2 3 4
y =
1 1 1 1
2 2 2 2
3 3 3 3
4 4 4 4
% yy :row index xx: column index
% applies the function mean to each subset of elements in imgBlock that have identical subscripts in [yy xx].
img2 = accumarray([yy xx], imgBlock, [], #mean);
img
img2
img =
6 2 2 7
5 8 7 8
1 4 3 5
4 6 7 1
img2 =
6 2 2 7
5 8 7 8
1 4 3 5
4 6 7 1
% [col,row,value]
a = [xx,yy,imgBlock]
a =
1 1 6
1 2 5
2 1 2
2 2 8
1 2 5
1 3 1
2 2 8
2 3 4
1 3 1
1 4 4
2 3 4
2 4 6
2 1 2
2 2 8
3 1 2
3 2 7
2 2 8
2 3 4
3 2 7
3 3 3
2 3 4
2 4 6
3 3 3
3 4 7
3 1 2
3 2 7
4 1 7
4 2 8
3 2 7
3 3 3
4 2 8
4 3 5
3 3 3
3 4 7
4 3 5
4 4 1
% The number of times that img(2,2) occurs in the matrix img
a(xx == 2 & yy == 2,:)
ans =
2 2 8
2 2 8
2 2 8
2 2 8