I need some help in converting a 2X2 matrix to a 4X4 matrix in the following manner:
A = [2 6;
8 4]
should become:
B = [2 2 6 6;
2 2 6 6;
8 8 4 4;
8 8 4 4]
How would I do this?
In newer versions of MATLAB (R2015a and later) the easiest way to do this is using the repelem function:
B = repelem(A, 2, 2);
For older versions, a short alternative to the other (largely) indexing-based solutions is to use the functions kron and ones:
>> A = [2 6; 8 4];
>> B = kron(A, ones(2))
B =
2 2 6 6
2 2 6 6
8 8 4 4
8 8 4 4
Can be done even easier than Jason's solution:
B = A([1 1 2 2], :); % replicate the rows
B = B(:, [1 1 2 2]); % replicate the columns
Here's one more solution:
A = [2 6; 8 4];
B = A( ceil( 0.5:0.5:end ), ceil( 0.5:0.5:end ) );
which uses indexing to do everything and doesn't rely on the size or shape of A.
This works:
A = [2 6; 8 4];
[X,Y] = meshgrid(1:2);
[XI,YI] = meshgrid(0.5:0.5:2);
B = interp2(X,Y,A,XI,YI,'nearest');
This is just two-dimensional nearest-neighbor interpolation of A(x,y) from x,y ∈ {1,2} to x,y ∈ {0.5, 1, 1.5, 2}.
Edit: Springboarding off of Jason S and Martijn's solutions, I think this is probably the shortest and clearest solution:
A = [2 6; 8 4];
B = A([1 1 2 2], [1 1 2 2]);
A = [2 6; 8 4];
% arbitrary 2x2 input matrix
B = repmat(A,2,2);
% replicates rows & columns but not in the way you want
B = B([1 3 2 4], :);
% swaps rows 2 and 3
B = B(:, [1 3 2 4]);
% swaps columns 2 and 3, and you're done!
Here's a method based on simple indexing that works for an arbitrary matrix. We want each element to be expanded to an MxN submatrix:
A(repmat(1:end,[M 1]),repmat(1:end,[N 1]))
Example:
>> A=reshape(1:6,[2,3])
A =
1 3 5
2 4 6
>> A(repmat(1:end,[3 1]),repmat(1:end,[4 1]))
ans =
1 1 1 1 3 3 3 3 5 5 5 5
1 1 1 1 3 3 3 3 5 5 5 5
1 1 1 1 3 3 3 3 5 5 5 5
2 2 2 2 4 4 4 4 6 6 6 6
2 2 2 2 4 4 4 4 6 6 6 6
2 2 2 2 4 4 4 4 6 6 6 6
To see how the method works, let's take a closer look at the indexing. We start with a simple row vector of consecutive numbers
>> m=3; 1:m
ans =
1 2 3
Next, we extend it to a matrix, by repeating it M times in the first dimension
>> M=4; I=repmat(1:m,[M 1])
I =
1 2 3
1 2 3
1 2 3
1 2 3
If we use a matrix to index an array, then the matrix elements are used consecutively in the standard Matlab order:
>> I(:)
ans =
1
1
1
1
2
2
2
2
3
3
3
3
Finally, when indexing an array, the 'end' keyword evaluates to the size of the array in the corresponding dimension. As a result, in the example the following are equivalent:
>> A(repmat(1:end,[3 1]),repmat(1:end,[4 1]))
>> A(repmat(1:2,[3 1]),repmat(1:3,[4 1]))
>> A(repmat([1 2],[3 1]),repmat([1 2 3],[4 1]))
>> A([1 2;1 2;1 2],[1 2 3;1 2 3;1 2 3;1 2 3])
>> A([1 1 1 2 2 2],[1 1 1 1 2 2 2 2 3 3 3 3])
There is a Reshape() function that allows you to do this...
For example:
reshape(array, [64, 16])
And you can find a great video tutorial here
Cheers
Related
If I have this vector:
x = [1 1 1 1 1 2 2 2 3 4 4 6 6 6 6]
I would like to get the position of each unique number according to itself.
y = [1 2 3 4 5 1 2 3 1 1 2 1 2 3 4]
At the moment I'm using:
y = sum(triu(x==x.')) % MATLAB 2016b and above
It's compact but obviously not memory efficient.
For the pure beauty of MATLAB programming I would avoid using a loop. Do you have a better simple implementation ?
Context:
My final goal is to sort the vector x but with the constraint that a number that appear N times has the priority over another number that has appeared more than N times:
[~,ind] = sort(y);
x_relative_sort = x(ind);
% x_relative_sort = 1 2 3 4 6 1 2 4 6 1 2 6 1 6 1
Assuming x is sorted, here's one vectorized alternative using unique, diff, and cumsum:
[~, index] = unique(x);
y = ones(size(x));
y(index(2:end)) = y(index(2:end))-diff(index).';
y = cumsum(y);
And now you can apply your final sorting:
>> [~, ind] = sort(y);
>> x_relative_sort = x(ind)
x_relative_sort =
1 2 3 4 6 1 2 4 6 1 2 6 1 6 1
If you have positive integers you can use sparse matrix:
[y ,~] = find(sort(sparse(1:numel(x), x, true), 1, 'descend'));
Likewise x_relative_sort can directly be computed:
[x_relative_sort ,~] = find(sort(sparse(x ,1:numel(x),true), 2, 'descend'));
Just for variety, here's a solution based on accumarray. It works for x sorted and containing positive integers, as in the question:
y = cell2mat(accumarray(x(:), x(:), [], #(t){1:numel(t)}).');
You can be more memory efficient by only comparing to unique(x), so you don't have a large N*N matrix but rather N*M, where N=numel(x), M=numel(unique(x)).
I've used an anonymous function syntax to avoid declaring an intermediate matrix variable, needed as it's used twice - this can probably be improved.
f = #(X) sum(cumsum(X,2).*X); y = f(unique(x).'==x);
Here's my solution that doesn't require sorting:
x = [1 1 1 1 1 2 2 2 3 4 4 6 6 6 6 1 1 1];
y = cell2mat( splitapply(#(v){cumsum(v)},x,cumsum(logical([1 diff(x)]))) ) ./ x;
Explanation:
% Turn each group new into a unique number:
t1 = cumsum(logical([1 diff(x)]));
% x = [1 1 1 1 1 2 2 2 3 4 4 6 6 6 6 1 1 1];
% t1 = [1 1 1 1 1 2 2 2 3 4 4 5 5 5 5 6 6 6];
% Apply cumsum separately to each group:
t2 = cell2mat( splitapply(#(v){cumsum(v)},x,t1) );
% t1 = [1 1 1 1 1 2 2 2 3 4 4 5 5 5 5 6 6 6];
% t2 = [1 2 3 4 5 2 4 6 3 4 8 6 12 18 24 1 2 3];
% Finally, divide by x to get the increasing values:
y = t2 ./ x;
% x = [1 1 1 1 1 2 2 2 3 4 4 6 6 6 6 1 1 1];
% t2 = [1 2 3 4 5 2 4 6 3 4 8 6 12 18 24 1 2 3];
I have a matrix as follows:
A= 1 2
3 4
5 6
7 8
I want to arrange the elements of this matrix in such a way that it will give me the following output:
B= 1
2
3
4
5
6
7
8
Any kind of suggestion will be helpful. Thanks!
Take the transpose of A and unroll it into a vector:
B = A.';
B = B(:);
Alternatively, you can use reshape:
B = reshape(A.', [], 1);
The reason why you transpose the matrix A first is because MATLAB does the unrolling in column-major format, which means that the columns are traversed first. You are trying to do this row-wise, and so you'd need to transpose the input to achieve the same effect.
Here's what the output looks like (using both):
>> A= [1 2
3 4
5 6
7 8];
>> B = A.';
>> B = B(:);
>> B
B =
1
2
3
4
5
6
7
8
Also:
>> A= [1 2
3 4
5 6
7 8];
>> B = reshape(A.', [], 1)
B =
1
2
3
4
5
6
7
8
I have a matrix, L, with two columns. I want to find its sub-matrices have equal values on their 2nd column. I want to do that using MATLAB without any for loop.
example:
L=[1 2;3 2;4 6;5 3;7 3;1 3;2 7;9 7]
then the sub-matrices are:
[1 2;3 2] , [4 6] , [5 3;7 3;1 3] and [2 7;9 7]
You can use a combination of arrayfun + unique to get that -
[~,~,labels] = unique(L(:,2),'stable')
idx = arrayfun(#(x) L(labels==x,:),1:max(labels),'Uniform',0)
Display output -
>> celldisp(idx)
idx{1} =
1 2
3 2
idx{2} =
4 6
idx{3} =
5 3
7 3
1 3
idx{4} =
2 7
9 7
You can use accumarray directly or with a sorted array, depending on you want the order of the rows to be stable, or the order of the submatricxes to be stable.
Say you want the rows to be stable:
>> [L2s,inds] = sort(L(:,2));
>> M = accumarray(L2s,inds,[],#(v){L(v,:)});
>> M(cellfun(#isempty,M)) = []; % remove empty cells
>> celldisp(M)
M{1} =
1 2
3 2
M{2} =
5 3
7 3
1 3
M{3} =
4 6
M{4} =
2 7
9 7
This question already has answers here:
Got confused with a vector indexed by a matrix, in Matlab
(2 answers)
Closed 8 years ago.
Suppose:
a =
1 2 3
4 5 6
2 3 4
and
b =
1 3 2
6 4 8
In MATLABa(b) gives:
>> a(b)
ans =
1 2 4
3 2 6
What is the reason for this output?
when you have a matrix a:
a =
1 2 3
4 5 6
7 8 9
and b:
b =
1 3 4
3 2 6
then a(b) is a way of adressing items in a and gives you:
>> a(b)
ans =
1 7 2
7 4 8
to understand this you have to think of a als a single column vector
>> a(:)
ans =
1
4
7
2
5
8
3
6
9
now the first row of b (1 3 4) addresses elements in this vector so the first, the 3rd and the forth element of that single column vector which are 1 7 and 2 are adressed. Next the secound row of b is used as adresses for a secound line in the output so the 3rd, the 2nd and the 6th elements are taken from a, those are 7 4 and 8.
It's just a kind of matrix indexing.
Matrix indexes numeration in 'a' matrix is:
1 4 7
2 5 8
3 6 9
This is a possible duplicate to this post where I gave an answer: Got confused with a vector indexed by a matrix, in Matlab
However, I would like to duplicate my answer here as I think it is informative.
That's a very standard MATLAB operation that you're doing. When you have a vector or a matrix, you can provide another vector or matrix in order to access specific values. Accessing values in MATLAB is not just limited to single indices (i.e. A(1), A(2) and so on).
For example, let's say we had a vector a = [1 2 3 4]. Let's also say we had b as a matrix such that it was b = [1 2 3; 1 2 3; 1 2 3]. By doing a(b) to access the vector, what you are essentially doing is a lookup. The output is basically the same size as b, and you are creating a matrix where there are 3 rows, and each element accesses the first, second and third element. Not only can you do this for a vector, but you can do this for a matrix as well.
Bear in mind that when you're doing this for a matrix, you access the elements in column major format. For example, supposing we had this matrix:
A = [1 2
3 4
5 6
7 8]
A(1) would be 1, A(2) would be 3, A(3) would be 5 and so on. You would start with the first column, and increasing indices will traverse down the first column. Once you hit the 5th index, it skips over to the next column. So A(5) would be 2, A(6) would be 4 and so on.
Here are some examples to further your understanding. Let's define a matrix A such that:
A = [5 1 3
7 8 0
4 6 2]
Here is some MATLAB code to strengthen your understanding for this kind of indexing:
A = [5 1 3; 7 8 0; 4 6 2]; % 3 x 3 matrix
B = [1 2 3 4];
C = A(B); % C should give [5 7 4 1]
D = [5 6 7; 1 2 3; 4 5 6];
E = A(D); % E should give [8 6 3; 5 7 4; 1 8 6]
F = [9 8; 7 6; 1 2];
G = A(F); % G should give [2 0; 3 6; 5 7]
As such, the output when you access elements this way is whatever the size of the vector or matrix that you specify as the argument.
In order to be complete, let's do this for a vector:
V = [-1 9 7 3 0 5]; % A 6 x 1 vector
B = [1 2 3 4];
C = V(B); % C should give [-1 9 7 3]
D = [1 3 5 2];
E = V(D); % E should give [-1 7 0 9]
F = [1 2; 4 5; 6 3];
G = V(F); % G should give [-1 9; 3 0; 5 7]
NB: You have to make sure that you are not providing indexes that would make the accessing out of bounds. For example if you tried to specify the index of 5 in your example, it would give you an error. Also, if you tried anything bigger than 9 in my example, it would also give you an error. There are 9 elements in that 3 x 3 matrix, so specifying a column major index of anything bigger than 9 will give you an out of bounds error.
I don't know how to do it. I have some matrices with repeated values and I'd like to sort them and get a matrix of sorted indices. For example:
a = [1 4 3 10 8 2];
b = sort(a);
% This doesn't work but I wish it did - that's what I'm looking for.
% idx = find(a==b); idx = [1 6 3 2 5 4];
[v idx] = ismember(b,a);
However, when there are repeated values and NaN this produces errors. Try this:
a = [1 NaN 4 2 10 8 2];
b=sort(a);
[v, i] = ismember(b,a);
Gives [1 7 7 3 6 5 0] which is valid but I need it to be [ 1 4 7 3 6 5 0].
I can deal with this later but it's going to be much more elegant if it were to return the above result.
sort has a two output value version that gives you exactly the indices you're looking for:
a = [1 4 3 10 8 2];
[b, idx] = sort(a);
Outputs
idx =
1 6 3 2 5 4
This works correctly with NaNs as well:
a = [1 NaN 4 2 10 8 2];
[b, idx] =sort(a);
>> idx
idx =
1 4 7 3 6 5 2
>> b
b =
1 2 2 4 8 10 NaN