I have genearalised the outer product between a 3vector and itself to take input from a collection (NxNx3 matrix) of 3-vectors.
At the moment, my function does what I want (see example output and input below), and it looks like this. I would like to make it faster by avoiding the two for loops which I currently use.
function [rr]=OuterVec(r)
N = size(r,1);
rr = zeros(N,N,3,3);
for i=1:N
for j=1:N
rr(i,j,:,:)=kron(reshape(r(i,j,:),[1,3]),permute(r(i,j,:),[3 2 1]));
end
end
end
I/O Examples
c = ones(2,2);
V(1,1,:)=[1 2 3];
u = c.*V;
OuterVec(u)
ans(:,:,1,1) =
1 1
1 1
ans(:,:,2,1) =
2 2
2 2
ans(:,:,3,1) =
3 3
3 3
ans(:,:,1,2) =
2 2
2 2
ans(:,:,2,2) =
4 4
4 4
ans(:,:,3,2) =
6 6
6 6
ans(:,:,1,3) =
3 3
3 3
ans(:,:,2,3) =
6 6
6 6
ans(:,:,3,3) =
9 9
9 9
You only need to permute dimensions and apply element-wise product with singleton expansion:
rr = bsxfun(#times, r, permute(r, [1 2 4 3]));
Or, in Matlab R2016b onwards,
rr = r .* permute(r, [1 2 4 3]);
Related
How to vectorize the following code in Matlab?
m = meshgrid(1:n);
for i = 1:n
for j = 1:n
m(i,j) = max(i,j);
end
end
Another way to think of the question would be: how to implement the Mathematica command:
Table[Max[i,j],{i,1,n},{j,1,n}]:
in Matlab.
With implicit expansion,
m = max([1:n].',[1:n]);
For n = 5:
m =
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
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];
How to effectively vectorize the following MATLAB code, which performs permutation of each row of matrix R by indices in corresponding row of matrix P?
for i = 1:size(P,1)
pP(i,:) = R(i,P(i,:));
end
example:
P = [3 2 1;
3 1 2;
2 3 1;
2 1 3;
1 2 3;
1 3 2]
R = [6 5 4;
6 4 5;
5 6 4;
5 4 6;
4 5 6;
4 6 5]
produce following matrix pR:
4 5 6
5 6 4
6 4 5
4 5 6
4 5 6
4 5 6
One approach with bsxfun -
nrows = size(R,1)
pP = R(bsxfun(#plus,[1:nrows]',(P-1)*nrows))
Or with ndgrid -
[m,n] = size(R)
pP = R(sub2ind([m n],ndgrid(1:m,1:n),P))
Or replace ndgrid(1:m,1:n) with repmat: repmat([1:m]',[1 n]) or with meshgrid:meshgrid(1:m,1:n).'.
This might not be the best way to do it, but you could do something like:
IND1 = P(:,1)
Q(:,1) = diag(R(:,IND));
and repeat for P(:,2), P(:,3) in a similar fashion.
You can use arrayfun to avoid the loop but probably won't gain in performance if that it is the reason for vectorizing it:
cell2mat(arrayfun(#(k) R(k, P(k,:)), (1:size(P,1)).', 'uni', 0))
In Matlab, I'm trying to transform a matrix A to another matrix B such that B's columns are made up of square submatrices of A. For example, if A is:
A = [1 1 2 2
1 1 2 2
3 3 4 4
3 3 4 4];
I'd like B to be:
B = [1 2 3 4
1 2 3 4
1 2 3 4
1 2 3 4]
A could be, say 16-by-16, and constructing B from 4-by-4 squares would result in B being 4-by-64.
Is there an efficient way to do this using reshape in combination with some other commands? Or some other approach? I am currently iterating in a loop, which is very slow with a large number of large source matrices.
Assume your matrix is a bit more general, and made of 3x2 blocks:
A = [1 1 2 2
1 1 2 2
1 1 2 2
3 3 4 4
3 3 4 4
3 3 4 4
5 5 6 6
5 5 6 6
5 5 6 6];
b = [3 2];
szA = size(A);
Transpose, reshape, permute, reshape.
nb = prod(szA./b); % Number of blocks
nelb = prod(b); % Number of elements per block
out1 = reshape(permute(reshape(A',szA(2),b(1),szA(1)/b(1)),[2,1,3]),nelb,nb)
Alternatively, slower and memory intensive but more readable:
d1 = repmat(b(1),1,szA(1)/b(1));
d2 = repmat(b(2),1,szA(2)/b(2));
out = reshape(mat2cell(A,d1,d2)',1,nelb);
out = reshape([out{:}],nelb,nb)
Now, if the blocks are square, simply set b = [2,2] or b = [3,3], etc..., or simplify the general formulation removing indexing of b and prod.
I have two vectors:
a = [1 3 5 7 9];
b = [2 4 6 8 10];
That I need to combine together element wise. Meaning that I need the first element of vector a, then the first element of vector b, second of a, second of b, and so forth until I get the following:
combined = [1 2 3 4 5 6 7 8 9 10]
How do I do this within MatLab?
Edit
I ran a test of the top three answers (Josh, Marc, & Kronos) and compared the time it took to run them. I ran each 100 times after doing a 10 iteration warmup. The vectors created were exactly the same size in length (16e+6) and were random values ranging from 1 to 100:
Test Results
Test: Total Time (100 runs): Avg Time Per Exec:
Josh B 21.3687 0.2137
Marc C 21.4273 0.2143
Kronos 31.1897 0.3119
It appears that both Josh's and Marc's solutions are similar in execution time.
a = [1 3 5 7 9];
b = [2 4 6 8 10];
temp = [a; b];
combined = temp(:)';
This can be done by the following:
a = [1 3 5 7 9];
b = [2 4 6 8 10];
combinedSize = size(a, 2) * 2;
combined(1:2:combinedSize) = a;
combined(2:2:combinedSize) = b;
This is obviously assuming that your vectors are exactly the same size. If by chance you want to merge two vectors that are not the same size then you can do the following:
combinedSize = max(size(a, 2), size(b, 2)) * 2;
combined = NaN(1,combinedSize);
combined(1:2:size(a,2)*2) = a;
combined(2:2:size(b,2)*2) = b;
This will place a NaN for the remaining elements of the smaller vector. For example, given the following sample vectors:
a = [1 3 5 7 9 11];
b = [2 4 6 8];
will result in the combined vector:
combined =
1 2 3 4 5 6 7 8 9 NaN 11 NaN
Place the vectors below eachother in a matrix and use reshape. For example:
>> A=[1 2 3]
A =
1 2 3
>> B=[4 5 6]
B =
4 5 6
>> C=reshape([A;B],1,size(A,2)+size(B,2))
C =
1 4 2 5 3 6
It's straightforward to generalize to more than 2 vectors.
You can also give a try to looping, for example:
a=[1 2 3 4 5];
b=[11 12 13 14 15];
for i = 1:N
{
if (i%2==0)
{ c[i] = b[i]; }
else
{ c[i] = a[i]; }
This shall work!
All the answers above only work if the two vectors have the same number of elements. The following will work even if they have different number of elements:
>>
A = [1 3 5];
B = [2 4 6 7 8];
C = [1 3 5 7 8];
D = [2 4 6];
AB = nan(1,2*max(numel(A),numel(B)));
CD = nan(1,2*max(numel(C),numel(D)));
AB(2*(1:length(A))) = A;
AB(1+2*(1:length(B))) = B;
CD(2*(1:length(C))) = C;
CD(1+2*(1:length(D))) = D;
>>
AB = AB(~isnan(AB))
CD = CD(~isnan(CD))
The result would be:
AB =
1 2 3 4 5 6 7 8
CD =
1 2 3 4 5 6 7 8