How to create a symmetric matrix where each row / column is a subset of a known vector
I have a 7 * 1 vector a = (1:7).'
. I want to form A
a 4 * 4 matrix from a vector A
so that the elements A
form the antidiagonal of the matrix A
like this:
A = [1 2 3 4;
2 3 4 5;
3 4 5 6;
4 5 6 7]
I would like this to work for general A
, not just when the elements are whole integers.
I appreciate any help.
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You can use hankel
:
n= 4;
A= hankel(a(1:n),a(n:2*n-1))
Another solution (extension / bsxfun):
In MATLAB r2016b / Octave It can be created as:
A = a((1:4)+(0:3).')
In pre r2016b, you can use bsxfun
:
A = a(bsxfun(@plus,1:4, (0:3).'))
I / O example
a = [4 6 2 7 3 5 1]
A =
4 6 2 7
6 2 7 3
2 7 3 5
7 3 5 1
Using the test provided by @Wolfie tested in Octave:
_____________________________________
|Method |memory peak(MB)|timing(Sec)|
|=========|===============|===========|
|bsxfun |2030 |1.50 |
|meshgrid |3556 |2.43 |
|repmat |2411 |2.64 |
|hankel |886 |0.43 |
|for loop |886 |0.82 |
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Setting up indexing
Adding two outputs meshgrid
can give indices:
[x, y] = meshgrid(1:4, 0:3);
x + y;
% ans = [1 2 3 4
% 2 3 4 5
% 3 4 5 6
% 4 5 6 7];
If it a
was the same as your example, you could stop there. Alternatively, use this to index a common vector a
. For comparison, I'll use the same input example as rahnema1 for my method:
a = [4 6 2 7 3 5 1]; [x, y] = meshgrid(1:4, 0:3); A = a(x + y); % A = [4 6 2 7 % 6 2 7 3 % 2 7 3 5 % 7 3 5 1]
There are many ways to create indexes instead of using them meshgrid
, see the benchmarking functions below for some examples!
Benchmarking and seven different methods.
Below are some points to run various methods, including using cumsum
, repmat
, hankel
and a simple cycle for
. This test was done in Matlab 2015b, so it takes advantage of Matlab optimizations, etc., which Octave tests in rahnema1's answer might not do. I also use a function timeit
that is more reliable than tic
/ toc
because it does multiple probes, etc.
function benchie()
n = 10000; % (large) square matrix size
a = 1:2*n-1; % array of correct size, could be anything this long
f1 = @() m1(a,n); disp(['bsxfun: ', num2str(timeit(f1))]);
f2 = @() m2(a,n); disp(['cumsum: ', num2str(timeit(f2))]);
f3 = @() m3(a,n); disp(['meshgrid: ', num2str(timeit(f3))]);
f4 = @() m4(a,n); disp(['repmat: ', num2str(timeit(f4))]);
f5 = @() m5(a,n); disp(['for loop: ', num2str(timeit(f5))]);
f6 = @() m6(a,n); disp(['hankel1: ', num2str(timeit(f6))]);
f7 = @() m7(a,n); disp(['hankel2: ', num2str(timeit(f7))]);
end
% Use bsxfun to do broadcasting of addition
function m1(a,n); A = a(bsxfun(@plus, (1:n), (0:n-1).')); end
% Use cumsum to do cumulative vertical addition to create indices
function m2(a,n); A = a(cumsum([(1:n); ones(n-1,n)])); end
% Add the two meshgrid outputs to get indices
function m3(a,n); [x, y] = meshgrid(1:n, 0:n-1); A = a(x + y); end
% Use repmat twice to replicate the meshgrid results, for equivalent one liner
function m4(a,n); A = a(repmat((1:n)',1,n) + repmat(0:n-1,n,1)); end
% Use a simple for loop. Initialise A and assign values to each row in turn
function m5(a,n); A = zeros(n); for ii = 1:n; A(:,ii) = a(ii:ii+n-1); end; end
% Create a Hankel matrix (constant along anti-diagonals) for indexing
function m6(a,n); A = a(hankel(1:n,n:2*n-1)); end
% Create a Hankel matrix directly from elements
function m7(a,n); A = hankel(a(1:n),a(n:2*n-1)); end
Output:
bsxfun: 1.4397 sec
cumsum: 2.0563 sec
meshgrid: 2.0169 sec
repmat: 1.8598 sec
for loop: 0.4953 sec % MUCH quicker!
hankel1: 2.6154 sec
hankel2: 1.4235 sec
So your best bet is to use rahnema1's suggestion bsxfun
or direct matrix creation hankel
if you want a single liner, here is a brilliant StackOverflow answer that explains some of the benefits bsxfun
: In Matlab, when is it optimal to use bsxfun?
However, the for loop is more than twice as fast! Conclusion: Matlab has many neat ways to achieve this kind of thing, sometimes a simple loop with some appropriate pre-placement and internal Matlab Optimization might be the fastest.
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