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performancematlabmatrixdiagonal
Alex R.
Alex R.

Reputation: 1437

Obtaining opposite diagonal of a matrix in Matlab

Let A be an matrix of size [n,n]. If I want to extract its diagonal, I do diag(A).

Actually, I want the opposite diagonal, which would be [A(n,1),A(n-1,2),A(n-2,3),...].

One way to do this is via diag(flipud(A)). However, flipud(A) is quite wasteful and multiplies the time it takes by a factor of 10 compared to finding the usual diagonal.

I'm looking for a fast way of obtaining the opposite diagonal. Naturally, for loops seem abysmally slow. Suggestions would be greatly appreciated.

Upvotes: 4

Views: 12800

Answers (4)

Firdous Ul Nazir
Firdous Ul Nazir

Reputation: 1

A = magic(6)

A =

35     1     6    26    19    24
 3    32     7    21    23    25
31     9     2    22    27    20
 8    28    33    17    10    15
30     5    34    12    14    16
 4    36    29    13    18    11

b = diag(A(1:length(A),length(A):-1:1))

b =

24
23
22
33
 5
 4

Upvotes: 0

Amro
Amro

Reputation: 124563

Below is a comparison of all the methods mentioned so far, plus a few other variations I could think of. This was tested on 64-bit R2013a using TIMEIT function.

function [t,v] = testAntiDiag()
    % data and functions
    A = magic(5000);
    f = {
        @() func0(A) ;
        @() func1(A) ;
        @() func2(A) ;
        @() func3(A) ;
        @() func4(A) ;
        @() func5(A) ;
        @() func6(A) ;
        @() func7(A) ;
    };

    % timeit and check results
    t = cellfun(@timeit, f, 'UniformOutput',true);
    v = cellfun(@feval, f, 'UniformOutput',false);
    assert( isequal(v{:}) )
end


function d = func0(A)
    d = diag(A(end:-1:1,:));
end

function d = func1(A)
    d = diag(flipud(A));
end

function d = func2(A)
    d = flipud(diag(fliplr(A)));
end

function d = func3(A)
    d = diag(rot90(A,3));
end

function d = func4(A)
    n = size(A,1);
    d = A(n:n-1:end-1).';
end

function d = func5(A)
    n = size(A,1);
    d = A(cumsum(n + [0,repmat(-1,1,n-1)])).';
end

function d = func6(A)
    n = size(A,1);
    d = A(sub2ind([n n], n:-1:1, 1:n)).';
end

function d = func7(A)
    n = size(A,1);
    d = zeros(n,1);
    for i=1:n
        d(i) = A(n-i+1,i);
    end
end

The timings (in the same order they are defined above):

>> testAntiDiag
ans =
   0.078635867152801
   0.077895631970976    % @AlexR.
   0.080368641824528
   0.195832501156751
   0.000074983294297    % @thefourtheye
   0.000143019460665    % @woodchips
   0.000174679680437
   0.000152488508547    % for-loop

The most suprising result to me is the last one. Apparently JIT compilation is very effective on such simple for-loops.

Upvotes: 6

thefourtheye
thefourtheye

Reputation: 607

Here is my matrix, produced by A = magic(5)

A =

17    24     1     8    15
23     5     7    14    16
 4     6    13    20    22
10    12    19    21     3
11    18    25     2     9


s = size(A,1)
A(s:s-1:end-1)

ans =
11    12    13    14    15

Upvotes: 10

user85109
user85109

Reputation:

The elements you want are easily obtained by indexing. For example, this should do the trick.

n = 4;
A = magic(n)
A =
    16     2     3    13
     5    11    10     8
     9     7     6    12
     4    14    15     1

A(cumsum(n + [0,repmat(-1,1,n-1)]))
ans =
     4     7    10    13

I could also have used sub2ind to get those element indexes, but this does it a bit more cleanly, though less obvious in how it works.

Upvotes: 0

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