フィルターのクリア

Exploiting symmetry in multidimensional arrays

2 ビュー (過去 30 日間)
Patrick Mboma
Patrick Mboma 2014 年 7 月 13 日
コメント済み: Roger Stafford 2014 年 7 月 14 日
Dear all, In a symmetric matrix A of size [n,n], for two indices i and j such that 1<=i<=n and 1<=j<=n we have that A(i,j)=A(j,i). Suppose I know A(i,j) for i>=j how can I efficiently set A(j,i)?
More generally, if I have a hypercube A of size n^m, how can I, on the basis of A(i1,i2,...,im) with i1>=i2>=i3....>=im, set all A for all permutations of i1,i2,...,im instead of recalculating the entry for each permutation.
Thanks,

採用された回答

Roger Stafford
Roger Stafford 2014 年 7 月 13 日
For a general m-dimensional n^m array, A, do this:
[I1,I2,...,Im] = ndgrid(1:n);
P = [I1(:),I2(:),...,Im(:)];
Q = sort(P,2,'descend');
A(sub2ind(size(P),P(:,1),P(:,2),...,P(:,m))) = ...
A(sub2ind(size(Q),Q(:,1),Q(:,2),...,Q(:,m)))
Note: The four three-dot ellipses (...) shown above (but not the one following the '=' sign) are to be filled in by the user.
  4 件のコメント
Patrick Mboma
Patrick Mboma 2014 年 7 月 14 日
編集済み: Patrick Mboma 2014 年 7 月 14 日
@Alfonso What you suggest would really be neat and seems to work well in the two-dimensional case that I tested. I still need to investigate higher dimensions and the speed.
Roger Stafford
Roger Stafford 2014 年 7 月 14 日
@Patrick. Yes, you are quite right. I should have used size(A). I must have had a mental black out at that point! :-)

サインインしてコメントする。

その他の回答 (1 件)

Roger Stafford
Roger Stafford 2014 年 7 月 13 日
For a 2D array it's easy:
B = tril(A,-1)
A = B+B.'+diag(A);

カテゴリ

Help Center および File ExchangeCreating and Concatenating Matrices についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by