How to improve the speed of computing trace in my code?

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Xiaohan Du
Xiaohan Du 2018-6-20
评论： Xiaohan Du 2018-6-20
Hi all,
In my code there is a key function which cost the majority of computational power due to the large number of repetitive computations. Imagine I have 2 cell arrays 'respi' and 'respj' which contains results of SVD vectors:
>> respi
respi =
1×10 cell array
Columns 1 through 9
{3×1 cell} {3×1 cell} {3×1 cell} {3×1 cell} {3×1 cell} {3×1 cell} {3×1 cell} {3×1 cell} {3×1 cell}
Column 10
{3×1 cell}
>> respi{1}
ans =
3×1 cell array
{242×5 double}
{ 5×5 double}
{ 3×5 double}
I have a function which takes respi and respj, compute and store the trace as follows
function otpt = uTuPartDemo(respi, respj)
otpt = zeros(length(respi), length(respj));
for iTr = 1:length(respi)
u1 = respi{iTr}; % find the ith cell element.
for jTr = 1:length(respj)
u2 = respj{jTr}; % find the jth cell element
otpt(iTr, jTr) = ... % compute the trace.
trace((u2{3}' * u1{3}) * u1{2}' * (u1{1}' * u2{1}) * u2{2});
end
end
And I need to run this function repetitively for many times as follows:
for i = 1:n % very large number, say n = 100000
otpt = uTuPartDemo(respi, respj);
end
How can I improve the speed and efficiency of function uTuPartDemo? Many thanks!

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回答（1 个）

Pieter Hamming
Pieter Hamming 2018-6-20
Presumably the trace calculation takes most time. Since you're only interested in the diagonals anyway, why not use the approach explained here. It should reduce your computational strain.
A minor other issue: you don't need to define u1 and u2. Instead of
u1 = respi{iTr};
u1{3}
Just use
respi{iTr}{3}
It will save you a minor amount of memory.
  1 个评论
Xiaohan Du
Xiaohan Du 2018-6-20
Hi Pieter,
Before trying this approach, I'd like to know if I need to recover the matrix before SVD (i.e. computing A = u1{1} * u1{2} * u1{3}', B = u2{1} * u2{2} * u2{3}'), then compute sum(A.*B',2)?

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