Vectorization of array comparison - matlab

Question

I have surjective functions created by matching one element in an array MatchesX.trainIdx to one or more elements in a second array MatchesX.queryIdx.
To obtain only the bijective elements of said funciton I run the same function forward
Matches1=Matcher.match(Descriptors1,Descriptors2);
and then backwards
Matches2=Matcher.match(Descriptors2,Descriptors1);
and then look for the elements occuring in both function in following fashion:
k=1;
DoubleMatches=Matches1;
for i=1:length(Matches1)
for j=1:length(Matches2)
if((Matches1(i).queryIdx==Matches2(j).trainIdx)&&(Matches1(i).trainIdx==Matches2(j).queryIdx))
DoubleMatches(k)=Matches1(i);
k=k+1;
end
end
end
DoubleMatches(k:end)=[];
This of course does the work, but it is rather unelegant and seems to bother the JIT accelerator (calc time with accel on and accel off is the same).
Can you think of a way to vectorize this expresion? Is there any other way of avoiding the JIT from "striking"?
Thanks a lot and sorry about the strange structs, I'm working with MEX-functions. Let me know if rewriting the code in "normal" arrays would help

Answer 1

Access to data in multi-dimensional structures is notoriously slow in MATLAB, so transforming your data to an ordinary array will certainly help:
kk = 1;
DoubleMatches = Matches1;
%// transform to regular array
Matches1queryIdx = [Matches1.queryIdx];
Matches1trainIdx = [Matches1.trainIdx];
Matches2queryIdx = [Matches2.queryIdx];
Matches2trainIdx = [Matches2.trainIdx];
%// loop through transformed data instead of structures
for ii = 1:length(Matches1queryIdx)
for jj = 1:length(Matches1queryIdx)
if((Matches1queryIdx(ii)==Matches2trainIdx(jj)) && ...
(Matches1trainIdx(ii)==Matches2queryIdx(jj)))
DoubleMatches(kk) = Matches1(ii);
kk = kk+1;
end
end
end
DoubleMatches(kk:end)=[];
There is also a solution that is almost entirely vectorized:
matches = sum(...
bsxfun(#eq, [Matches1.queryIdx], [Matches2.trainIdx].') & ...
bsxfun(#eq, [Matches1.trainIdx], [Matches2.queryIdx].'));
contents = arrayfun(#(x)..
repmat(Matches1(x),1,matches(x)), 1:numel(matches), ...
'Uniformoutput', false);
DoubleMatches2 = [contents{:}]';
Note that this can be a lot more memory intensive (it has O(N²) peak memory footprint, as opposed to O(N) for the others, although the data type at peak memory is logical and thus 8x smaller than double...). Better do some checks beforehand which one you should use.
A little test. I used the following dummy data:
Matches1 = struct(...
'queryIdx', num2cell(randi(25,1000,1)),...
'trainIdx', num2cell(randi(25,1000,1))...
);
Matches2 = struct(...
'queryIdx', num2cell(randi(25,1000,1)),...
'trainIdx', num2cell(randi(25,1000,1))...
);
and the following test:
%// Your original method
tic
kk = 1;
DoubleMatches = Matches1;
for ii = 1:length(Matches1)
for jj = 1:length(Matches2)
if((Matches1(ii).queryIdx==Matches2(jj).trainIdx) && ...
(Matches1(ii).trainIdx==Matches2(jj).queryIdx))
DoubleMatches(kk) = Matches1(ii);
kk = kk+1;
end
end
end
DoubleMatches(kk:end)=[];
toc
DoubleMatches1 = DoubleMatches;
%// Method with data transformed into regular array
tic
kk = 1;
DoubleMatches = Matches1;
Matches1queryIdx = [Matches1.queryIdx];
Matches1trainIdx = [Matches1.trainIdx];
Matches2queryIdx = [Matches2.queryIdx];
Matches2trainIdx = [Matches2.trainIdx];
for ii = 1:length(Matches1queryIdx)
for jj = 1:length(Matches1queryIdx)
if((Matches1queryIdx(ii)==Matches2trainIdx(jj)) && ...
(Matches1trainIdx(ii)==Matches2queryIdx(jj)))
DoubleMatches(kk) = Matches1(ii);
kk = kk+1;
end
end
end
DoubleMatches(kk:end)=[];
toc
DoubleMatches2 = DoubleMatches;
% // Vectorized method
tic
matches = sum(...
bsxfun(#eq, [Matches1.queryIdx], [Matches2.trainIdx].') & ...
bsxfun(#eq, [Matches1.trainIdx], [Matches2.queryIdx].'));
contents = arrayfun(#(x)repmat(Matches1(x),1,matches(x)), 1:numel(matches), 'Uniformoutput', false);
DoubleMatches3 = [contents{:}]';
toc
%// Check if all are equal
isequal(DoubleMatches1,DoubleMatches2, DoubleMatches3)
Results:
Elapsed time is 6.350679 seconds. %// ( 1×) original method
Elapsed time is 0.636479 seconds. %// (~10×) method with regular array
Elapsed time is 0.165935 seconds. %// (~40×) vectorized
ans =
1 %// indeed, outcomes are equal

Answer 2

Assuming Matcher.match returns array of the same objects as passed to it as arguments you can solve this like this
% m1 are all d1s which have relation to d2
m1 = Matcher.match(d1,d2);
% m2 are all d2s, which have relation to m1
% and all m1 already have backward relation
m2 = Matcher.match(d2,m1);