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How to speed up pairwise difference calculation between vectors under certain condition?

Asked by Raouf Amara on 26 May 2019
Latest activity Answered by Jan
on 8 Jun 2019
Accepted Answer by Jan
I have two double arrays A (size: Nx1) and B (Mx1) both with large increasing values (from ~1e9 to ~1e13; not sure this is relevant).
I am trying to compute the pairwise differences between the two vectors but only for pairs that respect a certain condition, that is the absolute value of the difference between the two values is smaller than t.
I first tried to brute force calculate all the pairwise differences but as you would expect a NxM array cannot be stored on a regular computer; it is around a few hundred of terabytes.
So far I have implemented a rather straightforward solution which works fine with small N and M.
Low = 1e9; % lower bound for values in A & B
Upp = 1e13; % upper bound for values in A & B
N = 1e4; % number of elements in A and B (assuming they are equal, which is not the case)
A = sort((Upp-Low)*rand(N,1)+Low,'ascend');
B = sort((Upp-Low)*rand(N,1)+Low,'ascend');
maxdif = 1000000; % maximum difference
PrWD = cell(1,N);
tic
for i = 1:N
% pairwise difference if |B-A(i)|<=maxdif
PrWD{i} = A(i) - B(abs(B-A(i))<=maxdif)';
end
toc
However, I have A and B arrays which are both very large (N and M is the order of a few tens of millions) such that this operation takes a few hours in practice.
I tried arrayfun or bsxfun on small subsets (N=M=1e5) but it does not speed up the process.
At this point, I understand that this is extremely demanding as calculating all the pairwise differences or even the indices verifying the condition seem to scale with N*M.
Is there any (Matlab) way to speed up this process? Maybe taking into acccount the fact that values are increasing, or that not all differences need to be calculated?

  1 Comment

You can improve the abs(B-A(i))<=maxdif part by locating the lower and upper indices iteratively. I will provide some code soon.
Dummy comment just to let me find this thread again: Jans magic message

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1 Answer

Answer by Jan
on 8 Jun 2019
 Accepted Answer

Instead of creating the huge index matrix abs(B-A(i))<=maxdif in each iteration, it is cheaper to search the lower and upper limits of the matching elements iteratively:
tic
PrWD2 = cell(1,N);
ini = 1;
fin = ini;
for i = 1:N
% Find initial and final index:
while B(ini) < A(i) - maxdif && ini < N
ini = ini + 1;
end
while B(fin) <= A(i) + maxdif && fin < N
fin = fin + 1;
end
if ini <= fin - 1
PrWD2{i} = A(i) - B(ini:fin - 1).';
end
end
toc
% PrWD2(cellfun('isempty', PrWD2)) = {zeros(1,0)};
% isequal(PrWD, PrWD2)
With your example and N=1e5 this is about 1100 times faster than the original code.
This code replies the empty matrix [], if no element is matching, while the original code creates a [1 x 0] matrix. For a comparison of the results, the empty matrices are adjusted, but this is not needed in the productive code most likely.

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