Problem using parfor for reading variable sized chunks of data into a larger pre-allocated container

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Hi,
I have an problem where I have pre-allocated a large matrix or vector, a, and where I will read data blocks from a large number of files that will be inserted at the right indexes in a. The different files, and resulting blocks, will typically have different size.
A simple example:
a = zeros(15,1); % pre-allocated vector
b = [1,10;11,15]; % each row contains the start to stop index for each block
parfor i = 1:size(b,1)
a(b(i,1):b(i,2),:) = i*ones((b(i,2)-b(i,1))+1,1);
end
With 'for' instead of 'parfor' it works as intended.
Any tips or solutions which don't reduce the performance I am trying to obtain by using parfor in the first place?
Thanks,
Oyvind

Akzeptierte Antwort

Edric Ellis
Edric Ellis am 21 Okt. 2020
Bearbeitet: Edric Ellis am 21 Okt. 2020
There's no simple way to do this without at least some duplication of data. With some duplication of data, you could do something simple like this:
aCell = cell(1, size(b,1));
parfor i = 1:size(b,1)
aCell{i} = <stuff>; % return each block in its entirety
end
a = vertcat(aCell{:}); % concatenate all cell entries into the final result
If that is not sufficiently performant, you could consider using parfeval to give you a little more control, but this is more difficult to code, and may not actually save you much. Here's an untested sketch though:
a = zeros(15,1);
for i = 1:size(b,1)
fut(i) = parfeval(@doStuff, 1, b(i,1), b(i,2)); % invoke doStuff(b(i,1),b(i,2))
end
for i = 1:size(b,1)
[idx, result] = fetchNext(fut); % collect the next result
% (note that 'idx' tells you the index into 'fut' that just
% completed)
a(b(idx,1):b(idx,2),:) = result; % push the result into 'a'
end
  4 Kommentare
Oyvind Heg
Oyvind Heg am 23 Okt. 2020
Thank you for the answer.
I'v done a quick profiler example:
parfor:
1 result_cell = cell(N,1);
2 parfor i = 1:N
3 result_cell{i} = readData(...);
4 end
parfeval:
5 fut(1:N) = parallel.FevalFuture;
6 for i = 1:N
7 fut(i) = parfeval(@readData,...)
8 end
9 for i = 1:N
10 [idx,result] = fetchNext(fut);
11 end
My observations when running the profiler are:
The 'for' loop on line 2 and line 9 take about the same amount of time (about 12 seconds in my example).
However line 7 takes as long as 8 seconds. Is that to be expected? Seems like a lot of overhead when the actual work takes 12 seconds.
Thanks,
Oyvind
Oyvind Heg
Oyvind Heg am 27 Okt. 2020
Hi,
Any thoughts on the overhead on line 7? Is is as expected, or am I doing something wrong?
Thanks,
Oyvind

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