how to apply horzcat to output arrayfun

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François Fabre
François Fabre 2020년 4월 30일
댓글: François Fabre 2020년 4월 30일
Here's a simplified example of my problem:
A = [2 7 11;
5 9 11]; % This is my starting array
I'd like to obtain
B = [2 3 4 5 7 8 9 11]; % in other words [2:5 7:9 11:11]
Right now, my solution is
B = arrayfun(@(x,y) colon(x,y), A(1,:), A(2,:), 'UniformOutput', false);
Then I use [B{:}] as a comma-separated-list to index another array.
Since I'm doing this in a for loop where A can have hundred of thousands of columns, I'd like to avoid to create a cell by horizontally concatenating output arguments from arrayfun to improve memory layout efficiency.
Do you know if it's possible or if there's a smarter approach?
Thanks in advance!
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François Fabre
François Fabre 2020년 4월 30일
Indeed, it's a good trick to know. Thank you!

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Guillaume 2020년 4월 30일
I like the functional programming aspect of arrayfun (it clearly says: apply this function to all the elements of these sequence) but note that if speed is critical then an explicit loop is likely to be faster.
With arrayfun you don't have a choice but going through an intermediate cell array. With your example, you don't have to split it in two lines:
B = cell2mat(arrayfun(@(x,y) colon(x,y), A(1,:), A(2,:), 'UniformOutput', false)); %works because the cell array coming out of arrayfun is a row vector of row vectors
With an explicit loop, in the case of your example you can easily calculate where the sequences land in the final array and avoid the cell array altogether:
seqlengths = A(2, :) - A(1, :) + 1;
seqstarts = cumsum([1, seqlengths(1:end-1)]);
B = zeros(1, seqstarts(end)-1);
for colidx = 1:size(A, 2)
B(seqstarts(colidx) + (0:seqlengths(colidx)-1)) = A(1, colidx) : A(2, colidx);
Whether the increase in code complexity is worth the gain is up to do. Of course the above may not work for your real use case.
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François Fabre
François Fabre 2020년 4월 30일
I compared the three methods when A has 20000 columns :
- arrayfun + [B{:}] - > 1.6s
- cell2mat(arrayfun()) - > 2.5-2.7s
- your explicite for loop - > 2.4-2.5s
However, your approach is the right one. I found here different ways to vectorize the notion of colon. In my case the fastest (0.8-0.9s with 20000 columns) is a slightly different implementation than yours:
seqlengths = A(2, :) - A(1, :) + 1;
% find end and beginning indices for each chunk
partialLength = cumsum(seqlengths);
cumStart = [1 partialLength(1:end-1)+1];
% preallocate output
% then loop through start/stop pairs, in order, and fill
numtot = sum(seqlengths);
B = zeros(1,numtot);
for colidx = 1:length(A(1, :))
B(cumStart(colidx):partialLength(colidx)) = A(1, colidx) : A(2, colidx);
Thank you for your help!

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