how to apply horzcat to output arrayfun
2 ビュー (過去 30 日間)
古いコメントを表示
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!
2 件のコメント
Stephen23
2020 年 4 月 30 日
Note that you don't need to define an anonymous function, just define a function handle to colon :
B = arrayfun(@colon, A(1,:), A(2,:), 'Uni',0);
% ^^^^^^ this is all you need
採用された回答
Guillaume
2020 年 4 月 30 日
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);
end
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.
その他の回答 (0 件)
参考
カテゴリ
Help Center および File Exchange で Matrix Indexing についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!