Trouble vectorising a loop

1 次查看（过去 30 天）
Rhys
Rhys 2013-7-9
I'm trying to tidy up and improve some of my code and have come across a loop that I am struggling to vectorise. I am calculating the mean value of specific parts of a matrix and replacing certain values within the matrix with the newly calculated values.
For example, if I have a matrix 'A' I would like to replace element A(3,1) with the mean of A(2,1) and A(3,1). Likewise, I would then like to replace A(3,2) with the mean of A(2,2) and A(3,2) and so on (although I don't want to do this for every single column, just specific ones). This process is also repeated at various other locations in the matrix (i.e. also replace A(6,1) with the mean of A(5,1) and A(6,1) and so on).
The looped version of my code is of the following form:
A = rand(7,6);
firstRows=[2;5];
lastRows=[3;6];
columnIndicies=[1 2 3];
A2=A;
for j=1:1:length(lastRows)
A2(lastRows(j),columnIndicies)= ...
mean(A((firstRows(j)):(lastRows(j)),columnIndicies),1);
end
A3 = A2(lastRows,:);
Any comments / assistance would be much appreciated.

请先登录，再回答此问题。

采纳的回答

Matt J
Matt J 2013-7-9
编辑：Matt J 2013-7-9
J=arrayfun(@(a,b) (a:b), firstRows,lastRows,'uni',0);
I=cellfun(@(a,b)ones(1,length(a))*b ,J.',num2cell(1:length(J)),'uni',0);
S=cellfun(@(a,b)ones(1,length(a))/length(a) ,J,'uni',0);
B=sparse([I{:}],[J{:}],[S{:}],length(lastRows),size(A,1));
A2=A;
A2(lastRows,columnIndices)=B*A(:,columnIndices);
  8 个评论
显示 6更早的评论
Matt J
Matt J 2013-7-10
Rhys, see my most recent version (and timing comparison) above. Chopping out the unused columns of A helped a lot. My version exceeded the for-loops signficantly, even including the cellfun calls.
Rhys
Rhys 2013-7-11
Thanks Matt, chopping the unused columns out of the calculation really does the trick!

请先登录，再进行评论。

更多回答（2 个）

Matt J
Matt J 2013-7-9
编辑：Matt J 2013-7-9
idx=false(size(A));
idx1=idx; idx1(firstRows,columnIndices)=1;
idx2=idx; idx2(lastRows,columnIndices)=1;
A2=A;
A2(idx2)=(A(idx1)+A(idx2))/2;
Jan
Jan 2013-7-11
编辑：Jan 2013-7-11
Another idea:
S = cumsum(A, 1);
B = S(lastRows, :) - S(firstRows - 1, :);
A0(lastRows, :) = bsxfun(@rdivide, B, lastRows - firstRows + 1);
Three problems: 1. I cannot test this currently, and I frequently confuse -1 and +1. 2. The exception firstRows==1 must be handled. 3. CUMSUM accumulates rounding errors and you have to check, if the accuracy of the results satisfies your needs.
Benefit: It avoid calculating the sum of neighboring elements multiple times in case of overlapping intervals.

类别

在 Help Center 和 File Exchange 中查找有关 Function Creation 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by