Clear Filters
Clear Filters

can this be done without a for loop?

5 views (last 30 days)
Luke
Luke on 20 Jan 2012
I have a huge matrix M such that each row is ascending ordered. I find if/where each row exceeds a certain threshold by looking at the last column of cumsum(M>threshold,2).
Rows that do not exceed threshold are left unchanged. For each row that does exceed the threshold, I find the first element on that row that's larger than the threshold and set it to -1. Note that these elements will be on different columns. Then, all elements between -1 and the end of the row are set to zero. For example,
M=[1 2 3 4; 1 4 6 8; 2 6 7 9]; threshold=5;
I need
result=[1 2 3 4; 1 4 -1 0; 2 -1 0 0];
Thanks!

Accepted Answer

Alex
Alex on 20 Jan 2012
The following will put all elements above your threshold value to 0.
[row, col, index] = find( M > threshold);
M(index) = 0;
This next part will put the -1 at the first element like you want.
for i = 1:length(row)
if(i > 1)
if(row(i) < row(i-1))
%the previous row was larger than the current row, break the loop
break;
end
end
%found the first element of the row that exceeds the threshold
M(row(i),col(i)) = -1;
end
Another option, is to do a loop and look at it row by row.
[row, col] = size(M);
for i = 1:row
index = find(M(i,:) > threshold, 1);
if(exists(index))
M(i,index) = -1;
M(i,index+1:end) = zeros(1,col-index);
end
end
  1 Comment
Luke
Luke on 20 Jan 2012
I already have implemented something similar to your loop solution. But since I figured out a way of setting the -1's without a loop, I was wondering if I could do everything without a loop.

Sign in to comment.

More Answers (2)

Sean de Wolski
Sean de Wolski on 20 Jan 2012
So something like:
A = sort(magic(5),2); %determinant sample data
thresh = 22; %greater than this
idx = A>thresh; %index
rows = any(idx,2); %which rows?
A(rows,:) = -1; %set 'em to -1
A(idx) = 0; %change back the parts that weren't

Luke
Luke on 20 Jan 2012
Although Alex's answer is not complete, he sent me on the right path, so I gave him credit for the solution.
[row, col] = find( M > threshold);
M(sub2ind(size(M),row,col))=0; clear row col
index=cumsum(M > threshold,2);
index=index(:,end); % the columns where to place -1's
row=1:size(M,1);
row=row(index>0); % only rows that exceed threshold
col=size(M,2)-index(index>0)+1;
M(sub2ind(size(M),row,col))=-1;
  1 Comment
Sean de Wolski
Sean de Wolski on 20 Jan 2012
for medium to large M, sub2ind is probably slower than a for loop.

Sign in to comment.

Categories

Find more on Multidimensional Arrays in Help Center and File Exchange

Community Treasure Hunt

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

Start Hunting!