Find maximum number of consecutive negative values

Hi,

I have a mxn matrix and I need to find the maximum number of consecutive negative values for each column.

So the output would be a 1xn vector, where the i-th element is the maximum number of consecutive negative values for column "i".

How would I do this without a loop?

[dave]

Thanks everybody for your great suggestions - I wish I could accept more than one answer...

Answer 1

[Andrei Bobrov]

A = randi([-6,3],20,15); % Let it your array

t = A < 0;
s = size(t);
tt = cumsum(diff([false(1,s(2));t]) == 1);
[jj,jj] = ndgrid(zeros(1,s(1)),1:s(2));
out = max(accumarray([tt(t),jj(t)],ones(nnz(t),1)));

[DARSHAN N KANNUR]

What to do, if I want to know the starting index of the maximum consecutive negative elements. Thank you in advance

Answer 2

[Iain]

 logi = x < 0;
 [bw n] = bwlabel(logi);
 A = regionprops(bw,'Area');
 Answer = max([A(:).Area]);

[Image Analyst]

regionprops() is in the Image Processing Toolbox. With later versions, it can do the labeling automatically internally:

x = array2D(:, columnNumber); % Extract just this column
negativeValues = x < 0; % Logical array
structureA = regionprops(negativeValues,'Area');
Answer = max([structureA.Area]);

Repeat for every column in your array.

[DARSHAN N KANNUR]

What to do, if I want to know the starting index of the maximum consecutive negative elements. Thank you in advance

Answer 3

[Roger Stafford]

If you want the code completely vectorized, let x be your matrix and do this:

 [m,n] = size(x);
 p = double([0;reshape([x;zeros(1,n)],[],1)]<0);
 f = find(diff(p)~=0);
 f2 = f(2:2:end);
 p(f2+1) = f(1:2:end-1)-f2;
 p = cumsum(p);
 p = reshape(p(2:end),[],n);
 mx = max(p); % mx is the required 1 x n row vector of column maxima

However, I suspect that a single for-loop, properly done, would be faster.

[DARSHAN N KANNUR]

What to do, if I want to know the starting index of the maximum consecutive negative elements. Thank you in advance