Detect where an array becomes negative

Hi everyone,

I have an array of size 401 x 381 which has positive and negative values. I would like to detect which row in each column this change of sign occurs for the purposes of comparison. If anyone has any ideas they will be appreciated. If you need to see the full code then just ask, I thought I would leave it out for the sake of simplicity.

Thanks, John.

[Azzi Abdelmalek]

How will you proceed? column by column or row by row?

[per isakson]

What should the result be for columns without negative values?

[John Draper]

Hi again everyone, I am still working on this problem but perhaps I wasn't clear in my explanation. As an end result I would like to have a row vector with each column in this row vector having a number which represents the row where the change from positive to negative takes place. Perhaps I forgot to mention that this change of sign only occurs ONCE per column.

Trying to incorporate some code suggested, here is my incomplete attempt:

if true
Brho_sign = sign(Brho);
Brho_diff = diff(Brho_sign);

Where 'Brho' is my initial array. So I am left with an array which has values 0, 2 and -2. As mentioned above each column only has one non zero element (either 2 or -2). I would like to detect the corresponding row number and store them in a row vector.

Thanks for your help thus far, any more would be appreciated!

[John Draper]

Azzi Abdelmalek I believe it would have to be column by column as some rows contain two non-zero values and thus are not uniquely defined.

[John Draper]

per isakson All columns have negative and positive values. The first 190 columns start with +ve values and as you move down the rows they eventually becomes -ve. The other 190 are essentially a mirror image of the first 190 and so start -ve and becomes +ve as you move down the rows.

[dpb]

Then for the latter columns there isn't a transition from + to -, it's t'other way 'round. So you're actually looking for the sign transition. Is zero a possible element in the array, too? What is the result in that column if so? Posted a solution that will give the zero location in that case.

Answer 1

[dpb]

Sometimes it's just as easy to loop...it's unfortunate in cases like this find isn't vectorized; a cleaner syntax doesn't pop into mind at the moment altho I suspect there is something I'm just not seeing at the moment...

nc=size(x,2);         % number of columns
ix=zeros(1,nc);       % preallocate for the index
for i=1:nc
  ix(i)=find(x(:,i)<0,1);  % the first negative location in ith column
end

ADDENDUM As Per notes, the above presumes there is a valid entry in every column, if it's possible that's not so then include test that the result from find isn't empty...

for i=1:nc
  idx=find(x(:,i)<0,1);  % the first negative location in ith column
  if ~isempty(idx)  % if none, find returns empty array
    ix(i)=idx;      % save the valid ones...
  end
end

As a design note, I chose the above so the resulting vector would have same length as number of columns in original array consistent with other Matlab behavior of things that operate over columns/rows. Since indices are always one-based, zero is unique indicator of missing value; one could also use NaN as the initial value as well for more obvious flag.

ADDENDUM 2

With the addition of the other conditional, the above works with a slight modification --

nc=size(x,2);          % number of columns
ix=zeros(1,nc);        % preallocate for the index
for i=1:nc
  ix(i)=find([0;diff(sign(m(:,i)))],1); 
end

With the caveat that, as described, there will be a sign change in each column so the empty condition doesn't occur...the defensive strategem has already been demonstrated above just in case.

ERRATUM I just note I left out the subscripting expression on the x array initially; what one gets for "keyboard debugging" instead of cut 'n paste. Fixed all instances I believe...if missed one, should be apparent.

Answer 2

[Star Strider]

This one approach:

[r,c] = find(A < 0);
Uc = unique(c);
R = {Uc accumarray(c, r, [], @(c) {r(c)})};

celldisp(R)

It produces ‘R’, a cell with the first column being the column numbers and the other cells being the corresponding rows with the negative values.

EDIT — Note that this outputs only columns with at least one negative number (my reason for using the unique function). A column with no negative number does not appear.

[John Draper]

Hi, I believe this did work for me but I was looking for a result that was a row vector. What you provided was i believe a linear indexed version of what I wanted. Sorry, it's my fault - I should have been clearer!

[Star Strider]

I’m not certain what you mean by ‘linear indexed’. See if this does what you want:

A = randi([-9 9], 5, 6);                            % Create Data
[r,c] = find(A < 0);
R = {[1:size(A,2)]'  accumarray(c, r, [], @(c) {r(c)})};
for k1 = 1:size(A,2)
    Rk(k1,:) = R{2}{k1}(1);
end
Result = [R{1}  Rk]'