Detect where an array becomes negative

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John Draper
John Draper 2016-7-25
编辑： dpb 2016-7-26
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.
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John Draper
John Draper 2016-7-26
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
dpb 2016-7-26
编辑：dpb 2016-7-26
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.

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回答（2 个）

dpb
dpb 2016-7-25
编辑：dpb 2016-7-26
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.
Star Strider
Star Strider 2016-7-25
编辑：Star Strider 2016-7-25
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.
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John Draper
John Draper 2016-7-26
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
Star Strider 2016-7-26
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]'

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