check alternation maxima minima in matlab - matlab

Question

I have written an algorithm that finds the local maxima and minima in a signal.
[id_max, id_min] = find_max_min(signal);
I would like now to check:
if the alterantion of maxima and minima is respected
i.e. id_max(1)<id_min(1)<id_max(2)<id_min(2)<...
we could start with a minimum..this is not known
Suppose that:
id_max = [1 3 5 7 10 14 20];
id_min = [2 4 6 8 16 19];
I would like to have 2 vectors missing_max missing_min indicating the location of the missing maxima and minima.
A missing maximum (minimum) occours when between two consecutive minima (maxima) in id_min (id_max) there is not a maximum (minimum).
In this example a maximum is missing in the 7th position of id_max because in id_min there are two consecutive values (16 19) without a maximum between.
Then we have
missing_max = [7]
missing_min = [5]
since
id_max = [1 3 5 7 10 14 X 20];
id_min = [2 4 6 8 X 16 19]; (with X I marked the missing values)
If the alternation is correct the vectors should be empty. Can you suggest an efficient way to do that without for loops?
Thanks in advance

Answer 1

Here's a script that you can adapt to a function if you want:
id_max = [1 3 5 7 10 14 20];
id_min = [2 4 6 8 16 19];
% Group all values, codify extremity (1-max, 0-min), and position
id_all = [ id_max, id_min ];
code_all = [ones(size(id_max)), zeros(size(id_min))];
posn_all = [ 1:numel(id_max), 1:numel(id_min) ];
% Reshuffle the codes and positions according to sorted IDs of min/max
[~, ix] = sort(id_all);
code_all = code_all(ix);
posn_all = posn_all(ix);
% Find adjacent IDs that have the same code, i.e. code diff = 0
code_diff = (diff(code_all)==0);
% Get the indices of same-code neighbors, and their original positions
ix_missing_min = find([code_diff,false] & (code_all==1));
ix_missing_max = find([code_diff,false] & (code_all==0));
missing_min = posn_all(ix_missing_min+1);
missing_max = posn_all(ix_missing_max+1);
Caveats on IDs:
Make sure your id_min and id_max are rows (even if empty);
Make sure that at least one of them is not empty;
While they need not to be sorted, their values must be unique (within the IDs and across).
Later edit:
New version of the code, based on new explanations about the definition:
id_max = [1 3 5 7 10 14 20];
id_min = [2 4 6 8 16 19];
%id_max = [12 14]
%id_min = [2 4 6 8 10];
id_min_ext = [-Inf, id_min];
id_max_ext = [-Inf, id_max];
% Group all values, and codify their extremity (1-max, 0-min), and position
id_all = [ id_max_ext, id_min_ext ];
code_all = [ones(size(id_max_ext)), zeros(size(id_min_ext))];
posn_all = [ 0:numel(id_max), 0:numel(id_min) ];
% Reshuffle the codes and position according to sorted positions of min/max
[~, ix] = sort(id_all);
code_all = code_all(ix);
posn_all = posn_all(ix);
% Find adjacent IDs that have the same code, i.e. code diff = 0
code_diff = (diff(code_all)==0);
% Get the indices of same-code neighbours, and their original positions
ix_missing_min = find([code_diff,false] & (code_all==1));
ix_missing_max = find([code_diff,false] & (code_all==0));
missing_min = unique(posn_all(ix_missing_min-1))+1;
missing_max = unique(posn_all(ix_missing_max-1))+1;
However, the code contains a subtle bug. The bug will be removed by either the person that asked the question, or by me after he/she improves the question in such a way that is really clear what's asked for. :-) Due the fact that we have 2 virtual extremums (one max and one min, at ID = −∞) is possible that the first missing extremum will be marked twice: once at −∞ and once at the first element of the ID list. unique() will take care of that (though is too much of a function call to check if the first 2 elements of an array have the same value)