From your explanation, I understand that if an event starts and ends with an index of 5, and there is a 2 sandwiched between two 5s, the entire sequence should be treated as a single event. However, if an event index of 2 is not followed by a 5, or if any index other than 2 or 5 appears, it breaks the event chain. and you want to determine the event duration for this logic.
This can be implemented in MATLAB using logical indexing, vectorized operations and tracking certain states, as shown in the following code:
data = [2024 8 1 5;
2024 8 2 2;
2024 8 3 5;
2024 8 4 5;
2024 8 5 2;
2024 8 6 3;
2024 8 7 5;
2024 8 8 5;
2024 8 9 4;
2024 8 10 3;
2024 8 11 5;
2024 8 12 5];
%4th column
event_type = data(:,4);
% Find indices where event starts and ends with 5
event_5_idx = find(event_type == 5);
% Initialize an array to store the event durations
event_durations = [];
start_idx = event_5_idx(1); % Set the first event start
for i = 2:length(event_5_idx)
% Check if the current index is consecutive or sandwiched by a 2
if all(event_type(event_5_idx(i-1)+1:event_5_idx(i)-1) == 2)
% It's part of the same event
continue;
else
% End the event and calculate the duration
event_end = event_5_idx(i-1);
duration = data(event_end, 3) - data(start_idx, 3) + 1;
event_durations = [event_durations; data(start_idx, 3), data(event_end, 3), duration];
% Start a new event
start_idx = event_5_idx(i);
end
end
%last event
event_end = event_5_idx(end);
duration = data(event_end, 3) - data(start_idx, 3) + 1;
event_durations = [event_durations; data(start_idx, 3), data(event_end, 3), duration];
% Display the event durations
disp('Event Start Day - Event End Day - Duration (Days)');
disp(event_durations);
I hope you find this helpful!
