calculate time from alarm (variabel x) to event (variable y)

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SRRellum 2022-4-22

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https://ww2.mathworks.cn/matlabcentral/answers/1702940-calculate-time-from-alarm-variabel-x-to-event-variable-y

评论： Les Beckham 2022-4-24

I want to calculate the time from an 'alarm' variable to the 'event' variable. In my dataset there are multiple alarms and multiple events. The time only needs to be calculated between an alarm and a direct subsequent event.

I am trying to go through my alarm column with a (for i) loop and try to find the event that comes after alarm(i). But I clearly need some help, this is where I am at:

time = [0;10;20;30;40;50;60;70;80;90;100]; % in sec; variables are generated every 10 seconds

alarm = [0;0;1;0;1;0;0;1;0;0;0];

event = [1;1;0;0;1;0;1;0;0;0;0];

dataset = table(time,alarm,event);

for i = 1:length(time)

if alarm(i) == 1 % looping through the alarms from i=1 to i=11

if (find(abs(dataset.event) > 0, 1, 'first') > i) % if the first event comes after alarm(i)

mark = find(abs(dataset.event) > 0, 1, 'first') > i);

time_event = dataset.time(mark); % return timepoint from corresponding row

else

% mark the first event after row alarm(i) && unless a new alarm occured before the next event

end

% time_to_event(i) = time_event - dataset.time;

end

% sum(time_to_event) = sum(time_to_event)

The line "find(abs(dataset.event) > 0, 1, 'first') > i)" is redundant if I knew how to target the row after alarm(i)..

So the output in my dataset should be as follows:

alarm = [0;0;1*;0;1^;0;0;1;0;0;0];

event = [1;1;0;0;1*;0;1^;0;0;0;0];

The time between the alarm* to event* and alarm^ to event^ should be calculated.

And after the loop is done, I want to sum all time to events.

Hope you can help

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Answer 1

Les Beckham 2022-4-22

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1702940-calculate-time-from-alarm-variabel-x-to-event-variable-y#answer_948790

在 MATLAB Online 中打开

Try this

t = [0;10;20;30;40;50;60;70;80;90;100]; % in sec; variables are generated every 10 seconds 
alarm = [0;0;1;0;1;0;0;1;0;0;0];
event = [1;1;0;0;1;0;1;0;0;0;0];
alarms = find([ false; (diff(alarm) > 0) ]); % detect rising edges
events = find([ false; (diff(event) > 0) ]);
% you have to have both an alarm and an event to measure the time - how
% many do we have?
num_timed_events = min(numel(alarms), numel(events)); 
delaytimes = zeros(1, num_timed_events); % preallocate
for i = 1:num_timed_events
    delaytimes(i) = t(events(i)) - t(alarms(i));
end
delaytimes
delaytimes = 1×2
    20    20

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Les Beckham 2022-4-23

在 MATLAB Online 中打开

Actually that doesn't seem to make much sense. Why is it only the first one in the alarm signal that should be discarded? What if there are more than three "alarms" in a row? What resets the logic to start looking for another pair?

Typically in signal processing you look for edges (rising or falling). Isn't one block of alarms considered to be a single alarm?

It would definitely be possible to throw away any negative delaytimes, like this.

% new dataset
t = [0;10;20;30;40;50;60;70;80;90;100]; 
alarm = [0;0;1;1;1;0;0;1;0;0;0];
event = [1;1;0;0;1;0;1;0;0;0;0];
alarms = find([ false; (diff(alarm) > 0) ]); % detect rising edges
events = find([ false; (diff(event) > 0) ]);
% you have to have both an alarm and an event to measure the time - how
% many do we have?
num_timed_events = min(numel(alarms), numel(events)); 
delaytimes = zeros(1, num_timed_events); % preallocate
for i = 1:num_timed_events
    delaytimes(i) = t(events(i)) - t(alarms(i));
    if delaytimes(i) < 0
        delaytimes(i) = []; % discard negative delay times
    end
end
delaytimes
delaytimes = 20

To me, that is how I would interpret this dataset.

Please clarify.

SRRellum 2022-4-24

Thanks, yeah you're right. Alarms in a row should be considered as the same alarm indeed.

However, in this dataset the 'alarms' are a predictive alarm that predict low blood pressure. Events are low blood pressure events. So if an alarm occurs I want to know how long before low pressure a prediction was provided. However, between a prediction and low blood pressure an intervention could have been given to the patient (the doctors were blinded to the alarms). That intervention prevents the low blood pressure from happening. So If an alarm occured in the dataset and then normalizes without a low blood pressure, then that alarm is not interesting to me, because it was altered. But the next alarm to an event is interesting.

But I will work with your solution for now. And see what it does in my real dataset.

Thanks a lot for your help!

Les Beckham 2022-4-24

You are quite welcome.

Glad that I was able to help.

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