How to identify the beginning of another hour and choose 3 rows per hour depending on the value of another metric in another column?

Hey everybody!

I want to analyse 24h ECG recordings (that are divided in several files) from several patients and I already have the metrics in struct files. The metrics are analysed each 5 minutes but I need to analyse statistically 10 min in each hour with an overlap of 50% (3 rows per hour). The rows need to be consecutive and depend on a threshold in column 3 (minimum of 50% in each row) so they don't always get chosen every N rows (attached picture). I need to find the mean, maximum and minimum of those 3 rows.

I also have a char event time with the time and date of the beginning of the file, the beginning and end of the intervals I am interested in (attached picture).

Could anyone help me with the easiest/efficient way to analyse all the data?

[Rishabh Mishra]

Hi,

Kindly attach the ECG readings file, it would enable me to come up with an efficient solution and help you in the best way possible.

[Márcia Nunes]

Hi Rishabh, I am afraid I can't attach the file you are are requesting since the data is confidential. If there is some way to help me it would be great, if not I understand it is complicated to help without all the information

Answer 1

[Rishabh Mishra]

Hi,

Consider the code given below. Few assumptions that I have made while writing this code are:

1. The readings are taken starting 12 at Midnight.
2. The ECG readings & it’s columns are stored in ‘M’ matrix.
3. The readings consist of 24 x 12 rows.
4. The number of readings taken every hour is 12 (as readings are taken every 5 minutes on average)

The code:

% assuming that ECG rreadings are taken starting 12 Midnight
start = datetime(2020,1,1,0,0,0)
 
% assuming that data is stored in 'M' matrix
 
% time of reading
time = M(:,1)
% reading value
reading = M(:,2)
% threshold value
threshold = M(:,3)
 
% traverse each row of the data
for k = 1:numel(time)
    i = 0
    
    if (time(k) >= start & threshold(k) >= 50.0)
        i = i+1
    else
        i = 0
    end
    
    % if 3 consecutive rows with given requirements are found
    if (i == 3)
        disp(mean(reading(k-2:k))
        disp(min(reading(k-2:k))
        disp(max(reading(k-2:k))
        start = start + duration(1,0,0)
        i = 0
    end
end

Hope this helps.