How do I integrate datetime data ?

10 vues (au cours des 30 derniers jours)
Slawomir Kania
Slawomir Kania le 8 Mai 2018
Commenté : Peter Perkins le 5 Juil 2018
I have a plot of datetime data with x axis being datetime (yyyy-MM-dd-HH:mm:ss) and y axis being my data.
my data file looks like this:
2018-03-12-17:50:43 836.00
2018-03-12-17:50:48 871.00
2018-03-12-17:50:54 1116.00
2018-03-12-17:50:59 906.00
2018-03-12-17:51:04 834.00
2018-03-12-17:51:10 927.00
2018-03-12-17:51:15 950.00
2018-03-12-17:51:21 827.00
2018-03-12-17:51:27 999.00
2018-03-12-17:51:33 1088.00
and I convert this data for plotting using this formula
fid = fopen('.\2018-04-02.txt','r');
c = textscan(fid,'%s%f');
fid = fclose(fid);
data = c{2};
%t=datetime(c{1},'Format','HH:mm:ss');
t=datetime(c{1},'Format','yyyy-MM-dd-HH:mm:ss');
figure;
plot(t,data);
Which gives me nice plot of my data depending on time.
I need to convert said data into an incrementing function portraing integration of continous changes adding up depending on time. I tried using the trapz function to conduct numerical integration to no success. The datetime format I'm using does not correlate with trapz function. Also changing the time format into double does not work either.
Basically the plot needs to portray incrementation of my data depending on time.
Any ideas ?
Many thanks in Advance

Réponse acceptée

Walter Roberson
Walter Roberson le 8 Mai 2018
You can use datenum() on datetime objects to get serial date numbers. Or you could subtract the first datetime entry from them all in order to get duration objects, which you could then use seconds() on to get a numeric version.

Plus de réponses (1)

Slawomir Kania
Slawomir Kania le 20 Juin 2018
So I found my solution!
What I did was making a table of small integrations using the datenum format and trapz function:
cz=datenum(t);
for i=2:1:size(t)
q1(i)=trapz([cz(i-1),cz(i)],[data1(i-1),data1(i)]);
end
then I transfered data into incrementing table:
for i=2:1:size(t)
q2(i)=q2(i-1)+q1(i);
end
end result gave me incrementing integration:
  1 commentaire
Peter Perkins
Peter Perkins le 5 Juil 2018
Walter's second suggestion, i.e.
>> t = datetime(2018,7,5,0,0,100*sort(rand(1,5)))
t =
1×5 datetime array
05-Jul-2018 00:00:03 05-Jul-2018 00:00:31 05-Jul-2018 00:00:43 05-Jul-2018 00:01:09 05-Jul-2018 00:01:35
>> dt = seconds(t - t(1))
dt =
0 28.265 40.43 66.038 91.578
is better from a numerical standpoint, because it avoids round-off due to datenum using "day" as the unit. May not matter much, hard to say.

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