How do I integrate datetime data ?
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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
采纳的回答
Walter Roberson
2018-5-8
0 个投票
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.
更多回答（1 个）
Slawomir Kania
2018-6-20
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 个评论
Peter Perkins
2018-7-5
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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