How can I modify negative values from two datasets based on a positive given one?
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Hi. I need some help.
I wrote this code to fix two datas(First_data,Second_data) based on another one(Third_data). The datasets have the same trend, so the graph should look similar(not identic).
The code looks that it works when the values are positive but not on negative ones.
DateO={'20-Apr-2020 12:06:00','20-Apr-2020 12:07:00','20-Apr-2020 12:08:00','20-Apr-2020 12:09:00','20-Apr-2020 12:10:00','20-Apr-2020 12:11:00','20-Apr-2020 12:12:00','20-Apr-2020 12:13:00'};
DateT={'20-Apr-2020 12:06:00','20-Apr-2020 12:07:00','20-Apr-2020 12:08:00','20-Apr-2020 12:09:00','20-Apr-2020 12:10:00','20-Apr-2020 12:11:00','20-Apr-2020 12:12:00','20-Apr-2020 12:13:00'};
DateTh={'20-Apr-2020 12:06:30','20-Apr-2020 12:07:30','20-Apr-2020 12:08:30','20-Apr-2020 12:09:30','20-Apr-2020 12:10:30','20-Apr-2020 12:11:30','20-Apr-2020 12:12:30','20-Apr-2020 12:13:30'}; % Picarro
A = datetime(DateO,'InputFormat','dd-MM-yyyy HH:mm:ss');
B = datetime(DateT,'InputFormat','dd-MM-yyyy HH:mm:ss');
C = datetime(DateTh,'InputFormat','dd-MM-yyyy HH:mm:ss');
First_data=[-216,-220,-225,-232,-233,-233,-244,-246];
Second_data=[89,66,30,7,9.6,-5.8,-12,-16];
Third_data=[7.59,3.52,3.8,3.21,2,3.42,2.3,4.36];
figure;
left={Third_data,Third_data};
right={First_data,Second_data};
right_time = {A, B};
xlab=["Third data"," Third data"];
ylab=["First data","Second data"];
for i=1:length(left)
x = left{i};
y = right{i};
t = right_time{i};
if i == 1
Coeff(i) = max(y)/max(x);
else
Coeff(i) = max(x)/max(y);
end
y(y>0) = y(y>0).*Coeff(i);
y(y<0) = y(y<0).*Coeff(i);
Clb_values{i} = y;
Clb = Clb_values{i};
subplot(2,1,1*i);
yyaxis left;
plot(C, x);
ylabel('Third_data')
yyaxis right;
plot(t,Clb);
ylabel(ylab(i))
end
Clb_1=Clb_values{1,1};
Clb_2=Clb_values{1,2};
Thank you
5 Comments
Nora Khaled
on 20 Nov 2020
Could you explain more about what is the modification you are trying to do given those data
First_data=[-216,-220,-225,-232,-233,-233,-244,-246];
Second_data=[89,66,30,7,9.6,-5.8,-12,-16];
Third_data=[7.59,3.52,3.8,3.21,2,3.42,2.3,4.36];
I am not sure but if it works with positive and not with negatives, that mean the first polt is wrong and the second is partially wrong ? case the first data contains just negatives and the second contains some negative and some positives.
Kofial
on 20 Nov 2020
Nora Khaled
on 20 Nov 2020
The coffee for the first data is negative... and when multiplying with the data you reveres the shape.
So maybe you should consider taking the abslute value of the coeff and then shifiting the graph upward.
Matt J
on 20 Nov 2020
The coffee for the first data is negative...
Even better than decaf!
Nora Khaled
on 20 Nov 2020
xD
Answers (1)
Nora Khaled
on 20 Nov 2020
maybe this works ?
if i == 1
y=y+(x-y);
Coeff(i) = abs(max(y)/max(x));
else
Coeff(i) = abs(max(x)/max(y));
end
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