how to create an index that keep tracking the hours in the data

Asked by Ida 1 on 30 Aug 2019

Latest activity Commented on by Ida 1 on 3 Sep 2019

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Is anyone helping me with a simple code ???

In the code below I want to find the peak load and average peak price in peak hour from 7-10 and then do it the same for each peak hours (11-15, 16-21 and 23-6am. I write the code below but I couldn't figure out how to calculate the peak load and average peak price for each peak with the same code

h=twomonth(:,1)
 p1= 7 <= h & h < 10
 price=twomonth(:,2);
 price=price.*p1  
 demand=twomonth(:,3);
 demand=demand.*p1
 Index1=(7:24:N)'; ss1=10-1
N1=length( Index1);
Data1=zeros(N1,2);
for k=1:N1
    kk=Index1(k); kkk=kk+ss1-1;
    dd=sum(d(kk:kkk));pp=sum(p(kk:kkk).*d(kk:kkk)); pp=pp/dd;
    Data1(k,:)=[pp dd];
end

2 Comments

Star Strider on 30 Aug 2019

What does ‘isn’t working’ mean when you try to run it?

I can’t run it.

The error:

d=load.*p1
Error using load
Unable to read file 'matlab.mat'. No such file or directory.

Please do not name a variable ‘load’. It overshadows the load function, necessary for reading .mat files and other files.

What do you want to do in that line? Note that you are overwriting ‘d’ that you defined in the line just before it.

Also, ’p’, and ‘N’ are not defined.

Ida 1 on 30 Aug 2019

Sorry Star Strider, you are right! I have fixed my mistakes in the code above.

Hope I was clear enough and you can help me!

1 Answer

Link

Answer by Guillaume on 2 Sep 2019

Edited by Guillaume on 2 Sep 2019

Accepted Answer

Here's how I'd do it:

%twomonth: a Nx3 matrix, whose first column is hour, 2nd is price, 3rd is demand
bin = discretize(mod(twomonth(:, 1)-7, 24), [7, 11, 16, 22, 23, Inf]-7);  %offset bin edges by 7 in order to put the 23-6 in the same bin (5)
%bin == 1 is [7-11), bin == 2 is [11-16), bin == 3 is [16-22), bin == 4 is [22], bin == 5 is [23-6]
avgprice = splitapply(@mean, twomonth(:, 2), bin);
peakdemand = splitapply(@max, twomonth(:, 3), bin);
result = table({'7-10'; '11-15'; '16-21'; '22'; '23-6'}, avgprice, peakdemand, 'VariableNames', {'period', 'avgprice', 'peakdemand'})

I've included the lone 22 in the result, you can always remove row 4 if you don't want it.

You have to do a little bit of arithmetic as discretize can't give you the same bin number for the start and end of a range, but once you've got your bin numbers it's trivial to calculate your mean and peak with splitapply.

14 Comments

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Guillaume on 2 Sep 2019

There's currently no day information in your matrix. If we assume a new day when the hours go from 23 to 0, then:

daybin = cumsum([1; diff(twomonth(:, 1)) < 0])

then, it's pretty much just the same, with an addition of a findgroups:

hourbin = discretize(mod(twomonth(:, 1)-7, 24), [7, 11, 16, 22, 23, Inf]-7);  %offset bin edges by 7 in order to put the 23-6 in the same bin (5)
hourranges = {'7-10'; '11-15'; '16-21'; '22'; '23-6'};
[bin, actualday, actualhour] = findgroups(daybin, hourbin);
actualhour = hourranges(actualhour);
avgprice = splitapply(@mean, twomonth(:, 2), bin);
peakdemand = splitapply(@max, twomonth(:, 3), bin);
weightedprice = splitapply(@(price, demand) mean(price .* demand), twomonth(:, 2), twomonth(:, 3), bin); 
result = table(actualday, actualhour, avgprice, peakdemand, weightedprice)

As for, "So hourRanges must be daily and another problem is that some days during the year has more/less hours due to daylight saving", that's vastly different from your original question and will require a different approach altogether. Currently, the necessary information is not in your data so I'm not sure what you want us to do about it. At the very least your data would need to contain date information (as datetime). A timetable would be more suitable for that instead of a matrix. We'd also need to know how you define the hour range for a particular day.

Ida 1 on 3 Sep 2019

Thank you Guillaume! I just run your code, and I have gotten this error only from the last line:

result = table(groupday, starthour,avgprice, peakdemand, weightedprice)

Error using table (line 231)

All input variables must have the same number of rows.

Guillaume on 3 Sep 2019

Oh yes, starthour probably needs transposing

result = table(groupday, starthour.', avgprice, peakdemand, weightedprice)

if it's not that then please give the size of each variables.

Ida 1 on 3 Sep 2019

Thank you very much Guillaume! Now the code is OK!

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