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# Multiply array if number is under certain value

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Im using a script to calculate power from two varaiables. Though i need to multiply my power (P_out) with a certain value (for example 2) if the value is under 1.000.000 to correct my data...

So i have 144 numbers, and all numbers beneath 1.000.000 needs to be multiplied by 2. Maybe 30 of these values are beneath 1.000.000...

How do i do this? Hope someone knows this

This is the code:

Data = readtable('Rejstrup Pyra .xlsx'); %Celcius

T_Pyra = table2array(Data(83:(223),8))

Radiation = table2array(Data(83:(223),3)) %kW/m^2

PV_radiation = Radiation./1000

T_modul = (T_Pyra)+(PV_radiation/20) %Celcius

T_korrektion = 1-(T_modul-T_Pyra).*(0.40/100)

I_korrektion = 0.988

P_peak = 23947000 %watt

P_out = P_peak.*T_korrektion.*I_korrektion.*PV_radiation %Watt

^^^^^^^^

This is my values for my power, where every value beneath 1.000.000 needs to be multiplied with 2.

Exporting it to excel (not relevant):

table_name=table(P_out1)

filnavn='Beregnet P_available fra Rejstrup.xlsx'

overskrift='P_available'

writetable(table_name,filnavn,'Sheet',overskrift)

Best regards

Peter

### Answers (1)

Mathieu NOE
on 22 Mar 2022

hello

this is very simple - see below :

P_out = 1000000 + 80000*randn(10,1); % dummy data

plot(P_out)

hold on

ind = P_out<=1000000; % find values below 1000000

P_out(ind) = P_out(ind)*2; % correction (put the right correction factor here)

plot(P_out,'-*')

hold off

##### 29 Comments

Peter Lyngbye
on 22 Mar 2022

So can i just copy the lower part into my script and it should multiply every value beneath 1.000.000 with 2?

I dont understand the top part... It is supposed to give me an excel were every value above 1.000.000 is not affected.

Best regards

Mathieu NOE
on 22 Mar 2022

ok

can you show the updated code and for my info , what are we comparing ? the picture shows 3 vectors (P_out, power , matlab) ... what is the supposed P_out before and after correction in your code ?

Peter Lyngbye
on 22 Mar 2022

This is the full code after adding yours:

Data = readtable('Rejstrup Pyra .xlsx'); %Celcius

T_Pyra = table2array(Data(87:(222),8));

Radiation = table2array(Data(87:(222),3)); %kW/m^2

PV_radiation = Radiation./1000;

T_modul = (T_Pyra)+(PV_radiation/20); %Celcius

T_korrektion = 1-(T_modul-T_Pyra).*(0.40/100);

I_korrektion = 0.988;

P_peak = 23947000; %watt

P_out = P_peak.*T_korrektion.*I_korrektion.*PV_radiation; %Watt

P_out = 1000000 + 80000*randn(10,1); % dummy data

plot(P_out)

hold on

ind = P_out<=1000000; % find values below 1000000

P_out(ind) = P_out(ind)*2; % correction (put the right correction factor here)

plot(P_out,'-*')

hold off

The script is supposed to add temperature and iradiation from a solar park, and via formulas give the power collected from them which is P_out. But becuase of shadows in the park at early ours i need to multiply these values with a correction factor. The data from the excel have 5 minute logs of the radiation and temperature, which i will run through the script and compare to actual pwer readings from the park. I have added one of the excels for iradtioan and temp... I just noticed that actually telling matlab to multiply the early hours instedet of values beneath 1.000.000 would be more correct. I hope u understand what im asking, but a script multiplying for example the readings between 6:30 and 9:30 would be absolutely perfect! it is 24 hour clock

Best regards!

Mathieu NOE
on 22 Mar 2022

hello again

the error came because you left my "dummy" data line active in your code

P_out = 1000000 + 80000*randn(10,1); % dummy data

that was of course to be removed

now it's working fine ,

but not yet to the point of being matching the new request for correction between 6:30 and 9:30

Data = readtable('Rejstrup Pyra .xlsx'); %Celcius

T_Pyra = table2array(Data(87:(222),8));

Radiation = table2array(Data(87:(222),3)); %kW/m^2

PV_radiation = Radiation./1000;

T_modul = (T_Pyra)+(PV_radiation/20); %Celcius

T_korrektion = 1-(T_modul-T_Pyra).*(0.40/100);

I_korrektion = 0.988;

P_peak = 23947000; %watt

P_out = P_peak.*T_korrektion.*I_korrektion.*PV_radiation; %Watt

plot(P_out)

hold on

ind = P_out<=1000000; % find values below 1000000

P_out(ind) = P_out(ind)*2; % correction (put the right correction factor here)

plot(P_out,'-*')

hold off

Peter Lyngbye
on 22 Mar 2022

Okay got it. But i dont understand the graph u show... I need to compare the graph to another graph which is why i need to be able to correct the data between 6:30 and 9:30 by a simple correction factor :/

i have added a picture so it is easier to understand my problem. The blue line is my calculations coming from matlab going into excel. As u can see i need correction in the start (6:30-9:30).

I am really appreciating ur help!

Peter Lyngbye
on 22 Mar 2022

I understand ur code, but i need it to get tranfered into my excel with the other data.

This is the bottom of my code, where the data gets into excel. It is here each data point beneath 1.000.000 needs to be corrected, so when i plot my function in excel it will have the corrrection

table_name=table(P_out);

filnavn='Beregnet P_available fra Rejstrup.xlsx';

overskrift='P_available';

writetable(table_name,filnavn,'Sheet',overskrift)

Mathieu NOE
on 22 Mar 2022

I was just finishing to modify your code for the correction in amplitude (x2 befoe 9:30 AM)

this is the code (a bit improved) and the result

but this is not what your last picture is telling me . It's a different correction method

so far the code is as pr your initial request

Data = readtable('Rejstrup Pyra .xlsx'); %Celcius

ind_start = 87;

ind_stop = 222;

Data = Data(ind_start:ind_stop,:); % retrict all table data between two time indexes from 7:10 AM to 18:25 PM

Data_time = datetime(Data.GeneratedOn);

T_Pyra = table2array(Data(:,8));

Radiation = table2array(Data(:,3)); %kW/m^2

PV_radiation = Radiation./1000;

T_modul = (T_Pyra)+(PV_radiation/20); %Celcius

T_korrektion = 1-(T_modul-T_Pyra).*(0.40/100);

I_korrektion = 0.988;

P_peak = 23947000; %watt

P_out = P_peak.*T_korrektion.*I_korrektion.*PV_radiation; %Watt

plot(Data_time,P_out)

hold on

ind = 116 - ind_start; % find time line corresponding 9:30 AM

P_out(1:ind) = P_out(1:ind)*2; % correction (put the right correction factor here)

plot(Data_time,P_out,'-*')

legend('before correction','after correction');

hold off

Mathieu NOE
on 22 Mar 2022

Please clarify

this is your picture a bit modified by myself (the red line is the extrapolation of the blue line for the first hours) : do you mean this is what you want to achieve ?

but the P_out graph I get even before any correction does not show this initial drop

this is what I get when I start the plot even earlier

Data = readtable('Rejstrup Pyra .xlsx'); %Celcius

ind_start = 87-40;

ind_stop = 222;

Data = Data(ind_start:ind_stop,:); % retrict all table data between two time indexes from 7:10 AM to 18:25 PM

Data_time = datetime(Data.GeneratedOn);

T_Pyra = table2array(Data(:,8));

Radiation = table2array(Data(:,3)); %kW/m^2

PV_radiation = Radiation./1000;

T_modul = (T_Pyra)+(PV_radiation/20); %Celcius

T_korrektion = 1-(T_modul-T_Pyra).*(0.40/100);

I_korrektion = 0.988;

P_peak = 23947000; %watt

P_out = P_peak.*T_korrektion.*I_korrektion.*PV_radiation; %Watt

plot(Data_time,P_out)

Mathieu NOE
on 22 Mar 2022

hmmm

this is how your Data table names are defined in my workspace (I run R2020b)

maybe the name of the first columns of Data is different in your matlab. Please correct on your side if you have a different name

Peter Lyngbye
on 22 Mar 2022

I want my calculation form matlab (the blue line) to be as close to the gray line (real life data). To do this i simply want to correct the data i get from my matlab script by correcting it primaryli in the start as the two lines are far apart here. I would think the script would look like my original script with something like:

- Between two points taking from my excel (showing radiation and temperature at some logging time) the calculations done by the script are multiplied by some factor for example 2. This way the blue and gray line will be closer to each other hence be more correct.

I

Peter Lyngbye
on 22 Mar 2022

Mathieu NOE
on 22 Mar 2022

Sorry but the correction you want is not a factor 2. it much less than that (and it varies with the time value) but I cannot guess it because the grey reference line is not supplied in your data ...

also I don't understand why the blue should match the grey curve simply for the appearance , you are introducing a bias on purpose in your code just to make things look better ??

In the mean time I fixed the previous problem (Unrecognized table variable name 'GeneratedOn'.)

Data = readtable('Rejstrup Pyra .xlsx','VariableNamingRule' ,'preserve'); %Celcius

ind_start = 87-40;

ind_stop = 222;

Data = Data(ind_start:ind_stop,:); % retrict all table data between two time indexes from 7:10 AM to 18:25 PM

Data_time = datetime(table2array(Data(:,1)));

T_Pyra = table2array(Data(:,8));

Radiation = table2array(Data(:,3)); %kW/m^2

PV_radiation = Radiation./1000;

T_modul = (T_Pyra)+(PV_radiation/20); %Celcius

T_korrektion = 1-(T_modul-T_Pyra).*(0.40/100);

I_korrektion = 0.988;

P_peak = 23947000; %watt

P_out = P_peak.*T_korrektion.*I_korrektion.*PV_radiation; %Watt

plot(Data_time,P_out)

Mathieu NOE
on 22 Mar 2022

ok

let's start again from the beginning

where are the data relative to the grey (reference) and your data (blue)

I cannot see anything like the blue line with the initial drop

I wonder if your matlab code is about is the grey curve and not the blue ??

Peter Lyngbye
on 22 Mar 2022

Peter Lyngbye
on 22 Mar 2022

Peter Lyngbye
on 22 Mar 2022

As u can see on the picture i now need to correct the later stages of the day around 16:00-18:30

Mathieu NOE
on 22 Mar 2022

okay

it's my turn to become nuts ... :) I still don't get what you want to correct from the P_out generated by the matlab code.

The top plot is your curves , the bottom one is the full 24 hours of P_out as it comes without any further corrections beside what you have already coded

so what correction in the lower plot do you want to do ? that is still unclear to me - make a handheld sketch if you want.

code used for the lower plot :

clc

clearvars

Data = readtable('Rejstrup Pyra .xlsx','VariableNamingRule' ,'preserve'); %Celcius

[m,n] = size(Data);

ind_start = 1; % 87-40

ind_stop = m; % 222

Data_time = datetime(table2array(Data(:,1)));

T_Pyra = table2array(Data(:,8));

Radiation = table2array(Data(:,3)); %kW/m^2

PV_radiation = Radiation./1000;

T_modul = (T_Pyra)+(PV_radiation/20); %Celcius

T_korrektion = 1-(T_modul-T_Pyra).*(0.40/100);

I_korrektion = 0.988;

P_peak = 23947000; %watt

P_out = P_peak.*T_korrektion.*I_korrektion.*PV_radiation; %Watt

plot(Data_time,P_out)

Mathieu NOE
on 22 Mar 2022

Peter Lyngbye
on 22 Mar 2022

Edited: Peter Lyngbye
on 22 Mar 2022

Ill try to explain step by step hehe,

- I have and excel were i have the 100% correct power is shown which is illustrated by the grey curve.
- The blue curve is my calculations done on temperature and radiation achieved over the same day the grey curve were formed. This means that if my script were perfect, the blue line would be exactly like the grey line.
- BUT the blue line i get is a bit shiftet in the early hours which is were i need to correct it by some 0.97 ISH factor so that it will compensator for the shadow which occours in the early mornings due to the sun being low in the sky.
- The shadows causes my power calculations to be slighty higher. This is exactly why i need this correction factor :)

Peter Lyngbye
on 22 Mar 2022

Also i understand why u are confussed since the curve which is corrected inside the matlab is the P_out calculated curve (the blue line) and not the grey curve which is plottet i my own excel... I have now added the excel.. This shows the true power generated over a day. This is what should be included inside the script instead of the: "P_out(1:ind)" it should be the excel data.

It is the collon B which has the power data in watt.. The other data is just inverter data which is not relevant. The collon B makes a curve which should simply be the curve which i need to correct my "P_out(1:ind)*0.97" too.

Mathieu NOE
on 22 Mar 2022

ok

this is where you should have started first

I took your last excel file (I removed the columns above B to speed up the code) into consideration

now I can draw the two lines (and understand the purpose of the code)

you can do all modifications you want up to force the blue to match the grey (simply compute the amplitude ratio at each time interval... or build a strategy around that)

so far it seems now you have reached your target...

clc

clearvars

%% reference data (grey)

Data_ref = readtable('Rejstrup Effekt.xlsx','VariableNamingRule' ,'preserve'); %Celcius

[m,n] = size(Data_ref);

ind_start = 60; %

ind_stop = 222; %

Data_ref = Data_ref(ind_start:ind_stop,:); % retrict all table data between two time indexes from 7:10 AM to 18:25 PM

Data_ref_time = datetime(table2array(Data_ref(:,1)));

P_out_ref= table2array(Data_ref(:,2));

plot(Data_ref_time,P_out_ref)

%% measurement (blue, to be corrected)

Data = readtable('Rejstrup Pyra .xlsx','VariableNamingRule' ,'preserve'); %Celcius

[m,n] = size(Data);

Data_time = datetime(table2array(Data(:,1)));

T_Pyra = table2array(Data(:,8));

Radiation = table2array(Data(:,3)); %kW/m^2

PV_radiation = Radiation./1000;

T_modul = (T_Pyra)+(PV_radiation/20); %Celcius

T_korrektion = 1-(T_modul-T_Pyra).*(0.40/100);

I_korrektion = 0.988;

P_peak = 23947000; %watt

P_out = P_peak.*T_korrektion.*I_korrektion.*PV_radiation; %Watt

% heavy correction : make the blue match exactly the grey

cor_factor = P_out_ref./(P_out+eps); % + eps to avoid division by zero !

P_out_corrected = P_out.*cor_factor;

plot(Data_ref_time,P_out_ref,'k',Data_time,P_out,'b',Data_time,P_out_corrected,'c*')

legend('ref (grey)','blue (uncorrected)','blue (100% corrected)');

Mathieu NOE
on 22 Mar 2022

Peter Lyngbye
on 22 Mar 2022

This is perfect!

Only last thing i need is this adjustment but for the other half of the curve (late hours):

ind = 130 - ind_start; % find time line corresponding 9:30 AM

P_out(1:ind) = P_out(1:ind)*0.97; % correction (put the right correction factor here)

Right now it only adjust the early hours :/

Mathieu NOE
on 22 Mar 2022

ok

I put back those lines in the code (see below) , but I have a hard time to see what the 0.97 makes as an improvement ?? or did I forgot another piece of code from the past ?

this is not doing any shift in the early hours...

see figure 1 (below)

figure 2 is simply the "heavy" correction method I already posted above

clc

clearvars

%% reference data (grey)

Data_ref = readtable('Rejstrup Effekt.xlsx','VariableNamingRule' ,'preserve'); %Celcius

% [m,n] = size(Data_ref);

ind_start = 60; %

ind_stop = 222; %

Data_ref = Data_ref(ind_start:ind_stop,:); % retrict all table data between two time indexes from 7:10 AM to 18:25 PM

Data_ref_time = datetime(table2array(Data_ref(:,1)));

P_out_ref= table2array(Data_ref(:,2));

plot(Data_ref_time,P_out_ref)

%% measurement (blue, to be corrected)

Data = readtable('Rejstrup Pyra .xlsx','VariableNamingRule' ,'preserve'); %Celcius

% [m,n] = size(Data);

Data_time = datetime(table2array(Data(:,1)));

T_Pyra = table2array(Data(:,8));

Radiation = table2array(Data(:,3)); %kW/m^2

PV_radiation = Radiation./1000;

T_modul = (T_Pyra)+(PV_radiation/20); %Celcius

T_korrektion = 1-(T_modul-T_Pyra).*(0.40/100);

I_korrektion = 0.988;

P_peak = 23947000; %watt

P_out = P_peak.*T_korrektion.*I_korrektion.*PV_radiation; %Watt

%% basic correction

P_out_corrected = P_out; % init P_out_corrected

ind = 116 - ind_start; % find time line corresponding 9:30 AM

Data_ref_time(ind)

P_out_corrected(1:ind) = P_out_corrected(1:ind)*0.97; % correction # 1 (put the right correction factor here)

figure(1),

plot(Data_ref_time,P_out_ref,'k',Data_time,P_out,'b',Data_time,P_out_corrected,'c*')

legend('ref (grey)','blue (uncorrected)','blue (corrected)');

%% full correction (remove code below if necessary)

% make the blue match exactly the grey

P_out_corrected = P_out; % init P_out_corrected

ind = P_out<eps;

P_out_ref(ind) = 0; % avoid very high (inf) cor_factor due to division by eps !

cor_factor = P_out_ref./(P_out+eps); % + eps to avoid division by zero if P_out = 0 !

P_out_corrected = P_out_corrected.*cor_factor;

figure(2),

plot(Data_ref_time,P_out_ref,'k',Data_time,P_out,'b',Data_time,P_out_corrected,'c*')

legend('ref (grey)','blue (uncorrected)','blue (corrected)');

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