Table and Array Indexing using loops

Hi everyone,

I have a challenge in indexing in a for loop.

I have a 696 by 2 table with a date identifier. This is as captured below. I would like to create a [648 4] from the [696 2] using the date identifier or otherwise.

I would like to create a new table such that

i=24;
n=numel(24:1:648);
H=zeros(n, 3);
for N=1:n
    for i=24:648
        i=[(i-days(1):days(1):i+days(2))] 
        % yields for me a new matrix for i=[GHI([i-day(1) i i+day(1)] 
        % which then gives me [648 3]
        i=i+1;
   
    end
    H(N,:)=table2array(hvalues(i,2:4));
end

Any help on the above is highly appreciated.

[dpb]

The indexing expression is simply

newdata(i-24,:)=data([i-24 i i+24],2);

[Guillaume]

Rather than giving us code that makes no sense at all, why don't you explain what it is you want.

[Image Analyst]

and attach a mat file with the table in it so we can code up something with your actual data. This will save you time in getting an answer from us.

[Walter Roberson]

Your table2array() call is using i after the end of the for i loop. The for loop would leave i with the last value it was set to during the loop. Each iteration of for i, any changes that were made to i during previous iterations are ignored and i will be set to the next value in sequence. The operation is like

hidden_i_start = 24;
hidden_i_increment = 1;
hidden_i_end = 648;
hidden_i = hidden_i_start;
if hidden_i_start > hidden_i_end
    i = [];
else
  while hidden_i <= hidden_i_end
      i = hidden_i;
      i = i-days(1):days(1):i+days(2);
      i = i + 1;
      hidden_i = hidden_i + hidden_i_increment;
  end
end

Observe that no matter what change you make to i in the loop, the next iteration i will get changed back to the next number in the sequence established by the initial for specification, but that on the last iteration, i will not be changed. So at the end of the for i loop with you changing i in that way, you will end up with i being set to (648-days(1)):days(1):648+days(2)) + 1

and then after that loop, you will use that final i as the row index into hvalues. Are you sure that is a good idea?

[Lui]

Given the table in the attached file create a new table with two more columns with data such that:

% row_date(i) = ([GHI((i)-days(1)) GHI(i) (GHI(i)+days(1))])

for example from the attached file

% values for (2004 02 02 10 0 0) = [137.42 118.9442 201.7317]. 

We initialize at i=25 to give for the one day allowance.

The aim is to increase the sample size of my data by sampling the data at 24 hours intervals.

Thank you.

[Walter Roberson]

Lui comments to dpb:

This works. I appreciate your help.

[Rik]

Comment posted as flag by Lui:

Would you please let me know why this does not work when I want to create n columns instead of 3.

 data =table2array(data(:,2));
 newdata(i-48, :)=data([i-48 i-24 i i+24 i+48], 1)

it returns an error. What could be the challenge? @dpb

[dpb]

What's the specific error text (all the red from the command line in context)???

I'd have to wrap my head around the original Q? again to remind myself but there's nothing specifically wrong with the expression other than one has to ensure that i-48 is >=1 and newdata hasn't been previously defined as a noncormant size that this vector won't fit into...but functionally, it's the same idea, yes, and "it will work" given the constraints of where it is attempted are legitimate.

[Walter Roberson]

We do not know which version of the code was being used, so we do not really have any useable context for thinking there is an error.

Answer 1

[Guillaume]

The aim is to increase the sample size of my data by sampling the data at 24 hours intervals.

why didn't you say that in the first place? That's a typical case of XY problem. You want to do something, don't know how to do it but think it can be done one way so ask about that way. Except that's not how it's done.

The simplest way to resample timed data is to convert your table into a timetable then use retime to resample it to whichever interval you want

hvalues = table2timetable(hvalues);
newvalues = retime(hvalues, 'daily', 'linear');  %resample at 24 hours interval using linear interpolation

However, I'm a bit confused, your data is hourly so I'm not sure how resampling in 24 hours interval is going to increase the sample size.

[Lui]

This gives me hourly values which I can easily retrieve. But I appreciate as this helps me elsewhere.