Function to fill an array given conditions.

Hi All,

I am trying to write a function to fill an array given conditions. basiclly the user inputs two variables a max and min and given these

values it creates two new row vectors, ph and date, which is a subset of an orignal vector.

My code is given below -

function [datenew,phnew] = subsetdata(date, ph, pressure, min, max)

%create empty array
datenew=cell(1,5000);
phnew=cell(1,5000);
%loop through each point
for i=1:size(pressure)
    %check condition
    if (pressure(i,:)>=min & pressure(i,:)<max)
        %fill array if condition met
        datenew{i}=date(i);
        phnew{i}=ph(i);
        
    end

end
end

[Andrew Czeizler]

date, ph and pressure are row vectors. we go through and check pressure againes max and min. Then if at that point its between the two values we take date and ph at that index and fill and new vector. Returning a new vector date and ph which meets the conditions.

[Adam]

What is your question?

Also, never use min and max as variable names. These are function names in Matlab and by creating variables with those names you hide the respective functions and make them inaccessible to use.

Answer 1

[Adam]

validIdx = pressure >= minVal & pressure < maxVal;
datenew = date( validIdx );
phnew = ph( validIdx );

should do this for you without needing the loop and using logical indexing instead. I don't know why you are putting results in a cell array though. Never use a cell array when a numeric array will do the job.

[Andrew Czeizler]

looking sharp!

Thank you :).

Will give it a try.

I was wondering, what should I use instead? nan()?

Best,

Andrew

[Adam]

Ah, well, if you want to keep positions with NaNs inbetween then you will need to do it slightly different:

datenew = nan( size( date ) );
phnew = nan( size( date ) );
validIdx = pressure >= minVal & pressure < maxVal;
datenew( validIdx ) = date( validIdx );
phnew( validIdx ) = ph( validIdx );

Answer 2

[Andrew Czeizler]

Hi Adam!

How would you define the numeric array. Im not sure of best practice here.

many thanks,

[Adam]

See my reply above.

[Andrew Czeizler]

so you would use a nan numeric array and then fill it based on conditions?

many thanks,

best,

anAndrew

[Adam]

If you want to retain the positions within the array then yes. You could use -1 if all true data should be positive, but NaN is more logical.