Function not processing dat for each element in array

Hi

I need to preform a simple temperture compensation.

I have variable called iot_temperture, which returns the temperture for 10K data points.

I then have a function

function per = tempcorrect(n)
%disp(n)
if n >= 50
    per = 115;
elseif n >=40
    per = 108;
elseif n >=30
    per = 145;
elseif n >=20
    per = 120;
% elseif temp >=10
%     per = 95;
% elseif temp >=0
%     per = 87;
% elseif temp >=-10
%     per = 85;
% elseif temp >=-20
%     per = 90;
% elseif temp >=-30    
%     per = 105;
else
    per = 100;
end
%     x=[-30,-20,-10,0,10,20,30,40,50]
         %    105    90    85    87    95   100   105   108   115
end

The function aims to return the the correction factor for a perticular

Therefore when i run the following code, i am expecting correction factor for all 10K points, but all i get is 100 ouput.

a=tempcorrect(iot_temperture)

What am i doing wrong?

Answer 1

[Voss]

The n given to tempcorrect is a vector (of length 10000). When you use a vector (or matrix, or any array) in an if statement, the if condition is considered true only if it is true for all elements of that vector (or matrix or array).

if [1 2 3] > 2
    disp('true');
else
    disp('false')
end

false

if [1 2 3] > 0
    disp('true');
else
    disp('false')
end

true

It turns out the n given to tempcorrect is such that all the if and elseif conditions checked in the function are false for at least one element of n, so the else block executes and you get per = 100 as output.

To have tempcorrect operate on one element of n at a time, you can give it one element of n at a time and collect the results in a vector:

a = zeros(size(iot_temperture))
for ii = 1:numel(iot_temperture)
    a(ii) = tempcorrect(iot_temperture(ii));
end

Or you can adapt tempcorrect to work for array inputs, either by putting the loop in the function itself, or by using logical indexing:

function per = tempcorrect(n)

per = 100*ones(size(n)); % initial value 100

per(n >= 50) = 115;
per(n < 50 & n >= 40) = 108;
per(n < 40 & n >= 30) = 145;
per(n < 30 & n >= 20) = 120;

% not necessary if using 100 as initial value:
% per(n < 20) = 100;

% per(n < 20 & n >= 10) = 95;
% per(n < 10 & n >= 0) = 87;
% per(n < 0 & n >= -10) = 85;
% per(n < -10 & n >= -20) = 90;
% per(n < -20 & n >= -30) = 105;

% not necessary if using 100 as initial value:
% per(n < -30) = 100;

end

[Dharmesh Joshi]

Thanks its working now.

[Voss]

You're welcome!

Answer 2

[Jan]

Remember, that Matlab's IF condition must be a scalar. If you provide n as a vector, the line:

if n >= 50    

implicitly performs:

if all(n >= 50) && ~isempty(n)    

You either need a loop over the elements of n or logical indexing:

per = zeros(size(n));
for k = 1:numel(n)
if n(k) >= 50
    per(k) = 115;
    ...        

or

per = repmat(100, size(n));
per(n >= 20) = 120;  % From small to large values
per(n >= 30) = 145;

or

Edge  = [-inf, 20,   30, 40,   50, inf];
Value = [100,  120, 145, 108, 115];
Group = discretize(n, Edge);
per   = Value(Group)