Function with 2 variables

Hi all,

I have a function y=f(x, Z), when I give inputs, it takes Z as a fix value. Would you please advise how can my function accept z as a variable? I need different inputs of X and Z to get different outputs of y.

Kind regards,

[James Tursa]

Please show the actual code you are using, not just a description of it. You can certainly pass two variables with the f(x,Z) syntax you mention, but only if the function is coded properly.

[Keen]

Hi James, I sent you my coding via email would you please check it and email back when you feel free.

Kind regards, Keen

Answer 1

[James Tursa]

If you are passing in arguments that are vectors, then everything about your calculations inside the function needs to be vectorized. You've got divisions in your calculations that are not vectorized. E.g., this

CA= 10.^(0.1482*(((log(ZP)+3.0166)/(-0.1782*log(SA)-1.0026)).^2)+...

should be this instead

CA= 10.^(0.1482*(((log(ZP)+3.0166)./(-0.1782*log(SA)-1.0026)).^2)+...  <-- changed / to ./

Also, your if-tests will not work with vectors since some of the elements might satisfy the condition while other elements do not. MATLAB isn't going to magically jump into the if-block for only those elements that satisfy the condition. So you need to rewrite that part of your code as well. (e.g., maybe using simple loops)

Answer 2

[Keen]

Hi James, I tried it and is the same error appearing. The equation is correct because the input of the first values of ZP and SA gave a correct value. The problem is my program accepts only one value of SA.

[James Tursa]

Please show us your current code.

Answer 3

[Keen]

function [CA]=myfunction2(ZP,SA)
CA= 10.^(0.1482*(((log(ZP)+3.0166)/(-0.1782*log(SA)-1.0026)).^2)+(0.8796*((log(ZP)+3.0166)/(-0.1782*log(SA)-1.0026)))+3.207);
if ZP>= 0 
CA= 10.^(0.1482*(((log(ZP)+3.0166)/(-0.1782*log(SA)-1.0026)).^2)+(0.8796*((log(ZP)+3.0166)/(-0.1782*log(SA)-1.0026)))+3.207);
elseif ZP< 0 
CA= 10.^(0.1482*(((-(log(-ZP))+3.0166)/(-0.1782*log(SA)-1.0026)).^2)+(0.8796*((-(log(-ZP))+3.0166)/(-0.1782*log(SA)-1.0026)))+3.207);
end
if (CA>= 0)&& (CA <= 10)
 CA=0 , disp('stronglywaterwet & CA=0')  %#ok<NOPRT>
elseif (CA >= 10)&& (CA <= 89) %#ok<*BDSCA>
    disp ('waterwet');
elseif (CA >= 90)&& (CA <= 180) %#ok<*BDSCA>
    disp ('oilwet');
plot(ZP,CA)
end

NB: my inputs are ZP=[-41 -45 -53], SA=[0.001 0.005 0.01] my outputs CA should be [79 93 110]

[James Tursa]

You haven't made the changes I suggested to you. You need to change the / matrix division operator to the ./ element-wise division operator (note the period). And you haven't changed your if-tests to be vectorized either (either use logical indexing or loop over the elements one-by-one). You will continue to have problems until you make these changes.

[Keen]

I MADE THE REQUESTED CHANGES, BUT I`M STILL STUCK.

Answer 4

[Keen]

Hi I made the changes and nothing happened to my results hence I removed it.

Answer 5

[Keen]

function [CA]=myfunction2(ZP,SA)
CA= 10.^(0.1482*(((log(ZP)+3.0166)./(-0.1782*log(SA)-1.0026)).^2)+(0.8796*((log(ZP)+3.0166)./(-0.1782*log(SA)-1.0026)))+3.207);
if ZP>= 0 
CA= 10.^(0.1482*(((log(ZP)+3.0166)./(-0.1782*log(SA)-1.0026)).^2)+(0.8796*((log(ZP)+3.0166)./(-0.1782*log(SA)-1.0026)))+3.207);
elseif ZP< 0 
CA= 10.^(0.1482*(((-(log(-ZP))+3.0166)./(-0.1782*log(SA)-1.0026)).^2)+(0.8796*((-(log(-ZP))+3.0166)./(-0.1782*log(SA)-1.0026)))+3.207);
end
for n=1:3
    ZP=ZP(n);
end
for n=1:3
    SA=SA(n);
end

[Stephen23]

@keen: you need to either vectorize your code or use loops properly, indexing into the output array. The loops you have written will always throw errors:

for n=1:3
    ZP=ZP(n);
end

On the first iteration (assuming that ZP non-empty) then your loop redefines ZP to be scalar (because you redefine ZP to be the first element of ZP). On the second iteration you try to access the second element of ZP, which does not exist because you just redefined ZP to be scalar (so it only has one element).

You need to learn how to debug your code, which means actually looking at what the code is really doing (not what you want or think it is doing) and to read about code vectorization and how to use loops:

https://www.mathworks.com/help/matlab/matlab_prog/debugging-process-and-features.html

https://www.mathworks.com/help/matlab/matlab_prog/vectorization.html

https://www.mathworks.com/help/matlab/matlab_prog/preallocating-arrays.html

Answer 6

[Keen]

My problem is my function cannot read the variable SA (2nd and 3rd value).

Kind regards,

Answer 7

[per isakson]

See log, Natural logarithm

and try

>> ZP=[-41 -45 -53]; SA=[0.001 0.005 0.01];
>> myfunction2( ZP, SA )
ans =
   79.7988   92.8386  109.8079

where

function  CA = myfunction2( ZP, SA )
    %
    fpos = @(z,s) 10.^(0.1482*(((log10(z)+3.0166)./(-0.1782*log10(s)-1.0026)).^2) ...
                  +(0.8796*((log10(z)+3.0166)./(-0.1782*log10(s)-1.0026)))+3.207) ;
    fneg = @(z,s) 10.^(0.1482*(((-(log10(-z))+3.0166)./(-0.1782*log10(s)-1.0026)).^2)...
                  +(0.8796*((-(log10(-z))+3.0166)./(-0.1782*log10(s)-1.0026)))+3.207);
    %
    ispos = ZP >= 0;
    isneg = ZP < 0;
    %
    CA = nan(size(ZP));
    CA( ispos ) = fpos( ZP(ispos), SA(ispos) );
    CA( isneg ) = fneg( ZP(isneg), SA(isneg) );
end

Answer 8

[Keen]

Many thanks,

Yes it works perfectly.

please advise about my errors.

Kind regards,