How to create function with name-value pair arguments?

Hello. I want to create the function like

function [x1, x2] = fun('name',varargin,'name',[..],'name',[..])

How i can do it?

[Simon Skillen]

I'd use the function argument method as described in https://de.mathworks.com/help/matlab/matlab_prog/function-argument-validation-1.html

Answer 1

[Walter Roberson]

https://www.mathworks.com/help/matlab/ref/inputparser.html

Note: in R2019b and later, see the new arguments keyword

https://www.mathworks.com/help/matlab/matlab_prog/function-argument-validation-1.html

Answer 2

[Voss]

fcn('name','booey','professional_life','noine','personal_life',2,'tooth_size','XXXL')

p = struct with fields:
                 name: 'booey'
    professional_life: 'noine'
        personal_life: 2
           tooth_size: 'XXXL'

fcn('name','Fred','home_planet','Mars','temper_quotient',Inf)

p = struct with fields:
               name: 'Fred'
        home_planet: 'Mars'
    temper_quotient: Inf

function fcn(varargin)

args = reshape(varargin,2,[]);
p = struct(args{:})

% do something with structure p

end

Answer 3

[Christopher Rowley]

To provide some sample code on how to do this with data in a structure, here is an example. We are have two variables 'x' and 'z' that are stored in structure 'params', and a test function looks to compute x+z +5. I have a simple check that the name is type character, but you can add more, or tailor to your needs. The function would look like:

function y = testfunction( params, varargin)
% set up name-value pairs for varargin.
for i = 1:2:length(varargin) % work for a list of name-value pairs
    if ischar(varargin{i}) % check if is character
        params.(varargin{i}) = varargin{i+1}; % override or add parameters to structure.
    end
end
y = params.z + params.x + 5;

And to try it, use:

params.x = 5;
params.z = 10;
testfunction( params) % try function
testfunction( params, 'x', 1) % try changing x from what it was set in the params structure.

In this example, the results should be 20, and 16, where the difference comes from the fact that x was decreased by 4 in the second function call.

Answer 4

[Black Woods]

function db=name_value_pairs(varargin)
if isempty(varargin) || mod(length(varargin),2)==1
    db={};
    return
end
jj=1;
for ii=1:2:length(varargin)
     if isa(varargin{ii},'char')
        db{jj,1}=varargin{ii};db{jj,2}=varargin{ii+1};jj=jj+1;
     else         
         db={};
         return
     end
end