name value pairs with variable input arguments

write a function called name_value_pair that has a variable no of input arguments representing name value pairs.The function must be called with even number of input arguments.The functiion returns a single cell array that has exactly two columns. First with names and another with values.If the functiion is called with no input arguments or odd number of input arguments or if the name is not of char type then it should return an empty cell array

function db = name_value_pairs(varargin)
if rem(nargin,2) ~= 0 || nargin < 2 
    db = {};
    return
end
if ~ischar(varargin(1:2:nargin))
    db = {};
    
end
    for i = 1:nargin/2
        db(i,1) = varargin(2*i-1);
        db(i,2) = varargin(2*i);
    end
end

this is code that i have written and ii cant understand where the error is. please help me with this.

[KSSV]

What error you are getting?

[M NAGA JAYANTH AVADHANI]

i am not getting an empty cell array if the name is not of type char

[hesham hany]

+1

but I didn't get the answer below

Answer 1

[Walter Roberson]

if ~ischar(varargin(1:2:nargin))
  db = {};

else   %changed

However you need to change what you are testing in the if. varargin() with () indices gives a cell array and ischar is false for cell array. You need to test the content of each cell. I suggest cellfun(@ischar, stuff)

Then you need to deal with the fact that the ischar will be true for some entries and false for others, but if is considered true only if all the entries are true. You need to be testing if ANY entry is not character, and equivalent to that would be to test whether ALL ischar is false.

[M NAGA JAYANTH AVADHANI]

thank you it worked!!

[Atif Penkar]

Still get the same error

[Walter Roberson]

Atif what is your code?

Answer 2

[Sovendo Talapatra]

for j = 1:2:nargin
    if ~ischar(varargin{j})
        db = {};
        return
    else
        continue
    end

By using for loop you can easily fix the code

Answer 3

[Abdul Quadir Khan]

function db = name_value_pairs(varargin)
if rem(nargin,2) ~= 0 || nargin < 2 
    db = {};
    return
end
if ~ischar(varargin(1:2:nargin))
  db = {};

else   %changed 
end
for i = 1:nargin/2
        db(i,1) = varargin(2*i-1);
        db(i,2) = varargin(2*i);
    end
    for j = 1:2:nargin
    if ~ischar(varargin{j})
        db = {};
        return
    else
        continue
    end
end

Answer 4

[Selman Baysal]

"varargin" is a cell array; this is why we cannot use something like varargin{1:2:nargin} and ~ischar(varargin(1:2:nargin)) gives 1 everytime. I think using two different for loops make this problem easier like:

function db = name_value_pairs(varargin)
l = nargin;
if rem(l,2) ~= 0 || l < 2
    db = {};
    return
end
for ii = 1:2:l
    if ~ischar(varargin{ii})
        db = {};
        return
    end
end
for ii = 1:l/2
    db(ii,1) = varargin(2*ii-1);
    db(ii,2) = varargin(2*ii);
end
end

Answer 5

[Yifan He]

function db = name_value_pairs(varargin)
if nargin == 0
    db = {};
elseif nargin/2 ~= fix(nargin/2)
    db = {};
elseif sum(cellfun(@ischar,varargin(1:2:end))) ~= nargin/2
    db = {};
else
    db(:,1) = varargin(1:2:end);
    db(:,2) = varargin(2:2:end);
end

Answer 6

[José Armando Velasco Villagomez]

Check this function it may solve your problem.

function db = name_value_pairs(varargin)
if nargin == 0 || rem(nargin,2) || sum(cellfun(@ischar,varargin(1:2:end))) ~= nargin/2
    db = {};
else
    db = [(varargin(1:2:end))',(varargin(2:2:end))'];
end
end