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name value pairs with variable input arguments

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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.
  3 Commenti
M NAGA JAYANTH AVADHANI
M NAGA JAYANTH AVADHANI il 17 Ago 2020
Modificato: M NAGA JAYANTH AVADHANI il 17 Ago 2020
i am not getting an empty cell array if the name is not of type char
hesham hany
hesham hany il 18 Ago 2020
+1
but I didn't get the answer below

Accedi per commentare.

Risposta accettata

Walter Roberson
Walter Roberson il 17 Ago 2020
Modificato: Walter Roberson il 17 Ago 2020
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.

Più risposte (5)

Sovendo Talapatra
Sovendo Talapatra il 20 Ago 2020
Modificato: Sovendo Talapatra il 20 Ago 2020
for j = 1:2:nargin
if ~ischar(varargin{j})
db = {};
return
else
continue
end
By using for loop you can easily fix the code

Abdul Quadir Khan
Abdul Quadir Khan il 7 Nov 2020
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

Selman Baysal
Selman Baysal il 11 Gen 2022
"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

Yifan He
Yifan He il 3 Ago 2022
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

José Armando Velasco Villagomez
Modificato: José Armando Velasco Villagomez il 8 Gen 2023
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

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