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 个评论
KSSV
2020-8-17
What error you are getting?
M NAGA JAYANTH AVADHANI
2020-8-17
编辑：M NAGA JAYANTH AVADHANI
2020-8-17
hesham hany
2020-8-18
+1
but I didn't get the answer below
采纳的回答
Walter Roberson
2020-8-17
编辑：Walter Roberson
2020-8-17
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.
3 个评论
M NAGA JAYANTH AVADHANI
2020-8-17
Atif Penkar
2020-8-19
Still get the same error
Walter Roberson
2020-8-20
Atif what is your code?
更多回答（5 个）
Sovendo Talapatra
2020-8-20
编辑：Sovendo Talapatra
2020-8-20
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
2020-11-7
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
2022-1-11
"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
2022-8-3
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
2023-1-8
编辑：José Armando Velasco Villagomez
2023-1-8
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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2020-8-17
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