Name-value pairs
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Can anyone point out the solution to the error shown below (in the end)?
Thank you in advance.
function output = name_value_pairs(varargin)
if nargin == 0 || rem(length(varargin),2) ~= 0
output = {};
return;
end
for i = 1:length(varargin)
if ~ischar(varargin{i})
if ~isdouble(varargin{i}) || isfloat(varargin{i})
output = {};
return;
end
end
end
output = cell(((length(varargin))/2),2);
n =1 ;
for j = 1: (length(varargin))/2
for k = 1:2
output{j,k} = varargin{n};
n = n+1;
end
end
end
采纳的回答
Ameer Hamza
2020-10-2
编辑：Ameer Hamza
2020-10-2
You can use reshape
function C = name_value_pairs(varargin)
if nargin == 0 || rem(numel(varargin),2) ~= 0
C = {};
return;
end
C = reshape(varargin, 2, []).';
if ~all(cellfun(@ischar, C))
C = {}
end
end
6 个评论
Milind Amga
2020-10-2
Milind Amga
2020-10-2
Ameer Hamza
2020-10-2
Thanks for pointing that out. I am glad to be of help!
Yigit Utku Er
2020-10-19
Hi @Amer i used reshape also however it didnt workout in the way i want.And i also see that u used dot transpose to solve the issue i been facing.How does dot transpose work can you explain ?
Ameer Hamza
2020-10-19
.' is just use to change the array shape from row to column. For cell array, both ' and .' are equivalent. For matrix ' means conjucate transpose, and .' means simple transpose.
Chem
2021-12-9
The code is great, it only has one small bug, when using the line ~all(cellfun(@ischar, C)) it requires all columns to be of type char, however, the problem stated says that only the first column must be of type char, so a convenient change is ~all(cellfun(@ischar, C(:,1))))
更多回答（1 个）
Steven Lord
2020-10-2
if ~isdouble(varargin{i}) || isfloat(varargin{i})
Assuming there were an isdouble function doing what you would expect from the name (Arthur Roué is correct, there is no such function in MATLAB though a few objects in various other MathWorks products have methods by that name) what could cause this condition to not pass?
If varargin{i} is a double, the first part of the if condition is false so we need to evaluate the second part. The second part is true so the whole condition is true.
If varargin{i} is a single, the first part of the if condition is true so we don't need to evaluate the second part.
If varargin{i} is an int8, the first part of the if condition is true so we don't need to evaluate the second part.
About the only condition I can think of offhand where this condition could fail would be an object that overloads isdouble and isfloat to return true and false respectively. If someone's willing to go to that length to trick MATLAB, they're likely to encounter other unusual behavior.
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