MATLAB doesn't recognise a string in a variable

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Lorenzo Chiarion Casoni
评论： Walter Roberson 2015-7-25
function q = edgeindex(edgenum, indexes, id)
length_index = length(indexes);
q = cell(1,edgenum);
b = 0;
for i = 1:length_index
l = indexes(i);
if i == 1
c = strsplit(id(l),'_');
b = b+1;
q{b} = l
else
d = strsplit(id(l),'_');
c{3}
d{3}
if srtcmp(c{3},d{3})
disp('equal')
q{b} = [q{b} (l)]
else
disp('not equal')
c = d;
b=b+1;
q{b} = l
end
end
end
end
Edgenum - an integer
Indexes - an array of integers
id - is a column vector made of text ex. OUT_218_46_[1][0].
At this point: c = strsplit(id(l),'_'); it gives me an error saying the first input has got to be a string. But it is a string? How do I overcome this problem?
I tried putting c = strsplit(char(id(l)),'_'); but a problem arises @ srtcmp(c{3},d{3})
Thanks in advance
  2 个评论
Arun Kumar
Arun Kumar 2015-7-25
id=OUT_218_46_[1][0];while l= 3(an integer) , id(l) gives 'T'.
Lorenzo Chiarion Casoni
Lorenzo Chiarion Casoni 2015-7-25
Sorry I meant to say, it's a column vector where each element is a string. These strings would be of format - 'OUT_218_46_[1]'.

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回答（1 个）

Walter Roberson
Walter Roberson 2015-7-25
When you have a character array and you index it with a single numeric index, the result is a single character.
Column vectors of text are not considered strings: strings are row vectors of text.
Is is possible that id is a cell array of strings? If so then to extract the l'th entry you need id{l} rather than id(l) . This possibility is consistent with your error message.
  2 个评论
Lorenzo Chiarion Casoni
Lorenzo Chiarion Casoni 2015-7-25
You are right, I should change it to id{l}. But even so it gives me this error at the strcmp() - Undefined function 'srtcmp' for input arguments of type 'char'.

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