Reputation: 1889
I have a cell described as the following:
mixed_values = {'jim', 89, [5 2 1; 1 2 3]};
mixed_values{1}
mixed_values{2}
mixed_values{3}
I loop it:
for k=1:length(mixed_values)
curstate=mixed_values{k};
% Check for the [5 2 1; 1 2 3]
if ismatrix(curstate)
disp('yes');
else
disp('no')
end
end
But it founds the matrix multiple times.
yes
yes
yes
How to check it by the way?
Upvotes: 1
Views: 41
Reputation: 2802
It really depends on what you define a matrix to be. In MathWorks case they decided that a matrix would be something with a valid size, which is certainly true. Notice that even scalars are matrices, of size 1x1. You can even have a matrix of characters. A = ['a' 'b';'c' 'd'];
. I gather that in your case you want a matrix to be a numerical collection of at least 2 dimensions. I would solve it this way:
function result = TestForMatrix(m)
t1 = isnumeric(m);
t2 = ~isvector(m);
result = all([t1 t2]);
end
Use it as if it was ismatrix
.
if (TestForMatrix(curstate))
disp('yes');
else
....
The way this works is the test for numerical numbers will eliminate character strings. The second test will eliminate vectors and scalars. As you find more things to include or eliminate you add those tests. For example, say you want to allow cells. t3 = iscell(m); result = all([t1 t2 t3]);
The are many logical tests that can be done on Matlab objects, see Matlab is*.
Upvotes: 1
Reputation: 1965
From Matlab help:
ismatrix(M) returns logical 1 (true) if SIZE(M) returns [m n] with nonnegative integer values m and n, and logical 0 (false) otherwise
so I checked size(curstate)
1 3 % 3 character string array
1 1 % of course you can do size of a single elements
2 3
so I modified your code
for k=1:length(mixed_values)
curstate=mixed_values{k};
% Check for the [5 2 1; 1 2 3]
if (size(curstate,1)) > 1 && (size(curstate,2)) > 1
disp('yes');
disp(size(curstate));
else
disp('no')
end
end
Upvotes: 1