How to determine if there is a value in a string and output it as a boolean?

 i.e,
A = [0 1 2 3 0 0 0]

 I want to write:
-----------------
 if % A(a)= 1%
      content
 end
 ---------
where 1<a<end.
so if at any index in A, there exists a A='1', the if statement is true and it is executed 

 Thanks

[Jan]

@Amar: Please do not post a question twice. Double posting confuses the ones, who want to help you.

Answer 1

[Mischa Kim]

Amar, you could use

 a = 3;
 if find(A==a)
    ...
 end

Note, that in your case A is not a string but numeric array. To compare (find) characters or sub-strings in a string you'd use strcmp.

[Amar]

I'm sorry I might have expressed myself incorrectly. Please read the question again for I have made further elaboration

[Mischa Kim]

Something like:

 A = {'0' '1' '2' '3' '0' '0'};    % cell array of char
 a = '1';

 if sum(strcmp(a,A))>0
    fprintf('Character ''%s'' found in A\n',a);
 else
    fprintf('Character ''%s'' not found in A\n',a);
 end

[Jan]

find(A==a) is tricky, because it can reply arrays without an element or withmultiple elements also. Then this is performed internally:

if all(find(A==a)) && ~isempty(find(A==a))

This implicit behaviour might be confusing, so I'd suggest to write explicitly:

if any(A==a)

While sum(strcmp(a,A))>0 is clear, the summation is less efficient than any().

Answer 2

[David Young]

I think you just want

if A(a) == 1
     % actions
end

[Amar]

I have no idea why I went full retarded lol. I am not a smart man.