How to determine if there is a value in a string and output it as a boolean?
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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
1 件のコメント
Jan
2014 年 2 月 23 日
@Amar: Please do not post a question twice. Double posting confuses the ones, who want to help you.
回答 (2 件)
Mischa Kim
2014 年 2 月 23 日
編集済み: Mischa Kim
2014 年 2 月 23 日
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.
3 件のコメント
Mischa Kim
2014 年 2 月 23 日
編集済み: Mischa Kim
2014 年 2 月 23 日
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
2014 年 2 月 23 日
編集済み: Jan
2014 年 2 月 23 日
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().
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