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Why does the || absolutely not work?

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Chris
Chris on 23 Dec 2014
Edited: John D'Errico on 23 Dec 2014
Hello,
My Code-Fragment is:
location = 'nswe';
if strfind(location,'o') > 0 || strfind(location,'e') > 0
disp('true');
end
when I try to execute it, the message
Operands to the || and && operators must be convertible to logical scalar values.
appears. I have no clue how on earth I'm supposed to make that work. can anybody help me? :/

  1 Comment

per isakson
per isakson on 23 Dec 2014
Because strfind returns empty
>> strfind(location,'o')
ans =
[]

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Answers (2)

John D'Errico
John D'Errico on 23 Dec 2014
Edited: John D'Errico on 23 Dec 2014
When something like this happens, take your expression apart. Look at the pieces. Then think about what MATLAB is telling you.
location = 'nswe';
strfind(location,'o') > 0
ans =
[]
strfind(location,'e') > 0
ans =
1
So the first fragment returns empty, the second returns true.
What does the operator do there?
[] || 1
Operands to the || and && operators must be convertible to logical scalar values.
Is an empty result a scalar, logical value? No. So you need to think about how to better write that test to not fail.
For example, since you know that strfind may return zero, one, or more than one solution depending on the string, you need to write code that will succeed in any case. Of course, I'm not terribly sure why you are testing to see if 'o' falls in the string 'nswe', but maybe you have a valid reason.
(I could show you a better way to write that test, but I don't know what strings you might have as possibilities. Only you know your real problem.)

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per isakson
per isakson on 23 Dec 2014
An alternate approach
switch lower( location )
case {'east','ost','öster'}
...
case {'south','süd','syd'}
...
otherwise
end
John D'Errico
John D'Errico on 23 Dec 2014
Empty is not > 0, nor is it less than or equal to zero. All it can be is empty. I'll admit that you can argue this either way, but that is how the > operator works.
So you might do this instead:
location = 'nswe';
if any([strfind(location,'o') > 0 ,strfind(location,'e') > 0])
disp('true');
end
which caters to the empty case as well as a true result by either subpart. This also works:
location = 'nswe';
if ~isempty(strfind(location,'o')) || ~isempty(strfind(location,'e'))
disp('true');
end
There are always many ways to solve a problem like this if you look. Sean had a nice alternative with ismember.

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Sean de Wolski
Sean de Wolski on 23 Dec 2014
How about just using ismember
if any(ismember('oe',location))
disp(true)
end

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