Switch/case in cell array of strings
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I have cell array of stings which I call it choice. The variable choice can take the following values:
choice={'Yes','Yes'};
choice={'Yes','No'};
choice={'No','Yes'};
choice={'No','No'};
I am using the switch/case to do other certain calculations. So..
switch choice
case {'Yes','Yes'}
...
case {'Yes','No'}
...
case {'No','Yes'}
...
case {'No','No'}
...
end
But switch/case gives an error:
SWITCH expression must be a scalar or string constant.
case {'pie','pie3'}
Why then I get an error?
Answers (3)
Azzi Abdelmalek
on 1 Jun 2013
Edited: Azzi Abdelmalek
on 1 Jun 2013
choice='Yes'
switch choice
case {'Yes','Yes'}
y=1
case {'Yes','No'}
y=2
case {'No','Yes'}
y=3
case {'No','No'}
y=4
otherwise
y=100
end
%Where is the problem?
Note that case {'Yes','No'} and case{'No','Yes') are the same.And case {'Yes','Yes'} is equivalent to case 'Yes'
3 Comments
Azzi Abdelmalek
on 1 Jun 2013
Edited: Azzi Abdelmalek
on 1 Jun 2013
Maybe you want this:
choice=['Yes' 'No']
%What you can do is
switch choice
case 'YesYes'
y=1
case 'YesNo'
y=2
case 'NoYes'
y=3
case 'NoNo'
y=4
otherwise
y=100
end
Giorgos Papakonstantinou
on 1 Jun 2013
Walter Roberson
on 1 Jun 2013
in a case label, the order of appearance of the values does not matter. Think of it as doing an ismember(expression, labels) and executing the case if it is true.
Walter Roberson
on 1 Jun 2013
An evaluated switch_expression is a scalar or string. An evaluated case_expression is a scalar, a string, or a cell array of scalars or strings.
Your switch expression is a cell array, not a scalar or string, so you get the error.
You may wish to use isequal() such as
isequal(choice, {'Yes', 'Yes'})
or you might want to
switch [choice{1} '/' choice{2}]
case 'Yes/Yes'
case 'Yes/No'
end
There are computational alternatives for the case of a fixed finite number of entries in the cell, each with a fixed finite set of possibilities, but once you get into that, you are probably better off going table-driven instead of using switch()
2 Comments
Giorgos Papakonstantinou
on 1 Jun 2013
Walter Roberson
on 1 Jun 2013
Computational:
[tf, idx] = ismember(choice, {'Yes', 'No'});
switch (idx(1)-1) * 2 + (idx(2)-1)
case 0 %yes yes
case 1 %yes no
case 2 %no yes
case 3 %no no
end
Table driven:
choice_routines = {@() disp('Yes/Yes'), @() disp('Yes/No'), @() disp('No/Yes'), @() disp('No/No') };
[tf, idx] = ismember(choice, {'Yes', 'No'});
offset = (idx(1)-1) * 2 + (idx(2)-1) + 1;
choice_routines{offset}(); %table lookup and call it
Giorgos Papakonstantinou
on 1 Jun 2013
0 votes
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