Switch/case in cell array of strings

Question

Giorgos Papakonstantinou 2013-6-1

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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.

In the documentation I find:

case {'pie','pie3'}

Why then I get an error?

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Answer 1

Azzi Abdelmalek 2013-6-1

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编辑：Azzi Abdelmalek 2013-6-1

在 MATLAB Online 中打开

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'

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Giorgos Papakonstantinou 2013-6-1

i didn't know that {'Yes','No'} and case{'No','Yes') were the same

Walter Roberson 2013-6-1

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.

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Answer 2

Walter Roberson 2013-6-1

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https://ww2.mathworks.cn/matlabcentral/answers/77672-switch-case-in-cell-array-of-strings#answer_87345

在 MATLAB Online 中打开

http://www.mathworks.com/help/matlab/ref/switch.html

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()

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Giorgos Papakonstantinou 2013-6-1

What do you mean by "going table driven"? Can you give an example of a computational alternative?

Walter Roberson 2013-6-1

在 MATLAB Online 中打开

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