find a string in a character array

A is an m x n character array (I think that is the right term. It says m x n char in the workspace under value).

I want to find a certain row depending on the characters. So say I have a string s of size 1 x n. I want something like this:

find(strcmp(s,A(1:m,:)))

I messed around with ismember instead of strcmp a little too. Can't get it right. I only want it to return an indicator if the row matches the string s.

help is appreciated. Thanks.

Answer 1

[Matt Fig]

A = ['asdf';'lelr';'wkre';'pope']
idx = all(ismember(A,'lelr'),2)

Now if you need linear indices rather than a logical index, use:

lidx = find(idx)

[Jonathan]

This solution will match all strings that contain any of the characters in the string, not unique matches. See my answer below for an example and a soluton to fix this issue.

Answer 2

[Jonathan]

The answer by Loginatorist was incorrect for my problem that is I believe the same as described by the OP. Below is an example of when the solution gives wrong results:

Occupations = ['educator ';'doctor   '];
all(ismember(Occupations,'doctor   '),2)
all(ismember(Occupations,'educator '),2)

Output:
ans =

  5×1 logical array

   0
   1

ans =

  5×1 logical array

   1
   1

Clearly, the second match is wrong, as doctor contains all the characters contained within educator.

A solution that fixes this is to use cellstr and strcomp:

OccupationsCell = cellstr(Occupations);
strcmp('doctor',OccupationsCell)
strcmp('educator',OccupationsCell)

Output:
ans =

  2×1 logical array

   0
   1


ans =

  2×1 logical array

   1
   0

[Stephen23]

Other options: use the 'rows' option:

>> Occupations = ['educator ';'doctor   '];
>> ismember(Occupations,'doctor   ','rows')
ans =
  0
  1  

Use a cell array:

>> ismember(cellstr(Occupations),'doctor')
ans =
  0
  1

Answer 3

[Edward Umpfenbach]

Great. Thanks.

Was just missing the "all" part.