find a string in a character array

218 次查看(过去 30 天)
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.

采纳的回答

Matt Fig
Matt Fig 2012-9-13
编辑:Matt Fig 2012-9-13
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)
  1 个评论
Jonathan
Jonathan 2019-1-29
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.

请先登录,再进行评论。

更多回答(2 个)

Jonathan
Jonathan 2019-1-29
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
  1 个评论
Stephen23
Stephen23 2019-1-29
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

请先登录,再进行评论。


Edward Umpfenbach
Edward Umpfenbach 2012-9-13
Great. Thanks.
Was just missing the "all" part.

类别

Help CenterFile Exchange 中查找有关 Characters and Strings 的更多信息

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by