Asked by pietro
on 11 Jul 2014

Hi all,

I need to set a cell as pattern for strfind without using a for. Here an example

a={'1','2','3'}; b={'1','2'}; c=strfind(a,b)

c=[1 1 [] ];

thanks

cheers

Answer by Image Analyst
on 12 Jul 2014

Edited by Image Analyst
on 12 Jul 2014

Accepted answer

Here is a simple, easy to understand way that will work:

clc; % Clear command window. % Initialize variables. a={'1','21','3'}; b={'2','1'}; % Initialize results. % Let's use a simple numerical array rather than a cell array! c = zeros(1, length(a)); % Scan each element of "a" for all the elements of "b". for k = 1 : length(a) for colb = 1 : length(b) if ismember(b{colb}, a{k}) c(k) = 1; break; end end end % Print out to command window. c

Answer by Jos (10584)
on 11 Jul 2014

% implicit for with CELLFUN c = cellfun(@(x) strfind(x,b), a, 'un', 0)

pietro
on 11 Jul 2014

Thanks for your reply. It doesn't work, I get the following error:

Error using cell/strfind (line 33) If any of the input arguments are cell arrays, the first must be a cell array of strings and the second must be a character array.

Error in @(x)strfind(x,b)

Answer by Titus Edelhofer
on 11 Jul 2014

Hi,

not exactly the same result but similar:

ismember(a, b)

Titus

Answer by Chris E.
on 11 Jul 2014

Hello! Well I think this answers your question, it does not have a for loop, however it uses "ismember" rather then 'strfind', but I think the output is the same as what you want.

a={'1','2','3'} b={'1','2'} val = ismember(a,b) val(val == 0)=[] c = val

Hope that helps!

pietro
on 11 Jul 2014

Unfortunately I cannot use ismember because it is not really equivalent to strfind since it works only when there is an exact match. I'm sorry for not being clearly enough, but it should work also with the following case:

a={'12','2','3'}; b={'1','2'};

Answer by Jos (10584)
on 11 Jul 2014

Then please explain the relationship between a,b,and c. Why is c{2} equal to 1?

pietro
on 11 Jul 2014

because:

strfind(a,b{2})

[] [1] []

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## 5 Comments

## Sara (view profile)

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/141451#comment_225168

Why not having a for loop???

## pietro (view profile)

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/141451#comment_225177

I would prefer a faster and more elegant solution but in case it doesn't exist, I'm up for a loop to.

## Sara (view profile)

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/141451#comment_225181

Elegant solutions aren't always faster; they are easier to read and save space and coding, I agree on that, but if performance is the issue, you have to test with and without the loop with the profiler (or tic toc). I was just trying to understand what the logic behind the question is.

## Image Analyst (view profile)

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/141451#comment_225296

What about this case:

Do you consider that the 1 and the 2 are found/matched, or not? They are not in the

same locations, but both are in both arrays.## pietro (view profile)

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/141451#comment_225315

the output should be {[1],[1],[]}.