strfind: how to set a cell for the pattern?

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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
  5 Comments
Image Analyst
Image Analyst on 11 Jul 2014
What about this case:
a={'1','2','3'};
b={'2','1'};
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
pietro on 12 Jul 2014
the output should be {[1],[1],[]}.

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Accepted Answer

Image Analyst
Image Analyst on 12 Jul 2014
Edited: Image Analyst on 12 Jul 2014
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

More Answers (4)

Jos (10584)
Jos (10584) on 11 Jul 2014
% implicit for with CELLFUN
c = cellfun(@(x) strfind(x,b), a, 'un', 0)
  1 Comment
pietro
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)

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Titus Edelhofer
Titus Edelhofer on 11 Jul 2014
Hi,
not exactly the same result but similar:
ismember(a, b)
Titus

Chris E.
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!
  1 Comment
pietro
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'};

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Jos (10584)
Jos (10584) on 11 Jul 2014
Then please explain the relationship between a,b,and c. Why is c{2} equal to 1?

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