strfind: how to set a cell for the pattern?

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pietro
pietro 2014년 7월 11일
편집: Image Analyst 2014년 7월 12일
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
  댓글 수: 5
Image Analyst
Image Analyst 2014년 7월 11일
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 2014년 7월 12일
the output should be {[1],[1],[]}.

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채택된 답변

Image Analyst
Image Analyst 2014년 7월 12일
편집: Image Analyst 2014년 7월 12일
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

추가 답변 (4개)

Jos (10584)
Jos (10584) 2014년 7월 11일
% implicit for with CELLFUN
c = cellfun(@(x) strfind(x,b), a, 'un', 0)
  댓글 수: 1
pietro
pietro 2014년 7월 11일
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 2014년 7월 11일
Hi,
not exactly the same result but similar:
ismember(a, b)
Titus

Chris E.
Chris E. 2014년 7월 11일
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!
  댓글 수: 1
pietro
pietro 2014년 7월 11일
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) 2014년 7월 11일
Then please explain the relationship between a,b,and c. Why is c{2} equal to 1?

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