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 Commenti
Image Analyst
Image Analyst il 11 Lug 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 il 12 Lug 2014
the output should be {[1],[1],[]}.

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Risposta accettata

Image Analyst
Image Analyst il 12 Lug 2014
Modificato: Image Analyst il 12 Lug 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

Più risposte (4)

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

Chris E.
Chris E. il 11 Lug 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 Commento
pietro
pietro il 11 Lug 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) il 11 Lug 2014
Then please explain the relationship between a,b,and c. Why is c{2} equal to 1?

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