strfind: how to set a cell for the pattern?

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 件のコメント

Sara
Sara 2014 年 7 月 11 日
Why not having a for loop???
pietro
pietro 2014 年 7 月 11 日
I would prefer a faster and more elegant solution but in case it doesn't exist, I'm up for a loop to.
Sara
Sara 2014 年 7 月 11 日
編集済み: Sara 2014 年 7 月 11 日
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
Image Analyst 2014 年 7 月 11 日
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 2014 年 7 月 12 日
the output should be {[1],[1],[]}.

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 採用された回答

Image Analyst
Image Analyst 2014 年 7 月 12 日
編集済み: Image Analyst 2014 年 7 月 12 日

0 投票

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

その他の回答 (4 件)

Jos (10584)
Jos (10584) 2014 年 7 月 11 日

0 投票

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

1 件のコメント

pietro
pietro 2014 年 7 月 11 日
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 2014 年 7 月 11 日

0 投票

Hi,
not exactly the same result but similar:
ismember(a, b)
Titus
Chris E.
Chris E. 2014 年 7 月 11 日

0 投票

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 件のコメント

pietro
pietro 2014 年 7 月 11 日
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) 2014 年 7 月 11 日

0 投票

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

1 件のコメント

pietro
pietro 2014 年 7 月 11 日
because:
strfind(a,b{2})
[] [1] []

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