Find value in cell array and return index

48 vues (au cours des 30 derniers jours)
MarionJ
MarionJ le 22 Sep 2017
Commenté : Alp Firkan le 2 Juin 2021
Hello
I have a cell array of the form: A={[1,2,3;4,5,6],[11,22],[33;55;66;7]}; Now I want the code to return the index of e.g. value "5", i.e. A{:,1}(2,2). How do I write this?
Thank you for any help LF

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Jan
Jan le 22 Sep 2017
Modifié(e) : Jan le 22 Sep 2017
What have you tried so far? How do you want to store the results?
The solution is more or less trivial with a loop:
A = {[1,2,3;4,5,6], [11,22], [33;55;66;7]}
s = 5;
nA = numel(A);
Match = cell(1, nA);
for iA = 1:nA
match{iA} = find(A{iA} == s);
end
Or perhaps:
nA = numel(A);
Match = cell(nA, 3); % Pre-allocate
iMatch = 0;
for iA = 1:nA
[i1, i2] = find(A{iA} == s);
if ~isempty(i1)
iMatch = iMatch + 1;
Match{iMatch, 1} = iA; % Index of the cell
Match{iMatch, 2} = i1; % Row and column indices
Match{iMatch, 3} = i2;
end
end
Match = Match(1:iMatch, :); % Crop unused elements
Now Match{:, 1} is the cell element with a match, and Match{:, 2:3} are the row and column indices.
Maybe this is enough:
Match = cellfun(@(c) find(c == s), A, 'uniform', false);
Then Match{k} contains the linear indices, if matching elements are found.
  4 commentaires
Stephen23
Stephen23 le 22 Sep 2017
Modifié(e) : Stephen23 le 22 Sep 2017
"Sorry, it seems not to work. The first code returns match-Array with [4][][]. Why 4 and not 5, since "5" has the linear indexing 5 in the first cell Array?"
Actually the code is correct, and 4 is the linear index of the value 5. This is the whole array and its corresponding linear indices:
>> [1,2,3;4,5,6] % your matrix
ans =
1 2 3
4 5 6
>> [1,3,5;2,4,6] % linear indices
ans =
1 3 5
2 4 6
Of course you don't have to believe us or the MATLAB documentation, you can simply try it yourself.
Alp Firkan
Alp Firkan le 2 Juin 2021
Would this method work to locate a string/number element in another array? I keep getting Arrays have incomptible sizes for this operation.

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