How to count the number of occurrences of each pair in a cell?
3 vues (au cours des 30 derniers jours)
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Md Shahidullah Kawsar
le 30 Oct 2018
Modifié(e) : Akira Agata
le 31 Oct 2018
Suppose I have a cell array
C = {[1; 2; 3]; [1; 2; 3; 4]; [1; 2]};
c{:}
ans =
1
2
3
ans =
1
2
3
4
ans=
1
2
% where any digit won't repeat in the individual cell.
I need to find out the number of occurrences of each pair. Expected output:
Pair(1,2) = 3 occurences;
Pair(1,3) = 0;
Pair(1,4) = 0;
Pair(2,1) = 0;
Pair(2,3) = 2;
Pair(2,4) = 0;
Pair(3,1) = 0;
Pair(3,2) = 0;
Pair(3,4) = 1;
How can I find it?
2 commentaires
Rik
le 30 Oct 2018
What code have you tried so far? It looks like there is a simple, naive approach with some loops that would solve it (not sure if there are some tricks you can pull to speed it up substantially).
Md Shahidullah Kawsar
le 31 Oct 2018
Modifié(e) : Md Shahidullah Kawsar
le 31 Oct 2018
Réponse acceptée
Akira Agata
le 31 Oct 2018
Modifié(e) : Akira Agata
le 31 Oct 2018
I think one possible way would be like this:
c = {[1; 2; 3]; [1; 2; 3; 4]; [1; 2]};
Pair = [repelem((1:4)',4,1),repmat((1:4)',4,1)];
Count = zeros(size(allPair,1),1);
for kk = 1:numel(c)
d = [c{kk}(1:end-1),c{kk}(2:end)];
[~,lo] = ismember(d,Pair,'rows');
Count = Count + (histcounts(lo,1:size(Pair,1)+1))';
end
T = table(Pair,Count);
The output is:
>> T
T =
16×2 table
Pair Count
______ _____
1 1 0
1 2 3
1 3 0
1 4 0
2 1 0
2 2 0
2 3 2
2 4 0
3 1 0
3 2 0
3 3 0
3 4 1
4 1 0
4 2 0
4 3 0
4 4 0
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