How to count the number of occurrences of each pair in a cell?
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Md Shahidullah Kawsar
am 30 Okt. 2018
Bearbeitet: Akira Agata
am 31 Okt. 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 Kommentare
Rik
am 30 Okt. 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
am 31 Okt. 2018
Bearbeitet: Md Shahidullah Kawsar
am 31 Okt. 2018
Akzeptierte Antwort
Akira Agata
am 31 Okt. 2018
Bearbeitet: Akira Agata
am 31 Okt. 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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