How to count the top five pairs (for each digit) which have the highest number of occurrences in a cell array?

Suppose I have a cell array

C = {[1; 3; 5; 4; 6]; [1; 2; 3; 4; 5; 6; 7]; [1; 4; 3; 6]};
c{:}
ans = 
1
3
5
4
6

ans =
1
2
3
4
5
6
7

ans=
1
4
3
6

 % where any digit won't repeat in the individual cell.

For each digit, I need find out the top five pairs which have the highest number of occurrences. Expected Output:

    Pair(1,2) = 1
    Pair(1,3) = 1
    Pair(1,4) = 1
    Pair(1,5) = 0
    Pair(1,6) = 0

Pair(2,3) = 1
Pair(2,1) = 0
Pair(2,4) = 0
Pair(2,5) = 0
Pair(2,6) = 0

Pair(3,4) = 1
Pair(3,5) = 1
Pair(3,6) = 1
Pair(3,1) = 0
Pair(3,2) = 0

Pair(4,3) = 1
Pair(4,5) = 1
Pair(4,6) = 1
Pair(4,1) = 0
Pair(4,2) = 0

Pair(5,4) = 1
Pair(5,6) = 1
Pair(5,1) = 0
Pair(5,2) = 0
Pair(5,3) = 0

Pair(6,7) = 1
Pair(6,1) = 0
Pair(6,2) = 0
Pair(6,3) = 0
Pair(6,4) = 0

Pair(7,1) = 0
Pair(7,2) = 0
Pair(7,3) = 0
Pair(7,4) = 0
Pair(7,5) = 0

[Image Analyst]

In that cell array, which two elements are you considering make up "a pair"?

[Stephen23]

See also earlier question:

https://www.mathworks.com/matlabcentral/answers/428517-how-to-count-the-total-number-of-occurrences-of-each-digit-in-cell-arrays

@Md Shahidullah Kawsar: please don't forget to accept answers, this shows your appreciation to the volunteers who help you on the forum.

[Md Shahidullah Kawsar]

done! Accepted :D sorry for late.

[Md Shahidullah Kawsar]

@Image Analyst for this cell array C = {[1; 3; 5; 4; 6]; [1; 2; 3; 4; 5; 6; 7]; [1; 4; 3; 6]}; total 13 pairs are possible.

1 3
3 5
5 4
4 6
1 2
2 3
3 4
4 5
5 6
6 7
1 4
4 3
3 6

Answer 1

[Bruno Luong]

C = {[1; 3; 5; 4; 6]; [1; 2; 3; 4; 5; 6; 7]; [1; 4; 3; 6]};

% generate list of all pairs from C elements
u = unique(cat(1,C{:})); 
[v,w] = ndgrid(u);
allpairs = [w(:) v(:)];

% Get existing Pairs from C
cp = cellfun(@(a) [a(1:end-1) a(2:end)], C, 'unif', 0);
cp = cat(1,cp{:});

% Count
[b,J]=ismember(cp,allpairs,'rows');
n = accumarray(J(b),1,[size(allpairs,1),1]);

% Select top 5 (==p)
p = 5;
m = length(u);
[n,is] = sort(reshape(n,[m m]),1,'descend');
n = n(1:p,:);
toppairs = reshape(allpairs(is(1:p,:)+(0:m-1)*m,:),[],2); 
count = n(:); 

% Display
T = table(toppairs,count)

[Md Shahidullah Kawsar]

Thanks but these are showing all the possible pairs and the number of occurrences. I need to find the only top five highest occurrences of pairs for each digit.

[Bruno Luong]

See edit code

Answer 2

[Stephen23]

Simpler:

>> C = {[1;3;5;4;6]; [1;2;3;4;5;6;7]; [1;4;3;6]};
>> H = cellfun(@(v)hankel(v(1:end-1),v(end-1:end)),C,'uni',0);
>> H = vertcat(H{:})
>> P = accumarray(H,1)
P =
 0   1   1   1   0   0   0
 0   0   1   0   0   0   0
 0   0   0   1   1   1   0
 0   0   1   0   1   1   0
 0   0   0   1   0   1   0
 0   0   0   0   0   0   1

For example:

 P(1,2)=1
 P(1,3)=1
 P(1,4)=1
 P(2,3)=1
 ... etc.