How to determine a matrix presenting the number of times that element j appears at position i, and the number of times that element j appears immediately after k?

Suppose, j=6 elements = 11, 12, 13, 21, 22, 23

X, Y, Z are three arrays containing the permutation of j elements.

X = [11 22 21 13 12 23]

Y = [11 13 21 23 22 12]

Z = [21 12 11 13 23 22]

How can I determine a matrix (A) from X, Y, Z representing the number of times that element j appears at position i (=1,2,3,4,5,6)?

A would be j*i matrix. The answer would be:

A = [A11 A12 A13 A14 A15 A16

A21 A22 A23 A24 A25 A26

A31 A32 A33 A34 A35 A36

A41 A42 A43 A44 A45 A46

A51 A52 A53 A54 A55 A56

A61 A62 A63 A64 A65 A66]

= [2 0 1 0 0 0

0 1 0 0 1 1

0 1 0 2 0 0

1 0 2 0 0 0

0 1 0 0 1 1

0 0 0 1 1 1]

Let j=k.

How can I determine a matrix (B) from X, Y, Z representing the number of times that element j appears immediately after k (where j ~= k)?

B = [- 1 0 1 0 0

2 - 2 3 2 1

3 1 - 2 1 0

2 0 1 - 1 0

3 1 2 2 - 2

3 2 3 3 1 -]

Answer 1

[Simon Chan]

Use function ismember and create a counter to accumulate the number of occurance

S = [11, 12, 13, 21, 22, 23];
X = [11 22 21 13 12 23];
Y = [11 13 21 23 22 12];
Z = [21 12 11 13 23 22];
K = [X;Y;Z];
index = arrayfun(@(t) ismember(K,t),S,'uni',0);         % Find the position index
A = cell2mat(cellfun(@sum,index,'uni',0)')      

A = 6×6

2 0 1 0 0 0 0 1 0 0 1 1 0 1 0 2 0 0 1 0 2 0 0 0 0 1 0 0 1 1 0 0 0 1 1 1

%
B = zeros(numel(S),numel(S));                           % Initialize counter B
for n = 1:numel(S)
    [row,col] = find(index{n});                         % Find the row and column number of the specific number
    for k = 1:size(K,1)
        data = K(row(k),1:col(k));                      % Extract the numbers in front of the specific number
        B(n,:) = B(n,:) + ismember(S,data);             % Add the number in the counter
    end
    B(n,n)=NaN;                                         % Set own counts to NaN
end
B

B = 6×6

NaN 1 0 1 0 0 2 NaN 2 3 2 1 3 1 NaN 2 1 0 2 0 1 NaN 1 0 3 1 2 2 NaN 2 3 2 3 3 1 NaN