Not sure how to ask this, but here's my scenario:

Ok, i'm not sure what this transformation is called, but here's what i want to do.

Given a matrix of categorical variables:

  x = ['a', 'b';
   'a', 'c';
   'a', 'b';
   'b', 'c';
   'c', 'a';
   'a', 'd' ]

Generate a matrix that has all of the combination counts:

  a  b  c  d
        -       -       -       -
a |   0  2  1  1
b |   0  0  1  0
c |   1  0  0  0
d |  0  0  0  0

Any thoughts?

Answer 1

[Sean de Wolski]

%%Data
x = {'a', 'b';
   'a', 'c';
   'a', 'b';
   'b', 'c';
   'c', 'a';
   'a', 'd'};

%%Engine
uv = unique(x(:));
[~,idx] = ismember(x,uv);
m = accumarray(idx,1,numel(uv)+[0 0]);

%%Formatting
T = array2table(m,'VariableNames',uv,'RowNames',uv)

Results:

T = 
       a    b    c    d
       _    _    _    _
  a    0    2    1    1
  b    0    0    1    0
  c    1    0    0    0
  d    0    0    0    0

[Walter]

Ahh! accumarray is the trick.. I knew I didn't need to loop over these.. thx!

[Image Analyst]

Be sure you comment it well, because, though it's 2 lines shorter than the for loop solution, it's much more cryptic and will be harder for others who maintain or look at your code later to understand what it's doing without comments.

Answer 2

[Image Analyst]

A simple, easy to understand for loop works fine:

x = ['a', 'b';
   'a', 'c';
   'a', 'b';
   'b', 'c';
   'c', 'a';
   'a', 'd' ]
 % Make a matrix for all possible 26 letters a-z,
 % (or however many you might have);
 matrix = zeros(26);
 for row = 1 : size(x, 1)
   thisRow = x(row, 1) - 'a' + 1;
    thisCol = x(row, 2) - 'a' + 1;
   matrix(thisRow, thisCol) = matrix(thisRow, thisCol) + 1;
 end
 % Display in command window:
 matrix

Answer 3

[Azzi Abdelmalek]

x = {'a', 'b'
  'a', 'c'
  'a', 'b'
  'b', 'c'
  'c', 'a'
  'a', 'd'}
v=regexp('a':'d','\S','match')
[~,jj]=ismember(x,v)
[xx,yy]=ndgrid(1:4,1:4)
freq=zeros(size(xx));
for k=1:numel(xx)
  freq(k)=sum(ismember(jj,[xx(k) yy(k)],'rows'));
end
freq