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

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Walter
Walter 2016-8-5
评论： Image Analyst 2016-8-5
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?

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采纳的回答

Sean de Wolski
Sean de Wolski 2016-8-5
%%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
  2 个评论
Walter
Walter 2016-8-5
Ahh! accumarray is the trick.. I knew I didn't need to loop over these.. thx!
Image Analyst
Image Analyst 2016-8-5
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.

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更多回答（2 个）

Image Analyst
Image Analyst 2016-8-5
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
Azzi Abdelmalek
Azzi Abdelmalek 2016-8-5
编辑：Azzi Abdelmalek 2016-8-5
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

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