{"id":195,"date":"2008-02-08T14:50:41","date_gmt":"2008-02-08T19:50:41","guid":{"rendered":"https:\/\/blogs.mathworks.com\/steve\/2008\/02\/08\/linear-indexing\/"},"modified":"2019-10-24T13:55:13","modified_gmt":"2019-10-24T17:55:13","slug":"linear-indexing","status":"publish","type":"post","link":"https:\/\/blogs.mathworks.com\/steve\/2008\/02\/08\/linear-indexing\/","title":{"rendered":"Linear indexing"},"content":{"rendered":"
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Last week I posted an introduction to logical indexing<\/a>. This week I want to continue with a brief discussion of linear indexing<\/i> and its connection to image processing in MATLAB.\r\n <\/p>\r\n

Lets start with a small matrix. Breaking with tradition, I'll use a Hilbert matrix instead of a magic square:<\/p>

A = hilb(5)<\/pre>
\r\nA =\r\n\r\n    1.0000    0.5000    0.3333    0.2500    0.2000\r\n    0.5000    0.3333    0.2500    0.2000    0.1667\r\n    0.3333    0.2500    0.2000    0.1667    0.1429\r\n    0.2500    0.2000    0.1667    0.1429    0.1250\r\n    0.2000    0.1667    0.1429    0.1250    0.1111\r\n\r\n<\/pre>
POP QUIZ: What does the MATLAB expression A(17)<\/tt> mean?\r\n   <\/p>
A(17)<\/pre>
\r\nans =\r\n\r\n    0.2000\r\n\r\n<\/pre>
MATLAB interprets a single subscript as an index into a vector containing all the values of A<\/tt> in column order.  So A(17)<\/tt> is the same as A(2, 4)<\/tt>.\r\n   <\/p>
A(2, 4)<\/pre>
\r\nans =\r\n\r\n    0.2000\r\n\r\n<\/pre>
This is called linear indexing<\/i>.\r\n   <\/p>\r\n   
So what's the connection to image processing?  Suppose A<\/tt> is our image, and some previous computation has told us that we are interested in the pixel values at these row and column\r\n      coordinates:\r\n   <\/p>
row = [2 1 5];\r\ncol = [1 3 4];<\/pre>
Can we extract the desired pixel values in a single expression? Let's try:<\/p>
A(row, col)<\/pre>
\r\nans =\r\n\r\n    0.5000    0.2500    0.2000\r\n    1.0000    0.3333    0.2500\r\n    0.2000    0.1429    0.1250\r\n\r\n<\/pre>
Well, we got nine values out, not three, so that certainly isn't what we are looking for.  MATLAB computes A(row, col)<\/tt> as the submatrix of A<\/tt> formed by the intersections of the specified rows and columns.\r\n   <\/p>\r\n   
Instead of indexing into A<\/tt> using row and column subscripts, we need to index using a single subscript.\r\n   <\/p>\r\n   
Here's how to convert from row-column subscripts into linear indices:<\/p>
M = size(A, 1);\r\nidx = M * (col - 1) + row<\/pre>
\r\nidx =\r\n\r\n     2    11    20\r\n\r\n<\/pre>
The expression A(idx)<\/tt> pulls out just the three values we want:\r\n   <\/p>
A(idx)<\/pre>
\r\nans =\r\n\r\n    0.5000    0.3333    0.1250\r\n\r\n<\/pre>
Or we can assign to just those elements:<\/p>
A(idx) = Inf<\/pre>
\r\nA =\r\n\r\n    1.0000    0.5000       Inf    0.2500    0.2000\r\n       Inf    0.3333    0.2500    0.2000    0.1667\r\n    0.3333    0.2500    0.2000    0.1667    0.1429\r\n    0.2500    0.2000    0.1667    0.1429    0.1250\r\n    0.2000    0.1667    0.1429       Inf    0.1111\r\n\r\n<\/pre>
The functions sub2ind<\/tt> (\"subscript\" to \"index\") and ind2sub<\/tt> convert back and forth between row-column subscripts and linear indices.\r\n   <\/p>
idx = sub2ind(size(A), row, col)<\/pre>
\r\nidx =\r\n\r\n     2    11    20\r\n\r\n<\/pre>
[r, c] = ind2sub(size(A), idx)<\/pre>
\r\nr =\r\n\r\n     2     1     5\r\n\r\n\r\nc =\r\n\r\n     1     3     4\r\n\r\n<\/pre>
For a couple of detailed image processing examples illustrating linear indexing, see these previous posts:<\/p>\r\n   
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