How do I index a different column for each row of a matrix without using a loop?

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SeanC
SeanC 2017-1-30
评论： Star Strider 2017-1-31
I have a 446,000 x 7 matrix. I have a vector 446,000 specifying the column of interest for each row in the matrix. I.e. from Row 1 I want the second column, from Row 2 I want the 3rd column etc.
How do I extract these values from the matrix without using a loop?
Thanks

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Star Strider
Star Strider 2017-1-30
编辑：Star Strider 2017-1-30
One approach:
A = randi(99, 10, 7); % Created Small Matrix
V = randi(7, 10, 1); % Create Similar Vector
idx = bsxfun(@eq, cumsum(ones(size(A)), 2), V);
Result = sum(A.*idx, 2);
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Star Strider
Star Strider 2017-1-31
@John Chilleri — I appreciate your interest!
I would have left the original, except that I re-read the original post and reconsidered it, since SeanC’s matrix is (446,000x7). My original idea would first create a 198916000000 element square matrix (the reason diag works with it) requiring 1.5913E+012 bytes. At best that’s computationally inefficient, and at most could cause memory problems.
My revised idea transiently duplicates the size of original matrix (in the bit-wise multiplication with the logical matrix), of 24976000 bytes, before ‘collapsing’ it into a vector with the sum call. While I haven’t compared the computation times with the simulated large matrix, I believe it’s more computationally efficient, and certainly more ‘friendly’ with respect to memory usage.

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