How to Obtain the Indeces of the Minimum Value of Each Row in a Matrix and then Apply These Indeces to a New Matrix of the Same Size

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Shawn
Shawn 2015-2-25
编辑： Sean de Wolski 2015-2-26
Say I have the following matrices:
x = [3 4; 1 3; 2 5; 7 4];
y = [1 2; 3 4; 5 6; 7 8];
If I want the minimum values by row in x, I can use
[M I] = min(x,[],2)
to obtain
M = [3; 1; 2; 4]
I = [1; 1; 1; 2]
but I am not sure how to apply I to the y matrix to obtain
y = [1; 3; 5; 8]
I have found that
diag(y(I,:))
works, but it is not efficient and will not work on the matrix that I need to apply this to, which is size(47*10^6,3). I also tried to use the find command, but I was unable to get that to work either.

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Sean de Wolski
Sean de Wolski 2015-2-25
编辑：Sean de Wolski 2015-2-26
Use sub2ind to build a linear index and extract with this.
The rows will be (1:size(y,1)) and the columns will be your I
doc sub2ind
Something like this (not tested)
idx = sub2ind(size(y),(1:size(y,1))',I);
ymn = y(idx)
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Shawn
Shawn 2015-2-25
In re-writing the size of the row, you wrote a "2" instead of "1", but it works once I change that. The code I use is:
ynew = y(sub2ind(size(x),(1:size(x,1))',I))
Thanks for the help!

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Matt J
Matt J 2015-2-25
编辑：Matt J 2015-2-25
[m,n]=size(x);
mask=sparse(1:m, I,true,m,n);
ymin=sum(mask.*y,2)
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Matt J
Matt J 2015-2-25
This is pretty close to what I want, but it returns a sparse logical.
I think you mean a sparse double, not a sparse logical
>> whos ymin
Name Size Bytes Class Attributes
ymin 4x1 80 double sparse
Is it possible to make this return just the vector?
The result you see is a perfectly valid vector, but if you want to see it in the full form that you are used to,
>> ymin=full(ymin)
ymin =
1
3
5
8
What is meant by true?
true() is a built in MATLAB command that returns a scalar or matrix of true logical values, see
http://www.mathworks.com/help/matlab/ref/true.html
Shawn
Shawn 2015-2-25
Ah, yes, I meant sparse double. mask was the sparse logical. Simplifying it down to one line, I used:
ymin=full(sum(sparse(1:size(x,1), I, true, size(x,1), size(x,2)).*y,2))
I must say that this was a really interesting and out of the box solution. I compared the 2 solutions given and found the above nearly 4x faster for random 100,000x2 matrices, which is very important to me since my matrices are very large, but I really appreciate your help and am very impressed by this method.

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