Use Indexing for more than 1 dimension of array simultaneously

I have a 10 by 3 matrix:
matrix = rand(10, 3);
I have an indexing array which selects a column number for each row:
idx = [1; 2; 2; 3; 1; 2; 1; 2; 3; 3];
I want to produce a 10x1 array which uses the above index array to extract the respective column number element at each row. Without using a for loop.
I have tried:
output = matrix(1:10, idx)
output = 10×10
0.2039 0.6450 0.6450 0.8811 0.2039 0.6450 0.2039 0.6450 0.8811 0.8811 0.3955 0.0347 0.0347 0.1645 0.3955 0.0347 0.3955 0.0347 0.1645 0.1645 0.5710 0.9536 0.9536 0.9736 0.5710 0.9536 0.5710 0.9536 0.9736 0.9736 0.5103 0.9297 0.9297 0.2534 0.5103 0.9297 0.5103 0.9297 0.2534 0.2534 0.0723 0.4962 0.4962 0.0039 0.0723 0.4962 0.0723 0.4962 0.0039 0.0039 0.1273 0.1957 0.1957 0.3339 0.1273 0.1957 0.1273 0.1957 0.3339 0.3339 0.0603 0.7878 0.7878 0.9416 0.0603 0.7878 0.0603 0.7878 0.9416 0.9416 0.0832 0.5866 0.5866 0.9793 0.0832 0.5866 0.0832 0.5866 0.9793 0.9793 0.5630 0.3352 0.3352 0.9332 0.5630 0.3352 0.5630 0.3352 0.9332 0.9332 0.7188 0.8864 0.8864 0.6835 0.7188 0.8864 0.7188 0.8864 0.6835 0.6835
This produces a 10x10 matrix.

Answers (3)

matrix(sub2ind(size(matrix),(1:10)',idx))
You can use linear indexing:
matrix = rand(10, 3);
idx = [1; 2; 2; 3; 1; 2; 1; 2; 3; 3];
linearIndex = sub2ind(size(matrix),1:10,idx');
matrix(linearIndex)
ans = 1×10
0.8064 0.6649 0.2393 0.9216 0.0166 0.3291 0.6957 0.8769 0.4766 0.2352
Yet another way using linear indexing (showing what sub2ind( ) does internally):
matrix = rand(10, 3);
idx = [1; 2; 2; 3; 1; 2; 1; 2; 3; 3];
m = size(matrix,1);
matrix( (1:m)' + m*(idx-1) )
ans = 10×1
0.8923 0.3166 0.8492 0.0101 0.2614 0.4642 0.7455 0.5160 0.9367 0.5664
And to show that it is the same as other solutions:
matrix(sub2ind(size(matrix),(1:m)',idx))
ans = 10×1
0.8923 0.3166 0.8492 0.0101 0.2614 0.4642 0.7455 0.5160 0.9367 0.5664

1 Comment

Note that linear indexing is what happens internally when you use sub2ind: sub2ind is a helper function to make it easier to calculate the correct linear indexing.

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R2023a

Asked:

on 25 Jul 2023

Edited:

on 25 Jul 2023

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