Use Indexing for more than 1 dimension of array simultaneously

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deathtime
deathtime el 25 de Jul. de 2023
Editada: James Tursa el 25 de Jul. de 2023
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.

Respuestas (3)

Fangjun Jiang
Fangjun Jiang el 25 de Jul. de 2023
Editada: Fangjun Jiang el 25 de Jul. de 2023
matrix(sub2ind(size(matrix),(1:10)',idx))

the cyclist
the cyclist el 25 de Jul. de 2023
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

James Tursa
James Tursa el 25 de Jul. de 2023
Editada: James Tursa el 25 de Jul. de 2023
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 comentario
Walter Roberson
Walter Roberson el 25 de Jul. de 2023
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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