Use Indexing for more than 1 dimension of array simultaneously

6 Ansichten (letzte 30 Tage)
deathtime
deathtime am 25 Jul. 2023
Bearbeitet: James Tursa am 25 Jul. 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.

Antworten (3)

Fangjun Jiang
Fangjun Jiang am 25 Jul. 2023
Bearbeitet: Fangjun Jiang am 25 Jul. 2023
matrix(sub2ind(size(matrix),(1:10)',idx))

the cyclist
the cyclist am 25 Jul. 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 am 25 Jul. 2023
Bearbeitet: James Tursa am 25 Jul. 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 Kommentar
Walter Roberson
Walter Roberson am 25 Jul. 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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