Use Indexing for more than 1 dimension of array simultaneously

Question

deathtime 2023-7-25

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2000537-use-indexing-for-more-than-1-dimension-of-array-simultaneously

编辑： James Tursa 2023-7-25

在 MATLAB Online 中打开

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
2039    0.6450    0.6450    0.8811    0.2039    0.6450    0.2039    0.6450    0.8811    0.8811
3955    0.0347    0.0347    0.1645    0.3955    0.0347    0.3955    0.0347    0.1645    0.1645
5710    0.9536    0.9536    0.9736    0.5710    0.9536    0.5710    0.9536    0.9736    0.9736
5103    0.9297    0.9297    0.2534    0.5103    0.9297    0.5103    0.9297    0.2534    0.2534
0723    0.4962    0.4962    0.0039    0.0723    0.4962    0.0723    0.4962    0.0039    0.0039
1273    0.1957    0.1957    0.3339    0.1273    0.1957    0.1273    0.1957    0.3339    0.3339
0603    0.7878    0.7878    0.9416    0.0603    0.7878    0.0603    0.7878    0.9416    0.9416
0832    0.5866    0.5866    0.9793    0.0832    0.5866    0.0832    0.5866    0.9793    0.9793
5630    0.3352    0.3352    0.9332    0.5630    0.3352    0.5630    0.3352    0.9332    0.9332
7188    0.8864    0.8864    0.6835    0.7188    0.8864    0.7188    0.8864    0.6835    0.6835

This produces a 10x10 matrix.

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Answer 1

Fangjun Jiang 2023-7-25

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https://ww2.mathworks.cn/matlabcentral/answers/2000537-use-indexing-for-more-than-1-dimension-of-array-simultaneously#answer_1278647

编辑：Fangjun Jiang 2023-7-25

matrix(sub2ind(size(matrix),(1:10)',idx))

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Answer 2

the cyclist 2023-7-25

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https://ww2.mathworks.cn/matlabcentral/answers/2000537-use-indexing-for-more-than-1-dimension-of-array-simultaneously#answer_1278657

在 MATLAB Online 中打开

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

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Answer 3

James Tursa 2023-7-25

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2000537-use-indexing-for-more-than-1-dimension-of-array-simultaneously#answer_1278747

编辑：James Tursa 2023-7-25

在 MATLAB Online 中打开

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
8923
3166
8492
0101
2614
4642
7455
5160
9367
5664

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Walter Roberson 2023-7-25

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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