How does indexing work when sorting a matrix?

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Darcy Cordell
Darcy Cordell 2024-8-20
评论： Stephen23 2024-8-20
My intuition is clearly off on this and any help is appreciated.
For a vector, if you sort the vector and find the indices of the sorted vector, you can reproduce the sorted vector by indexing the original vecor using the sorted indices:
r = rand(10,1);
[rsort,indsort] = sort(r);
r(indsort) % This will equal rsort exactly
rsort - r(indsort) % this will equal zero everywhere
However, strangely, this same logic doesn't seem to work when sorting matrices:
R = rand(10,10);
[Rsort,indsort] = sort(R);
R(indsort) %This does NOT match Rsort
Rsort - R(indsort) % This does NOT equal zero everyone, only in the first column
Obviously there is something really simply going on here, but I can't figure it out. Any help is appreciated.
  1 个评论
Stephen23
Stephen23 2024-8-20
Ran in:
"However, strangely, this same logic doesn't seem to work when sorting matrices"
Perhaps this helps understand what is going on:
R = rand(10,10);
[Rsort,indsort] = sort(R);
for k = 1:size(R,2)
X = indsort(:,k);
R(X,k) - Rsort(:,k)
end
ans = 10x1
0 0 0 0 0 0 0 0 0 0
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ans = 10x1
0 0 0 0 0 0 0 0 0 0
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ans = 10x1
0 0 0 0 0 0 0 0 0 0
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ans = 10x1
0 0 0 0 0 0 0 0 0 0
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ans = 10x1
0 0 0 0 0 0 0 0 0 0
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ans = 10x1
0 0 0 0 0 0 0 0 0 0
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ans = 10x1
0 0 0 0 0 0 0 0 0 0
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ans = 10x1
0 0 0 0 0 0 0 0 0 0
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ans = 10x1
0 0 0 0 0 0 0 0 0 0
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ans = 10x1
0 0 0 0 0 0 0 0 0 0
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回答（3 个）

Voss
Voss 2024-8-20
Ran in:
For matrix R, sort(R) sorts each column and indsort are row indices.
To reproduce the sorted matrix, you can convert those row indices into linear indices. Here are a few ways to do that:
R = rand(10,10)
R = 10x10
0.3822 0.3039 0.3901 0.6018 0.8021 0.5926 0.0603 0.1671 0.2621 0.5227 0.2526 0.2543 0.4592 0.6546 0.6726 0.1193 0.8220 0.7939 0.6572 0.2299 0.5452 0.0107 0.6738 0.6339 0.6675 0.6142 0.0388 0.5753 0.4326 0.4067 0.1317 0.1817 0.2517 0.5425 0.2303 0.3802 0.7924 0.7668 0.8333 0.8137 0.3915 0.1373 0.5168 0.4309 0.8889 0.0126 0.7957 0.2650 0.3544 0.9291 0.5511 0.8038 0.7838 0.3338 0.9130 0.1119 0.7811 0.4130 0.3063 0.2598 0.2918 0.7141 0.2002 0.9318 0.4224 0.6488 0.9882 0.9189 0.7345 0.1722 0.4971 0.7946 0.9597 0.7124 0.5456 0.8197 0.0367 0.0661 0.6479 0.7692 0.0686 0.0876 0.2461 0.0902 0.5496 0.0117 0.1257 0.7602 0.9223 0.7158 0.2146 0.2000 0.5664 0.7364 0.0102 0.9625 0.8238 0.8660 0.8919 0.6220
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[Rsort,indsort] = sort(R)
Rsort = 10x10
0.0686 0.0107 0.2002 0.0902 0.0102 0.0117 0.0367 0.0661 0.2621 0.1722 0.1317 0.0876 0.2461 0.3338 0.2303 0.0126 0.0388 0.1671 0.3063 0.2299 0.2146 0.1373 0.2517 0.4309 0.4224 0.1119 0.0603 0.2650 0.3544 0.2598 0.2526 0.1817 0.3901 0.5425 0.5456 0.1193 0.1257 0.4130 0.4326 0.4067 0.2918 0.2000 0.4592 0.6018 0.5496 0.3802 0.7811 0.5753 0.6479 0.5227 0.3822 0.2543 0.5168 0.6339 0.6675 0.5926 0.7924 0.7602 0.6572 0.6220 0.3915 0.3039 0.5664 0.6546 0.6726 0.6142 0.7957 0.7668 0.7345 0.7158 0.4971 0.7141 0.6738 0.7124 0.8021 0.6488 0.8220 0.7939 0.8333 0.7692 0.5452 0.7946 0.7838 0.7364 0.8889 0.8197 0.8238 0.8660 0.8919 0.8137 0.5511 0.8038 0.9597 0.9318 0.9130 0.9625 0.9882 0.9189 0.9223 0.9291
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indsort = 10x10
9 3 7 9 10 9 8 8 1 7 4 9 9 6 4 5 3 1 6 2 10 5 4 5 7 6 1 5 5 6 2 4 1 4 8 2 9 6 3 3 7 10 2 1 9 4 6 3 8 1 1 2 5 3 3 1 4 9 2 10 5 1 10 2 2 3 5 4 7 9 8 7 3 8 1 7 2 2 4 8 3 8 6 10 5 8 10 10 10 4 6 6 8 7 6 10 7 7 9 5
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[m,n] = size(R);
% one way to get the linear indices from the row indices:
idx = indsort+(0:n-1)*m
idx = 10x10
9 13 27 39 50 59 68 78 81 97 4 19 29 36 44 55 63 71 86 92 10 15 24 35 47 56 61 75 85 96 2 14 21 34 48 52 69 76 83 93 7 20 22 31 49 54 66 73 88 91 1 12 25 33 43 51 64 79 82 100 5 11 30 32 42 53 65 74 87 99 8 17 23 38 41 57 62 72 84 98 3 18 26 40 45 58 70 80 90 94 6 16 28 37 46 60 67 77 89 95
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% another way:
% idx = sub2ind([m,n],indsort,repmat(1:n,m,1))
% another way:
% idx = sub2ind([m,n],indsort,(1:n)+zeros(m,1))
% with linear indexing, the sorted matrix is reproduced:
Rsort-R(idx)
ans = 10x10
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
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ScottB
ScottB 2024-8-20
For matrices, ‘sort’ orders the elements within columns.
Mario Malic
Mario Malic 2024-8-20
Ran in:
Hi,
It is explained here - https://uk.mathworks.com/company/technical-articles/matrix-indexing-in-matlab.html find linear indexing section.
R = rand(10,10)
R = 10x10
0.8288 0.3864 0.9078 0.3396 0.0713 0.9826 0.4747 0.7265 0.3330 0.3868 0.4524 0.4739 0.1924 0.6879 0.0571 0.8137 0.2290 0.3599 0.4431 0.2071 0.3368 0.7460 0.9924 0.0108 0.5101 0.5442 0.8826 0.9589 0.4309 0.1135 0.9472 0.6128 0.1313 0.4831 0.0055 0.6986 0.2818 0.4825 0.5604 0.4890 0.3230 0.9389 0.3916 0.7357 0.4607 0.0714 0.1820 0.1908 0.6452 0.4623 0.3826 0.5543 0.5639 0.4370 0.8388 0.6972 0.7228 0.4804 0.8008 0.2282 0.8786 0.9003 0.7716 0.9740 0.0975 0.7926 0.2955 0.6008 0.6380 0.0111 0.1152 0.0771 0.6003 0.9861 0.0473 0.2355 0.3360 0.8173 0.3133 0.6045 0.6821 0.1103 0.1930 0.3148 0.6608 0.1082 0.4856 0.1991 0.7621 0.9948 0.4198 0.7858 0.9622 0.5720 0.5160 0.8098 0.6810 0.4578 0.5295 0.0804
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[Rsort,indsort] = sort(R);
indsort
indsort = 10x10
8 8 4 3 4 5 5 5 8 7 5 9 2 9 8 9 2 9 1 10 3 1 9 1 2 8 4 2 3 3 6 2 5 6 1 3 7 10 2 2 10 6 6 4 7 6 8 6 10 6 2 4 8 10 5 4 1 4 4 1 9 3 7 2 3 7 9 7 7 5 1 10 1 5 10 10 10 1 5 4 7 7 10 7 9 2 6 8 9 8 4 5 3 8 6 1 3 3 6 9
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So, when you do the the
Rsorted = R(indsort)
Rsorted = 10x10
0.1152 0.1152 0.9472 0.3368 0.9472 0.3230 0.3230 0.3230 0.1152 0.8786 0.3230 0.6821 0.4524 0.6821 0.1152 0.6821 0.4524 0.6821 0.8288 0.4198 0.3368 0.8288 0.6821 0.8288 0.4524 0.1152 0.9472 0.4524 0.3368 0.3368 0.3826 0.4524 0.3230 0.3826 0.8288 0.3368 0.8786 0.4198 0.4524 0.4524 0.4198 0.3826 0.3826 0.9472 0.8786 0.3826 0.1152 0.3826 0.4198 0.3826 0.4524 0.9472 0.1152 0.4198 0.3230 0.9472 0.8288 0.9472 0.9472 0.8288 0.6821 0.3368 0.8786 0.4524 0.3368 0.8786 0.6821 0.8786 0.8786 0.3230 0.8288 0.4198 0.8288 0.3230 0.4198 0.4198 0.4198 0.8288 0.3230 0.9472 0.8786 0.8786 0.4198 0.8786 0.6821 0.4524 0.3826 0.1152 0.6821 0.1152 0.9472 0.3230 0.3368 0.1152 0.3826 0.8288 0.3368 0.3368 0.3826 0.6821
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because of linear indexing
% Rsorted(1, 1) will be R(8)
% Rsorted(2, 1) will be R(5)
% Rsorted(8, 8) will be R(1)
Even though it's a matrix, you will see how to use one index to access the element in the link above.
As you see, your sort operated on columns, and it gave sorted indices for each column. You can fix this by adding the number of elements from the columns to the left which is second to last line
R = rand(10, 10);
[Rsort,indsort] = sort(R);
vec = 0 : height(R) : (height(R) * width(R) - 1);
indLinExp = indsort + vec; % care, linear expansion in MATLAB
Rsort - R(indLinExp)
ans = 10x10
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
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