Index a Y = MxN, with X=Mx5 (where elements in X are column IDs for values to extract from Y)

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Andy
Andy 2023-12-6
编辑： Voss 2023-12-6
Having a Y = MxN matrix of data, and a X = Mx5 (or other fixed nr. < than N) matrix obtained as:
[~, X] = maxk(Z,5,2); % where Z has the same size as Y
is there a non-FOR way to extract the values from Y coresponding to positions as expressed in X? Current solution is:
for line = 1:size(Y,1)
X1(line,:) = Y(line, X(line,:))
end

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Voss
Voss 2023-12-6
编辑：Voss 2023-12-6
Ran in:
You can use sub2ind for that.
https://www.mathworks.com/help/matlab/ref/sub2ind.html
Example:
Y = randi(50,6,10)
Y = 6×10
16 15 5 1 3 27 32 25 19 34 48 32 21 13 39 24 10 10 21 43 19 42 38 40 24 37 32 1 2 23 41 41 13 26 46 24 1 31 9 48 4 39 17 29 4 43 5 27 42 38 5 17 18 9 14 29 41 23 32 49
Z = Y/100;
[~,X] = maxk(Z,5,2)
X = 6×5
10 7 6 8 9 1 10 5 2 6 2 4 3 6 7 10 5 1 2 8 6 9 2 10 4 10 7 9 6 8
% non for-loop method:
[M,N] = size(Y);
rows = repmat((1:M).',1,size(X,2));
cols = X;
X1 = Y(sub2ind([M N],rows,cols))
X1 = 6×5
34 32 27 25 19 48 43 39 32 24 42 40 38 37 32 48 46 41 41 31 43 42 39 38 29 49 41 32 29 23
% for-loop method:
X1_for = zeros(size(Y,1),size(X,2));
for line = 1:size(Y,1)
X1_for(line,:) = Y(line, X(line,:));
end
X1_for
X1_for = 6×5
34 32 27 25 19 48 43 39 32 24 42 40 38 37 32 48 46 41 41 31 43 42 39 38 29 49 41 32 29 23
% both methods produce the same result:
isequal(X1,X1_for)
ans = logical
1
  2 个评论
Andy
Andy 2023-12-6
编辑：Andy 2023-12-6
yes, sub2ind & repmat do show up in other links ... and it's a tad better than a for - still, not a one liner ... surprised there isn't one word function for this
Voss
Voss 2023-12-6
编辑：Voss 2023-12-6
Ran in:
Y = randi(50,6,10)
Y = 6×10
43 10 39 13 47 41 21 29 3 45 12 33 12 18 49 22 38 7 47 31 46 13 14 13 27 23 8 2 22 17 12 33 35 15 7 17 42 32 33 1 19 2 44 46 32 18 7 32 19 5 5 9 25 25 35 10 2 42 32 24
Z = Y/100;
[~,X] = maxk(Z,5,2)
X = 6×5
5 10 1 6 3 5 9 7 2 10 1 5 6 9 10 7 3 2 9 8 4 3 5 8 1 8 5 9 3 4
% one-liner:
X1 = Y((X-1)*size(Y,1)+(1:size(Y,1)).')
X1 = 6×5
47 45 43 41 39 49 47 38 33 31 46 27 23 22 17 42 35 33 33 32 46 44 32 32 19 42 35 32 25 25
% for-loop method:
X1_for = zeros(size(Y,1),size(X,2));
for line = 1:size(Y,1)
X1_for(line,:) = Y(line, X(line,:));
end
X1_for
X1_for = 6×5
47 45 43 41 39 49 47 38 33 31 46 27 23 22 17 42 35 33 33 32 46 44 32 32 19 42 35 32 25 25
% both methods produce the same result:
isequal(X1,X1_for)
ans = logical
1

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