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

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

Answer 1

[Voss]

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

[Andy]

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]

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