Replacing values in a matrix based on values in a column vector

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Digaamber Dhamija 2020-6-21

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编辑： Digaamber Dhamija 2020-6-22

I have a column vector (with dimensions m x 1) which contains values from 1 to 10.

I want to make a matrix yMat with dimension m x 10, such that, in corresponding rows of y and yMat, the value at the column of yMat corresponding to the value in y is equal to 1.

All other values in yMat should be 0.

For example, if

y = [2; 4; 1; 8]

then

yMat = [0 1 0 0 0 0 0 0 0 0; 0 0 0 1 0 0 0 0 0 0; 1 0 0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 1 0 0]

The following code works, but is there any way to do this without using a for loop?

yMat = zeros(m, 10);    % m is the number of rows in column vector y
for i = 1:m
    yMat(i, y(i)) = 1;
end

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Answer 1

Digaamber Dhamija 2020-6-22

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编辑：Digaamber Dhamija 2020-6-22

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Just in case anyone else has a similar problem, I found an even simpler solution.

y = [2; 4; 1; 8];
yMat = 1:10 == y;

which produces a logical array yMat.

You can get numeric values by doing:

yMat = double(1:10 == y);

producing

yMat =
   1     0     0     0     0     0     0     0     0
   0     0     1     0     0     0     0     0     0
   0     0     0     0     0     0     0     0     0
   0     0     0     0     0     0     1     0     0

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Answer 2

Mara 2020-6-21

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I came up with this. But not sure if this the easiest way to do it.

yMat = zeros(length(y)*10, 1);
yMat((y-1)*length(y)+ (1:length(y))') =1;
yMat = reshape(yMat, length(y), []);

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Digaamber Dhamija 2020-6-21

编辑：Digaamber Dhamija 2020-6-21

在 MATLAB Online 中打开

The formula

(y-1)*length(y) + (1:length(y))'

seems to be doing something similar to calculating the linear indices for the original dimensions of yMat.

It would also have worked if you did this

yMat = zeros(length(y), 10);
yMat((y-1)*length(y) + (1:length(y))') = 1;

Mara 2020-6-21

True, it does not have to be a vector for the indexing to work. Thanks for the feedback! I did not see that you received another answer already, it seems to be more intuitive than this one, anyways.

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