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

2020 6 21
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更新时间：2020 6 22
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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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Digaamber Dhamija
Digaamber Dhamija 2020-6-22
编辑：Digaamber Dhamija 2020-6-22

0 个投票

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 =
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

更多回答（1 个）

Mara
Mara 2020-6-21

1 个投票

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
Digaamber Dhamija 2020-6-21
编辑：Digaamber Dhamija 2020-6-21
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
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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