Creating a matrix of ones and zeros with location of ones based on array value

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Melissa Moore
Melissa Moore 2019-4-8
编辑： madhan ravi 2019-4-9
I need to acheive the following without a loop (for speed):
I have a vertical array where the value of each row tells me which column in a matrix of zeros should be replaced with a one.
For example, if I input:
y = [2; 3];
Then I want MATLAB to produce output:
Y = [0,1,0; 0, 0, 1];
I can acheive the desired output with the following for loop:
Y = zeros(2,3);
for j = 1:3
Y(:,j) = (y==j);
end
In reality, my matrices are much larger and I want to avoid using for loops for speed. Is there anyway I can do this without a for loop?

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采纳的回答

Kevin Rawson
Kevin Rawson 2019-4-8
One possible way is:
Y = zeros(length(y), max(y));
Y(sub2ind(size(Y), (1:length(y))', y)) = 1;

更多回答（2 个）

A. Sawas
A. Sawas 2019-4-8
编辑：A. Sawas 2019-4-8
The best solution I can think of is to use a for loop but will be faster than the one you showed because it will do one scalar assignment in each iteration compared to a vector assignment.
for j=1:2
Y(j,y(j)) = 1;
end
Timewise this is faster than the method propsed above by Kevin Rawson
  1 个评论
Kevin Rawson
Kevin Rawson 2019-4-8
编辑：Kevin Rawson 2019-4-8
A. Sawas, your method is faster for small arrays, but not for large arrays. I'm using your 10,000x10,000 example with y assigned to the diagonal (see code above).
Sawas method (run 10000 times):
tic;
for i=1:10000
for j=1:length(y)
Y(j,y(j)) = 1;
end
end;
toc;
%Elapsed time is 1.682664 seconds.
Rawson method (run 10000 times):
tic;
for i=1:10000
Y(ind2sub(1:length(y), y)) = 1;
end
toc;
%Elapsed time is 0.596671 seconds.

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Melissa Moore
Melissa Moore 2019-4-8
I ended up doing this and it worked:
I = eye(3);
Y = I(y,:);
  3 个评论
显示 1更早的评论
Kevin Phung
Kevin Phung 2019-4-8
Is Melissa's solution a valid solution?... IMO, I is a diagonal matrix where there are already existing 1's. And this is not what you want.
A. Sawas's is valid and more efficient.
Kevin Rawson
Kevin Rawson 2019-4-8
编辑：Kevin Rawson 2019-4-9
Kevin Phung, as long as I = eye(max(y)), then Y = I(y,:) will return a valid solution.
However, as A. Sawas pointed out, this is an extremely slow operation, ~700 times slower than A. Sawas' method, and ~1988 times slower than my method for a 10,000x10,000 array.

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