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

Question

Melissa Moore 2019-4-8

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/455160-creating-a-matrix-of-ones-and-zeros-with-location-of-ones-based-on-array-value

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?

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

madhan ravi 2019-4-9

编辑：madhan ravi 2019-4-9

@ Melissa Moore :Are you interested in a faster solution still? If yes another solution can be proposed.

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Kevin Rawson 2019-4-8

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/455160-creating-a-matrix-of-ones-and-zeros-with-location-of-ones-based-on-array-value#answer_369624

One possible way is:

    Y = zeros(length(y), max(y));
    Y(sub2ind(size(Y), (1:length(y))', y)) = 1;

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

Answer 2

A. Sawas 2019-4-8

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/455160-creating-a-matrix-of-ones-and-zeros-with-location-of-ones-based-on-array-value#answer_369625

编辑：A. Sawas 2019-4-8

在 MATLAB Online 中打开

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
显示 -1更早的评论隐藏 -1更早的评论

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.

请先登录，再进行评论。

Answer 3

Melissa Moore 2019-4-8

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/455160-creating-a-matrix-of-ones-and-zeros-with-location-of-ones-based-on-array-value#answer_369626

在 MATLAB Online 中打开

I ended up doing this and it worked:

I = eye(3);
Y = I(y,:);

3 个评论
显示 1更早的评论隐藏 1更早的评论

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 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.

请先登录，再进行评论。