将向量转换成逻辑矩阵? [英] Convert vector into logical matrix?

问题描述

我有一个长度为n的向量y. y(i)是1..m中的整数.有没有一种简单的方法可以将y转换为n x m逻辑矩阵yy，如果y(i)= j，则yy(i，j)= 1，否则为0?这是我一直在做的事情:

I have a vector y of length n. y(i) is an integer in 1..m. Is there a simpler way to convert y into an n x m logical matrix yy, where yy(i, j) = 1 if y(i) = j, but 0 otherwise? Here's how I've been doing it:

% If m is known (m = 3 here), you could write it out all at once
yy = [y == 1; y== 2; y == 3];
yy = reshape(yy, n, 3);

或

% if m is not known ahead of time
yy = [ y == 1 ];
for i = 2:m;
    yy = [ yy; y == i ];
end
yy = reshape(yy, n, m);

推荐答案

您可以使用 bsxfun 为此

yy = bsxfun(@eq,y(:),[1,2,3])

y被转换(如有必要)为列向量，而另一个向量为行向量. bsxfun隐式扩展m-by-1和1-by-n数组，使结果变为m-by-n.

y is transformed (if necessary) to a column-vector, while the other vector is a row vector. bsxfun implicitly expands the m-by-1 and 1-by-n arrays so that the result becomes m-by-n.

这篇关于将向量转换成逻辑矩阵?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！